Related
I am trying to get this result: 'Summer-is-here'. Why does the code below generate extra spaces? (Current result: '-Summer--Is- -Here-').
function spinalCase(str) {
var newA = str.split(/([A-Z][a-z]*)/).join("-");
return newA;
}
spinalCase("SummerIs Here");
You are using a variety of split where the regexp contains a capturing group (inside parentheses), which has a specific meaning, namely to include all the splitting strings in the result. So your result becomes:
["", "Summer", "", "Is", " ", "Here", ""]
Joining that with - gives you the result you see. But you can't just remove the unnecessary capture group from the regexp, because then the split would give you
["", "", " ", ""]
because you are splitting on zero-width strings, due to the * in your regexp. So this doesn't really work.
If you want to use split, try splitting on zero-width or space-only matches looking ahead to a uppercase letter:
> "SummerIs Here".split(/\s*(?=[A-Z])/)
^^^^^^^^^ LOOK-AHEAD
< ["Summer", "Is", "Here"]
Now you can join that to get the result you want, but without the lowercase mapping, which you could do with:
"SummerIs Here" .
split(/\s*(?=[A-Z])/) .
map(function(elt, i) { return i ? elt.toLowerCase() : elt; }) .
join('-')
which gives you want you want.
Using replace as suggested in another answer is also a perfectly viable solution. In terms of best practices, consider the following code from Ember:
var DECAMELIZE_REGEXP = /([a-z\d])([A-Z])/g;
var DASHERIZE_REGEXP = /[ _]/g;
function decamelize(str) {
return str.replace(DECAMELIZE_REGEXP, '$1_$2').toLowerCase();
}
function dasherize(str) {
return decamelize(str).replace(DASHERIZE_REGEXP, '-');
}
First, decamelize puts an underscore _ in between two-character sequences of lower-case letter (or digit) and upper-case letter. Then, dasherize replaces the underscore with a dash. This works perfectly except that it lower-cases the first word in the string. You can sort of combine decamelize and dasherize here with
var SPINALIZE_REGEXP = /([a-z\d])\s*([A-Z])/g;
function spinalCase(str) {
return str.replace(SPINALIZE_REGEXP, '$1-$2').toLowerCase();
}
You want to separate capitalized words, but you are trying to split the string on capitalized words that's why you get those empty strings and spaces.
I think you are looking for this :
var newA = str.match(/[A-Z][a-z]*/g).join("-");
([A-Z][a-z]*) *(?!$|[a-z])
You can simply do a replace by $1-.See demo.
https://regex101.com/r/nL7aZ2/1
var re = /([A-Z][a-z]*) *(?!$|[a-z])/g;
var str = 'SummerIs Here';
var subst = '$1-';
var result = str.replace(re, subst);
var newA = str.split(/ |(?=[A-Z])/).join("-");
You can change the regex like:
/ |(?=[A-Z])/ or /\s*(?=[A-Z])/
Result:
Summer-Is-Here
I want to use str_replace or its similar alternative to replace some text in JavaScript.
var text = "this is some sample text that i want to replace";
var new_text = replace_in_javascript("want", "dont want", text);
document.write(new_text);
should give
this is some sample text that i dont want to replace
If you are going to regex, what are the performance implications in
comparison to the built in replacement methods.
You would use the replace method:
text = text.replace('old', 'new');
The first argument is what you're looking for, obviously. It can also accept regular expressions.
Just remember that it does not change the original string. It only returns the new value.
More simply:
city_name=city_name.replace(/ /gi,'_');
Replaces all spaces with '_'!
All these methods don't modify original value, returns new strings.
var city_name = 'Some text with spaces';
Replaces 1st space with _
city_name.replace(' ', '_'); // Returns: "Some_text with spaces" (replaced only 1st match)
Replaces all spaces with _ using regex. If you need to use regex, then i recommend testing it with https://regex101.com/
city_name.replace(/ /gi,'_'); // Returns: Some_text_with_spaces
Replaces all spaces with _ without regex. Functional way.
city_name.split(' ').join('_'); // Returns: Some_text_with_spaces
You should write something like that :
var text = "this is some sample text that i want to replace";
var new_text = text.replace("want", "dont want");
document.write(new_text);
The code that others are giving you only replace one occurrence, while using regular expressions replaces them all (like #sorgit said). To replace all the "want" with "not want", us this code:
var text = "this is some sample text that i want to replace";
var new_text = text.replace(/want/g, "dont want");
document.write(new_text);
The variable "new_text" will result in being "this is some sample text that i dont want to replace".
To get a quick guide to regular expressions, go here:
http://www.cheatography.com/davechild/cheat-sheets/regular-expressions/
To learn more about str.replace(), go here:
https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/String/replace
Good luck!
that function replaces only one occurrence.. if you need to replace
multiple occurrences you should try this function:
http://phpjs.org/functions/str_replace:527
Not necessarily.
see the Hans Kesting answer:
city_name = city_name.replace(/ /gi,'_');
Using regex for string replacement is significantly slower than using a string replace.
As demonstrated on JSPerf, you can have different levels of efficiency for creating a regex, but all of them are significantly slower than a simple string replace. The regex is slower because:
Fixed-string matches don't have backtracking, compilation steps, ranges, character classes, or a host of other features that slow down the regular expression engine. There are certainly ways to optimize regex matches, but I think it's unlikely to beat indexing into a string in the common case.
For a simple test run on the JS perf page, I've documented some of the results:
<script>
// Setup
var startString = "xxxxxxxxxabcxxxxxxabcxx";
var endStringRegEx = undefined;
var endStringString = undefined;
var endStringRegExNewStr = undefined;
var endStringRegExNew = undefined;
var endStringStoredRegEx = undefined;
var re = new RegExp("abc", "g");
</script>
<script>
// Tests
endStringRegEx = startString.replace(/abc/g, "def") // Regex
endStringString = startString.replace("abc", "def", "g") // String
endStringRegExNewStr = startString.replace(new RegExp("abc", "g"), "def"); // New Regex String
endStringRegExNew = startString.replace(new RegExp(/abc/g), "def"); // New Regexp
endStringStoredRegEx = startString.replace(re, "def") // saved regex
</script>
The results for Chrome 68 are as follows:
String replace: 9,936,093 operations/sec
Saved regex: 5,725,506 operations/sec
Regex: 5,529,504 operations/sec
New Regex String: 3,571,180 operations/sec
New Regex: 3,224,919 operations/sec
From the sake of completeness of this answer (borrowing from the comments), it's worth mentioning that .replace only replaces the first instance of the matched character. Its only possible to replace all instances with //g. The performance trade off and code elegance could be argued to be worse if replacing multiple instances name.replace(' ', '_').replace(' ', '_').replace(' ', '_'); or worse while (name.includes(' ')) { name = name.replace(' ', '_') }
var new_text = text.replace("want", "dont want");
hm.. Did you check replace() ?
Your code will look like this
var text = "this is some sample text that i want to replace";
var new_text = text.replace("want", "dont want");
document.write(new_text);
JavaScript has replace() method of String object for replacing substrings. This method can have two arguments. The first argument can be a string or a regular expression pattern (regExp object) and the second argument can be a string or a function. An example of replace() method having both string arguments is shown below.
var text = 'one, two, three, one, five, one';
var new_text = text.replace('one', 'ten');
console.log(new_text) //ten, two, three, one, five, one
Note that if the first argument is the string, only the first occurrence of the substring is replaced as in the example above. To replace all occurrences of the substring you need to provide a regular expression with a g (global) flag. If you do not provide the global flag, only the first occurrence of the substring will be replaced even if you provide the regular expression as the first argument. So let's replace all occurrences of one in the above example.
var text = 'one, two, three, one, five, one';
var new_text = text.replace(/one/g, 'ten');
console.log(new_text) //ten, two, three, ten, five, ten
Note that you do not wrap the regular expression pattern in quotes which will make it a string not a regExp object. To do a case insensitive replacement you need to provide additional flag i which makes the pattern case-insensitive. In that case the above regular expression will be /one/gi. Notice the i flag added here.
If the second argument has a function and if there is a match the function is passed with three arguments. The arguments the function gets are the match, position of the match and the original text. You need to return what that match should be replaced with. For example,
var text = 'one, two, three, one, five, one';
var new_text = text.replace(/one/g, function(match, pos, text){
return 'ten';
});
console.log(new_text) //ten, two, three, ten, five, ten
You can have more control over the replacement text using a function as the second argument.
In JavaScript, you call the replace method on the String object, e.g. "this is some sample text that i want to replace".replace("want", "dont want"), which will return the replaced string.
var text = "this is some sample text that i want to replace";
var new_text = text.replace("want", "dont want"); // new_text now stores the replaced string, leaving the original untouched
You can use
text.replace('old', 'new')
And to change multiple values in one string at once, for example to change # to string v and _ to string w:
text.replace(/#|_/g,function(match) {return (match=="#")? v: w;});
There are already multiple answers using str.replace() (which is fair enough for this question) and regex but you can use combination of str.split() and join() together which is faster than str.replace() and regex.
Below is working example:
var text = "this is some sample text that i want to replace";
console.log(text.split("want").join("dont want"));
If you really want a equivalent to PHP's str_replace you can use Locutus. PHP's version of str_replace support more option then what the JavaScript String.prototype.replace supports.
For example tags:
//PHP
$bodytag = str_replace("%body%", "black", "<body text='%body%'>");
//JS with Locutus
var $bodytag = str_replace(['{body}', 'black', '<body text='{body}'>')
or array's
//PHP
$vowels = array("a", "e", "i", "o", "u", "A", "E", "I", "O", "U");
$onlyconsonants = str_replace($vowels, "", "Hello World of PHP");
//JS with Locutus
var $vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"];
var $onlyconsonants = str_replace($vowels, "", "Hello World of PHP");
Also this doesn't use regex instead it uses for loops. If you not want to use regex but want simple string replace you can use something like this ( based on Locutus )
function str_replace (search, replace, subject) {
var i = 0
var j = 0
var temp = ''
var repl = ''
var sl = 0
var fl = 0
var f = [].concat(search)
var r = [].concat(replace)
var s = subject
s = [].concat(s)
for (i = 0, sl = s.length; i < sl; i++) {
if (s[i] === '') {
continue
}
for (j = 0, fl = f.length; j < fl; j++) {
temp = s[i] + ''
repl = r[0]
s[i] = (temp).split(f[j]).join(repl)
if (typeof countObj !== 'undefined') {
countObj.value += ((temp.split(f[j])).length - 1)
}
}
}
return s[0]
}
var text = "this is some sample text that i want to replace";
var new_text = str_replace ("want", "dont want", text)
document.write(new_text)
for more info see the source code https://github.com/kvz/locutus/blob/master/src/php/strings/str_replace.js
You have the following options:
Replace the first occurrence
var text = "this is some sample text that i want to replace and this i WANT to replace as well.";
var new_text = text.replace('want', 'dont want');
// new_text is "this is some sample text that i dont want to replace and this i WANT to replace as well"
console.log(new_text)
Replace all occurrences - case sensitive
var text = "this is some sample text that i want to replace and this i WANT to replace as well.";
var new_text = text.replace(/want/g, 'dont want');
// new_text is "this is some sample text that i dont want to replace and this i WANT to replace as well
console.log(new_text)
Replace all occurrences - case insensitive
var text = "this is some sample text that i want to replace and this i WANT to replace as well.";
var new_text = text.replace(/want/gi, 'dont want');
// new_text is "this is some sample text that i dont want to replace and this i dont want to replace as well
console.log(new_text)
More info -> here
In Javascript, replace function available to replace sub-string from given string with new one.
Use:
var text = "this is some sample text that i want to replace";
var new_text = text.replace("want", "dont want");
console.log(new_text);
You can even use regular expression with this function. For example, if want to replace all occurrences of , with ..
var text = "123,123,123";
var new_text = text.replace(/,/g, ".");
console.log(new_text);
Here g modifier used to match globally all available matches.
Method to replace substring in a sentence using React:
const replace_in_javascript = (oldSubStr, newSubStr, sentence) => {
let newStr = "";
let i = 0;
sentence.split(" ").forEach(obj => {
if (obj.toUpperCase() === oldSubStr.toUpperCase()) {
newStr = i === 0 ? newSubStr : newStr + " " + newSubStr;
i = i + 1;
} else {
newStr = i === 0 ? obj : newStr + " " + obj;
i = i + 1;
}
});
return newStr;
};
RunMethodHere
If you don't want to use regex then you can use this function which will replace all in a string
Source Code:
function ReplaceAll(mystring, search_word, replace_with)
{
while (mystring.includes(search_word))
{
mystring = mystring.replace(search_word, replace_with);
}
return mystring;
}
How to use:
var mystring = ReplaceAll("Test Test", "Test", "Hello");
Use JS String.prototype.replace first argument should be Regex pattern or String and Second argument should be a String or function.
str.replace(regexp|substr, newSubStr|function);
Ex:
var str = 'this is some sample text that i want to replace';
var newstr = str.replace(/want/i, "dont't want");
document.write(newstr); // this is some sample text that i don't want to replace
ES2021 / ES12
String.prototype.replaceAll()
is trying to bring the full replacement option even when the input pattern is a string.
const str = "Backbencher sits at the Back";
const newStr = str.replaceAll("Back", "Front");
console.log(newStr); // "Frontbencher sits at the Front"
1- String.prototype.replace()
We can do a full **replacement** only if we supply the pattern as a regular expression.
const str = "Backbencher sits at the Back";
const newStr = str.replace(/Back/g, "Front");
console.log(newStr); // "Frontbencher sits at the Front"
If the input pattern is a string, replace() method only replaces the first occurrence.
const str = "Backbencher sits at the Back";
const newStr = str.replace("Back", "Front");
console.log(newStr); // "Frontbencher sits at the Back"
2- You can use split and join
const str = "Backbencher sits at the Back";
const newStr = str.split("Back").join("Front");
console.log(newStr); // "Frontbencher sits at the Front"
function str_replace($old, $new, $text)
{
return ($text+"").split($old).join($new);
}
You do not need additional libraries.
In ECMAScript 2021, you can use replaceAll can be used.
const str = "string1 string1 string1"
const newStr = str.replaceAll("string1", "string2");
console.log(newStr)
// "string2 string2 string2"
simplest form as below
if you need to replace only first occurrence
var newString = oldStr.replace('want', 'dont want');
if you want ot repalce all occurenace
var newString = oldStr.replace(want/g, 'dont want');
Added a method replace_in_javascript which will satisfy your requirement. Also found that you are writing a string "new_text" in document.write() which is supposed to refer to a variable new_text.
let replace_in_javascript= (replaceble, replaceTo, text) => {
return text.replace(replaceble, replaceTo)
}
var text = "this is some sample text that i want to replace";
var new_text = replace_in_javascript("want", "dont want", text);
document.write(new_text);
In my code I split a string based on _ and grab the second item in the array.
var element = $(this).attr('class');
var field = element.split('_')[1];
Takes good_luck and provides me with luck. Works great!
But, now I have a class that looks like good_luck_buddy. How do I get my javascript to ignore the second _ and give me luck_buddy?
I found this var field = element.split(new char [] {'_'}, 2); in a c# stackoverflow answer but it doesn't work. I tried it over at jsFiddle...
Use capturing parentheses:
'good_luck_buddy'.split(/_(.*)/s)
['good', 'luck_buddy', ''] // ignore the third element
They are defined as
If separator contains capturing parentheses, matched results are returned in the array.
So in this case we want to split at _.* (i.e. split separator being a sub string starting with _) but also let the result contain some part of our separator (i.e. everything after _).
In this example our separator (matching _(.*)) is _luck_buddy and the captured group (within the separator) is lucky_buddy. Without the capturing parenthesis the luck_buddy (matching .*) would've not been included in the result array as it is the case with simple split that separators are not included in the result.
We use the s regex flag to make . match on newline (\n) characters as well, otherwise it would only split to the first newline.
What do you need regular expressions and arrays for?
myString = myString.substring(myString.indexOf('_')+1)
var myString= "hello_there_how_are_you"
myString = myString.substring(myString.indexOf('_')+1)
console.log(myString)
I avoid RegExp at all costs. Here is another thing you can do:
"good_luck_buddy".split('_').slice(1).join('_')
With help of destructuring assignment it can be more readable:
let [first, ...rest] = "good_luck_buddy".split('_')
rest = rest.join('_')
A simple ES6 way to get both the first key and remaining parts in a string would be:
const [key, ...rest] = "good_luck_buddy".split('_')
const value = rest.join('_')
console.log(key, value) // good, luck_buddy
Nowadays String.prototype.split does indeed allow you to limit the number of splits.
str.split([separator[, limit]])
...
limit Optional
A non-negative integer limiting the number of splits. If provided, splits the string at each occurrence of the specified separator, but stops when limit entries have been placed in the array. Any leftover text is not included in the array at all.
The array may contain fewer entries than limit if the end of the string is reached before the limit is reached.
If limit is 0, no splitting is performed.
caveat
It might not work the way you expect. I was hoping it would just ignore the rest of the delimiters, but instead, when it reaches the limit, it splits the remaining string again, omitting the part after the split from the return results.
let str = 'A_B_C_D_E'
const limit_2 = str.split('_', 2)
limit_2
(2) ["A", "B"]
const limit_3 = str.split('_', 3)
limit_3
(3) ["A", "B", "C"]
I was hoping for:
let str = 'A_B_C_D_E'
const limit_2 = str.split('_', 2)
limit_2
(2) ["A", "B_C_D_E"]
const limit_3 = str.split('_', 3)
limit_3
(3) ["A", "B", "C_D_E"]
This solution worked for me
var str = "good_luck_buddy";
var index = str.indexOf('_');
var arr = [str.slice(0, index), str.slice(index + 1)];
//arr[0] = "good"
//arr[1] = "luck_buddy"
OR
var str = "good_luck_buddy";
var index = str.indexOf('_');
var [first, second] = [str.slice(0, index), str.slice(index + 1)];
//first = "good"
//second = "luck_buddy"
You can use the regular expression like:
var arr = element.split(/_(.*)/)
You can use the second parameter which specifies the limit of the split.
i.e:
var field = element.split('_', 1)[1];
Replace the first instance with a unique placeholder then split from there.
"good_luck_buddy".replace(/\_/,'&').split('&')
["good","luck_buddy"]
This is more useful when both sides of the split are needed.
I need the two parts of string, so, regex lookbehind help me with this.
const full_name = 'Maria do Bairro';
const [first_name, last_name] = full_name.split(/(?<=^[^ ]+) /);
console.log(first_name);
console.log(last_name);
Non-regex solution
I ran some benchmarks, and this solution won hugely:1
str.slice(str.indexOf(delim) + delim.length)
// as function
function gobbleStart(str, delim) {
return str.slice(str.indexOf(delim) + delim.length);
}
// as polyfill
String.prototype.gobbleStart = function(delim) {
return this.slice(this.indexOf(delim) + delim.length);
};
Performance comparison with other solutions
The only close contender was the same line of code, except using substr instead of slice.
Other solutions I tried involving split or RegExps took a big performance hit and were about 2 orders of magnitude slower. Using join on the results of split, of course, adds an additional performance penalty.
Why are they slower? Any time a new object or array has to be created, JS has to request a chunk of memory from the OS. This process is very slow.
Here are some general guidelines, in case you are chasing benchmarks:
New dynamic memory allocations for objects {} or arrays [] (like the one that split creates) will cost a lot in performance.
RegExp searches are more complicated and therefore slower than string searches.
If you already have an array, destructuring arrays is about as fast as explicitly indexing them, and looks awesome.
Removing beyond the first instance
Here's a solution that will slice up to and including the nth instance. It's not quite as fast, but on the OP's question, gobble(element, '_', 1) is still >2x faster than a RegExp or split solution and can do more:
/*
`gobble`, given a positive, non-zero `limit`, deletes
characters from the beginning of `haystack` until `needle` has
been encountered and deleted `limit` times or no more instances
of `needle` exist; then it returns what remains. If `limit` is
zero or negative, delete from the beginning only until `-(limit)`
occurrences or less of `needle` remain.
*/
function gobble(haystack, needle, limit = 0) {
let remain = limit;
if (limit <= 0) { // set remain to count of delim - num to leave
let i = 0;
while (i < haystack.length) {
const found = haystack.indexOf(needle, i);
if (found === -1) {
break;
}
remain++;
i = found + needle.length;
}
}
let i = 0;
while (remain > 0) {
const found = haystack.indexOf(needle, i);
if (found === -1) {
break;
}
remain--;
i = found + needle.length;
}
return haystack.slice(i);
}
With the above definition, gobble('path/to/file.txt', '/') would give the name of the file, and gobble('prefix_category_item', '_', 1) would remove the prefix like the first solution in this answer.
Tests were run in Chrome 70.0.3538.110 on macOSX 10.14.
Use the string replace() method with a regex:
var result = "good_luck_buddy".replace(/.*?_/, "");
console.log(result);
This regex matches 0 or more characters before the first _, and the _ itself. The match is then replaced by an empty string.
Javascript's String.split unfortunately has no way of limiting the actual number of splits. It has a second argument that specifies how many of the actual split items are returned, which isn't useful in your case. The solution would be to split the string, shift the first item off, then rejoin the remaining items::
var element = $(this).attr('class');
var parts = element.split('_');
parts.shift(); // removes the first item from the array
var field = parts.join('_');
Here's one RegExp that does the trick.
'good_luck_buddy' . split(/^.*?_/)[1]
First it forces the match to start from the
start with the '^'. Then it matches any number
of characters which are not '_', in other words
all characters before the first '_'.
The '?' means a minimal number of chars
that make the whole pattern match are
matched by the '.*?' because it is followed
by '_', which is then included in the match
as its last character.
Therefore this split() uses such a matching
part as its 'splitter' and removes it from
the results. So it removes everything
up till and including the first '_' and
gives you the rest as the 2nd element of
the result. The first element is "" representing
the part before the matched part. It is
"" because the match starts from the beginning.
There are other RegExps that work as
well like /_(.*)/ given by Chandu
in a previous answer.
The /^.*?_/ has the benefit that you
can understand what it does without
having to know about the special role
capturing groups play with replace().
if you are looking for a more modern way of doing this:
let raw = "good_luck_buddy"
raw.split("_")
.filter((part, index) => index !== 0)
.join("_")
Mark F's solution is awesome but it's not supported by old browsers. Kennebec's solution is awesome and supported by old browsers but doesn't support regex.
So, if you're looking for a solution that splits your string only once, that is supported by old browsers and supports regex, here's my solution:
String.prototype.splitOnce = function(regex)
{
var match = this.match(regex);
if(match)
{
var match_i = this.indexOf(match[0]);
return [this.substring(0, match_i),
this.substring(match_i + match[0].length)];
}
else
{ return [this, ""]; }
}
var str = "something/////another thing///again";
alert(str.splitOnce(/\/+/)[1]);
For beginner like me who are not used to Regular Expression, this workaround solution worked:
var field = "Good_Luck_Buddy";
var newString = field.slice( field.indexOf("_")+1 );
slice() method extracts a part of a string and returns a new string and indexOf() method returns the position of the first found occurrence of a specified value in a string.
This should be quite fast
function splitOnFirst (str, sep) {
const index = str.indexOf(sep);
return index < 0 ? [str] : [str.slice(0, index), str.slice(index + sep.length)];
}
console.log(splitOnFirst('good_luck', '_')[1])
console.log(splitOnFirst('good_luck_buddy', '_')[1])
This worked for me on Chrome + FF:
"foo=bar=beer".split(/^[^=]+=/)[1] // "bar=beer"
"foo==".split(/^[^=]+=/)[1] // "="
"foo=".split(/^[^=]+=/)[1] // ""
"foo".split(/^[^=]+=/)[1] // undefined
If you also need the key try this:
"foo=bar=beer".split(/^([^=]+)=/) // Array [ "", "foo", "bar=beer" ]
"foo==".split(/^([^=]+)=/) // [ "", "foo", "=" ]
"foo=".split(/^([^=]+)=/) // [ "", "foo", "" ]
"foo".split(/^([^=]+)=/) // [ "foo" ]
//[0] = ignored (holds the string when there's no =, empty otherwise)
//[1] = hold the key (if any)
//[2] = hold the value (if any)
a simple es6 one statement solution to get the first key and remaining parts
let raw = 'good_luck_buddy'
raw.split('_')
.reduce((p, c, i) => i === 0 ? [c] : [p[0], [...p.slice(1), c].join('_')], [])
You could also use non-greedy match, it's just a single, simple line:
a = "good_luck_buddy"
const [,g,b] = a.match(/(.*?)_(.*)/)
console.log(g,"and also",b)
I have the following string:
",'first string','more','even more'"
I want to transform this into an Array but obviously this is not valid due to the first comma. How can I remove the first comma from my string and make it a valid Array?
I’d like to end up with something like this:
myArray = ['first string','more','even more']
To remove the first character you would use:
var myOriginalString = ",'first string','more','even more'";
var myString = myOriginalString.substring(1);
I'm not sure this will be the result you're looking for though because you will still need to split it to create an array with it. Maybe something like:
var myString = myOriginalString.substring(1);
var myArray = myString.split(',');
Keep in mind, the ' character will be a part of each string in the split here.
In this specific case (there is always a single character at the start you want to remove) you'll want:
str.substring(1)
However, if you want to be able to detect if the comma is there and remove it if it is, then something like:
if (str[0] == ',') {
str = str.substring(1);
}
One-liner
str = str.replace(/^,/, '');
I'll be back.
var s = ",'first string','more','even more'";
var array = s.split(',').slice(1);
That's assuming the string you begin with is in fact a String, like you said, and not an Array of strings.
Assuming the string is called myStr:
// Strip start and end quotation mark and possible initial comma
myStr=myStr.replace(/^,?'/,'').replace(/'$/,'');
// Split stripping quotations
myArray=myStr.split("','");
Note that if a string can be missing in the list without even having its quotation marks present and you want an empty spot in the corresponding location in the array, you'll need to write the splitting manually for a robust solution.
var s = ",'first string','more','even more'";
s.split(/'?,'?/).filter(function(v) { return v; });
Results in:
["first string", "more", "even more'"]
First split with commas possibly surrounded by single quotes,
then filter the non-truthy (empty) parts out.
To turn a string into an array I usually use split()
> var s = ",'first string','more','even more'"
> s.split("','")
[",'first string", "more", "even more'"]
This is almost what you want. Now you just have to strip the first two and the last character:
> s.slice(2, s.length-1)
"first string','more','even more"
> s.slice(2, s.length-2).split("','");
["first string", "more", "even more"]
To extract a substring from a string I usually use slice() but substr() and substring() also do the job.
s=s.substring(1);
I like to keep stuff simple.
You can use directly replace function on javascript with regex or define a help function as in php ltrim(left) and rtrim(right):
1) With replace:
var myArray = ",'first string','more','even more'".replace(/^\s+/, '').split(/'?,?'/);
2) Help functions:
if (!String.prototype.ltrim) String.prototype.ltrim = function() {
return this.replace(/^\s+/, '');
};
if (!String.prototype.rtrim) String.prototype.rtrim = function() {
return this.replace(/\s+$/, '');
};
var myArray = ",'first string','more','even more'".ltrim().split(/'?,?'/).filter(function(el) {return el.length != 0});;
You can do and other things to add parameter to the help function with what you want to replace the char, etc.
this will remove the trailing commas and spaces
var str = ",'first string','more','even more'";
var trim = str.replace(/(^\s*,)|(,\s*$)/g, '');
remove leading or trailing characters:
function trimLeadingTrailing(inputStr, toRemove) {
// use a regex to match toRemove at the start (^)
// and at the end ($) of inputStr
const re = new Regex(`/^${toRemove}|{toRemove}$/`);
return inputStr.replace(re, '');
}
I'm trying to extract a substring from a file with JavaScript Regex. Here is a slice from the file :
DATE:20091201T220000
SUMMARY:Dad's birthday
the field I want to extract is "Summary". Here is the approach:
extractSummary : function(iCalContent) {
/*
input : iCal file content
return : Event summary
*/
var arr = iCalContent.match(/^SUMMARY\:(.)*$/g);
return(arr);
}
function extractSummary(iCalContent) {
var rx = /\nSUMMARY:(.*)\n/g;
var arr = rx.exec(iCalContent);
return arr[1];
}
You need these changes:
Put the * inside the parenthesis as
suggested above. Otherwise your matching
group will contain only one
character.
Get rid of the ^ and $. With the global option they match on start and end of the full string, rather than on start and end of lines. Match on explicit newlines instead.
I suppose you want the matching group (what's
inside the parenthesis) rather than
the full array? arr[0] is
the full match ("\nSUMMARY:...") and
the next indexes contain the group
matches.
String.match(regexp) is
supposed to return an array with the
matches. In my browser it doesn't (Safari on Mac returns only the full
match, not the groups), but
Regexp.exec(string) works.
You need to use the m flag:
multiline; treat beginning and end characters (^ and $) as working
over multiple lines (i.e., match the beginning or end of each line
(delimited by \n or \r), not only the very beginning or end of the
whole input string)
Also put the * in the right place:
"DATE:20091201T220000\r\nSUMMARY:Dad's birthday".match(/^SUMMARY\:(.*)$/gm);
//------------------------------------------------------------------^ ^
//-----------------------------------------------------------------------|
Your regular expression most likely wants to be
/\nSUMMARY:(.*)$/g
A helpful little trick I like to use is to default assign on match with an array.
var arr = iCalContent.match(/\nSUMMARY:(.*)$/g) || [""]; //could also use null for empty value
return arr[0];
This way you don't get annoying type errors when you go to use arr
This code works:
let str = "governance[string_i_want]";
let res = str.match(/[^governance\[](.*)[^\]]/g);
console.log(res);
res will equal "string_i_want". However, in this example res is still an array, so do not treat res like a string.
By grouping the characters I do not want, using [^string], and matching on what is between the brackets, the code extracts the string I want!
You can try it out here: https://www.w3schools.com/jsref/tryit.asp?filename=tryjsref_match_regexp
Good luck.
(.*) instead of (.)* would be a start. The latter will only capture the last character on the line.
Also, no need to escape the :.
You should use this :
var arr = iCalContent.match(/^SUMMARY\:(.)*$/g);
return(arr[0]);
this is how you can parse iCal files with javascript
function calParse(str) {
function parse() {
var obj = {};
while(str.length) {
var p = str.shift().split(":");
var k = p.shift(), p = p.join();
switch(k) {
case "BEGIN":
obj[p] = parse();
break;
case "END":
return obj;
default:
obj[k] = p;
}
}
return obj;
}
str = str.replace(/\n /g, " ").split("\n");
return parse().VCALENDAR;
}
example =
'BEGIN:VCALENDAR\n'+
'VERSION:2.0\n'+
'PRODID:-//hacksw/handcal//NONSGML v1.0//EN\n'+
'BEGIN:VEVENT\n'+
'DTSTART:19970714T170000Z\n'+
'DTEND:19970715T035959Z\n'+
'SUMMARY:Bastille Day Party\n'+
'END:VEVENT\n'+
'END:VCALENDAR\n'
cal = calParse(example);
alert(cal.VEVENT.SUMMARY);