Javascript Remove strings in beginning and end - javascript

base on the following string
...here..
..there...
.their.here.
How can i remove the . on the beginning and end of string like the trim that removes all spaces, using javascript
the output should be
here
there
their.here

These are the reasons why the RegEx for this task is /(^\.+|\.+$)/mg:
Inside /()/ is where you write the pattern of the substring you want to find in the string:
/(ol)/ This will find the substring ol in the string.
var x = "colt".replace(/(ol)/, 'a'); will give you x == "cat";
The ^\.+|\.+$ in /()/ is separated into 2 parts by the symbol | [means or]
^\.+ and \.+$
^\.+ means to find as many . as possible at the start.
^ means at the start; \ is to escape the character; adding + behind a character means to match any string containing one or more that character
\.+$ means to find as many . as possible at the end.
$ means at the end.
The m behind /()/ is used to specify that if the string has newline or carriage return characters, the ^ and $ operators will now match against a newline boundary, instead of a string boundary.
The g behind /()/ is used to perform a global match: so it find all matches rather than stopping after the first match.
To learn more about RegEx you can check out this guide.

Try to use the following regex
var text = '...here..\n..there...\n.their.here.';
var replaced = text.replace(/(^\.+|\.+$)/mg, '');

Here is working Demo
Use Regex /(^\.+|\.+$)/mg
^ represent at start
\.+ one or many full stops
$ represents at end
so:
var text = '...here..\n..there...\n.their.here.';
alert(text.replace(/(^\.+|\.+$)/mg, ''));

Here is an non regular expression answer which utilizes String.prototype
String.prototype.strim = function(needle){
var first_pos = 0;
var last_pos = this.length-1;
//find first non needle char position
for(var i = 0; i<this.length;i++){
if(this.charAt(i) !== needle){
first_pos = (i == 0? 0:i);
break;
}
}
//find last non needle char position
for(var i = this.length-1; i>0;i--){
if(this.charAt(i) !== needle){
last_pos = (i == this.length? this.length:i+1);
break;
}
}
return this.substring(first_pos,last_pos);
}
alert("...here..".strim('.'));
alert("..there...".strim('.'))
alert(".their.here.".strim('.'))
alert("hereagain..".strim('.'))
and see it working here : http://jsfiddle.net/cettox/VQPbp/

Slightly more code-golfy, if not readable, non-regexp prototype extension:
String.prototype.strim = function(needle) {
var out = this;
while (0 === out.indexOf(needle))
out = out.substr(needle.length);
while (out.length === out.lastIndexOf(needle) + needle.length)
out = out.slice(0,out.length-needle.length);
return out;
}
var spam = "this is a string that ends with thisthis";
alert("#" + spam.strim("this") + "#");
Fiddle-ige

Use RegEx with javaScript Replace
var res = s.replace(/(^\.+|\.+$)/mg, '');

We can use replace() method to remove the unwanted string in a string
Example:
var str = '<pre>I'm big fan of Stackoverflow</pre>'
str.replace(/<pre>/g, '').replace(/<\/pre>/g, '')
console.log(str)
output:
Check rules on RULES blotter

Related

JavaScript remove a character from a string and remove the previous character

How do I remove a character from a string and remove the previous character as well?
Example:
"ABCXDEXFGHXIJK"
I want to split the string by "X" and remove the previous character which returns
"ABDFGIJK" // CX, EX, HX are removed
I found this thread but it removes everything before rather than a specific amount of characters: How to remove part of a string before a ":" in javascript?
I can run a for loop but I was wondering if there was a better/simpler way to achieve this
const remove = function(str){
for(let i = 0; i < str.length; i++){
if(str[i] === "X") str = str.slice(0, i - 1) + str.slice(i + 1);
}
return str
}
console.log(remove("ABCXDEXFGHXIJK")) // ABDFGIJK
You can use String.prototype.replace and regex.
"ABCXDEXFGHXIJK".replace(/.X/g, '')
The g at the end is to replace every occurrence of .X. You can use replaceAll as well, but it has less support.
"ABCXDEXFGHXIJK".replaceAll(/.X/g, '')
If you want it to be case insensitive, use the i flag as well.
"ABCXDEXFGHXIJK".replace(/.x/gi, '')
The simplest way is to use a regular expression inside replace.
"ABCXDEXFGHXIJK".replace(/.X/g, "")
.X means "match the combination of X and any single character before it, g flag after the expression body repeats the process globally (instead of doing it once).
While not the most computationally efficient, you could use the following one-liner that may meet your definition of "a better/simpler way to achieve this":
const remove = str => str.split("X").map((ele, idx) => idx !== str.split("X").length - 1 ? ele.slice(0, ele.length - 1) : ele).join("");
console.log(remove("ABCXDEXFGHXIJK"));
Maybe you can use recursion.
function removeChar(str, char){
const index = str.indexOf(char);
if(index < 0) return str;
// removes 2 characters from string
return removeChar(str.split('').splice(index - 2, index).join());
}
Try this way (Descriptive comments are added in the below code snippet itself) :
// Input string
const str = "ABCXDEXFGHXIJK";
// split the input string based on 'X' and then remove the last item from each element by using String.slice() method.
const splittedStrArr = str.split('X').map(item => item = item.slice(0, -1));
// Output by joining the modified array elements.
console.log(splittedStr.join(''))
By using RegEx :
// Input string
const str = "ABCXDEXFGHXIJK";
// Replace the input string by matching the 'X' and one character before that with an empty string.
const modifiedStr = str.replace(/.X/g, "")
// Output
console.log(modifiedStr)

Regex split comma except escaped [duplicate]

I have this string:
a\,bcde,fgh,ijk\,lmno,pqrst\,uv
I need a JavaScript function that will split the string by every , but only those that don't have a \ before them
How can this be done?
Here's the shortest thing I could come up with:
'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv'.replace(/([^\\]),/g, '$1\u000B').split('\u000B')
The idea behind is to find every place where comma isn't prefixed with a backslash, replace those with string that is uncommon to come up in your strings and then split by that uncommon string.
Note that backslashes before commas have to be escaped using another backslash. Otherwise, javascript treats form \, as escaped comma and produce simply a comma out of it! In other words if you won't escape the backslash, javascript sees this: a\,bcde,fgh,ijk\,lmno,pqrst\,uv as this a,bcde,fgh,ijk,lmno,pqrst,uv.
Since regular expressions in JavaScript does not support lookbehinds, I'm not going to cook up a giant hack to mimic this behavior. Instead, you can just split() on all commas (,) and then glue back the pieces that shouldn't have been split in the first place.
Quick 'n' dirty demo:
var str = 'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv'.split(','), // Split on all commas
out = []; // Output
for (var i = 0, j = str.length - 1; i < j; i++) { // Iterate all but last (last can never be glued to non-existing next)
var curr = str[i]; // This piece
if (curr.charAt(curr.length - 1) == '\\') { // If ends with \ ...
curr += ',' + str[++i]; // ... glue with next and skip next (increment i)
}
out.push(curr); // Add to output
}
Another ugly hack around the lack of look-behinds:
function rev(s) {
return s.split('').reverse().join('');
}
var s = 'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv';
// Enter bizarro world...
var r = rev(s);
// Split with a look-ahead
var rparts = r.split(/,(?!\\)/);
// And put it back together with double reversing.
var sparts = [ ];
while(rparts.length)
sparts.push(rev(rparts.pop()));
for(var i = 0; i < sparts.length; ++i)
$('#out').append('<pre>' + sparts[i] + '</pre>');
Demo: http://jsfiddle.net/ambiguous/QbBfw/1/
I don't think I'd do this in real life but it works even if it does make me feel dirty. Consider this a curiosity rather than something you should really use.
In case if need remove backslashes also:
var test='a\\.b.c';
var result = test.replace(/\\?\./g, function (t) { return t == '.' ? '\u000B' : '.'; }).split('\u000B');
//result: ["a.b", "c"]
In 2022 most of browsers support lookbehinds:
https://caniuse.com/js-regexp-lookbehind
Safari should be your only concern.
With a lookbehind you can split your string this way:
"a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv".split(/(?<!\\),/)
// => ['a\\,bcde', 'fgh', 'ijk\\,lmno', 'pqrst\\,uv']
You can use regex to do the split.
Here is the link to regex in javascript http://www.w3schools.com/jsref/jsref_obj_regexp.asp
Here is the link to other post where the author have used regex for split Javascript won't split using regex
From the first link if you note you can create a regular expression using
?!n Matches any string that is not followed by a specific string n
[,]!\\

String replace a url with part of the same url in javascript

I have string that contains a random url:
http://google.com/vocab/prefix#Billy
That needs to be transformed so that everything up to, and including the first # is replaced with the value between the last / and the first # followed by a :.
The result would be:
prefix:Billy
More examples:
http://some.url/a/path#elephant --> path:elephant
http://random.com/cool/black/beans/thing#Bob --> thing:bob
I understand how to capture the prefix part /([^\/]+(?=#))/, but I'm struggling to do a string replace because I can't figure out how to capture the part I need to replace.
myString.replace(/([^\/]+(?=#))/, '$1:')
I would prefer to use string.replace with regex if at all possible
When using replace method you need to match all the patterns you want to replace instead of just the part you need to keep; Here are two options:
let s = 'http://google.com/vocab/prefix#Billy'
// using greedy regex
console.log(s.replace(/.*\/([^#]+)#/, '$1:'))
// adapted from OP's attempt
console.log(s.replace(/.*?([^\/]+?)#/, '$1:'))
Note .* part to match the substring you want to discard, () to capture the pattern you want to keep, then reformat the output.
Try this code:
var myString = "http://google.com/vocab/prefix#Billy";
var hashIndex = myString.indexOf("#"); // find where the "#" is
for(var i = hashIndex; i > 0; i--) { // loop from "#" index *back* to closest "/" symbol
if(myString[i] == "/") break; // break loop when "/" is found
}
myString = myString.replace("#", ":"); // replace "#" with ":" as in your example
console.log(myString.substring(i, hashIndex); // output
Shortened:
var myString = "http://google.com/vocab/prefix#Billy".replace("#",":");
for(var i = myString.indexOf(":"); i > 0; i--) { if(myString[i] == "/") break; }
console.log(myString.substring(i, myString.indexOf(":");

Why is my RegExp ignoring start and end of strings?

I made this helper function to find single words, that are not part of bigger expressions
it works fine on any word that is NOT first or last in a sentence, why is that?
is there a way to add "" to regexp?
String.prototype.findWord = function(word) {
var startsWith = /[\[\]\.,-\/#!$%\^&\*;:{}=\-_~()\s]/ ;
var endsWith = /[^A-Za-z0-9]/ ;
var wordIndex = this.indexOf(word);
if (startsWith.test(this.charAt(wordIndex - 1)) &&
endsWith.test(this.charAt(wordIndex + word.length))) {
return wordIndex;
}
else {return -1;}
}
Also, any improvement suggestions for the function itself are welcome!
UPDATE: example: I want to find the word able in a string, I waht it to work in cases like [able] able, #able1 etc.. but not in cases that it is part of another word like disable, enable etc
A different version:
String.prototype.findWord = function(word) {
return this.search(new RegExp("\\b"+word+"\\b"));
}
Your if will only evaluate to true if endsWith matches after the word. But the last word of a sentence ends with a full stop, which won't match your alphanumeric expression.
Did you try word boundary -- \b?
There is also \w which match one word character ([a-zA-Z_]) -- this could help you too (depends on your word definition).
See RegExp docs for more details.
If you want your endsWith regexp also matches the empty string, you just need to append |^$ to it:
var endsWith = /[^A-Za-z0-9]|^$/ ;
Anyway, you can easily check if it is the beginning of the text with if (wordIndex == 0), and if it is the end with if (wordIndex + word.length == this.length).
It is also possible to eliminate this issue by operating on a copy of the input string, surrounded with non-alphanumerical characters. For example:
var s = "#" + this + "#";
var wordIndex = this.indexOf(word) - 1;
But I'm afraid there is another problems with your function:
it would never match "able" in a string like "disable able enable" since the call to indexOf would return 3, then startsWith.test(wordIndex) would return false and the function would exit with -1 without searching further.
So you could try:
String.prototype.findWord = function (word) {
var startsWith = "[\\[\\]\\.,-\\/#!$%\\^&\*;:{}=\\-_~()\\s]";
var endsWith = "[^A-Za-z0-9]";
var wordIndex = ("#"+this+"#").search(new RegExp(startsWith + word + endsWith)) - 1;
if (wordIndex == -1) { return -1; }
return wordIndex;
}

How can I remove all characters up to and including the 3rd slash in a string?

I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?
To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"
var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex
var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);
You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";
You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"
let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);

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