What would be the cleanest way of doing this that would work in both IE and Firefox?
My string looks like this sometext-20202
Now the sometext and the integer after the dash can be of varying length.
Should I just use substring and index of or are there other ways?
How I would do this:
// function you can use:
function getSecondPart(str) {
return str.split('-')[1];
}
// use the function:
alert(getSecondPart("sometext-20202"));
A solution I prefer would be:
const str = 'sometext-20202';
const slug = str.split('-').pop();
Where slug would be your result
var testStr = "sometext-20202"
var splitStr = testStr.substring(testStr.indexOf('-') + 1);
var the_string = "sometext-20202";
var parts = the_string.split('-', 2);
// After calling split(), 'parts' is an array with two elements:
// parts[0] is 'sometext'
// parts[1] is '20202'
var the_text = parts[0];
var the_num = parts[1];
With built-in javascript replace() function and using of regex (/(.*)-/), you can replace the substring before the dash character with empty string (""):
"sometext-20202".replace(/(.*)-/,""); // result --> "20202"
AFAIK, both substring() and indexOf() are supported by both Mozilla and IE. However, note that substr() might not be supported on earlier versions of some browsers (esp. Netscape/Opera).
Your post indicates that you already know how to do it using substring() and indexOf(), so I'm not posting a code sample.
myString.split('-').splice(1).join('-')
I came to this question because I needed what OP was asking but more than what other answers offered (they're technically correct, but too minimal for my purposes). I've made my own solution; maybe it'll help someone else.
Let's say your string is 'Version 12.34.56'. If you use '.' to split, the other answers will tend to give you '56', when maybe what you actually want is '.34.56' (i.e. everything from the first occurrence instead of the last, but OP's specific case just so happened to only have one occurrence). Perhaps you might even want 'Version 12'.
I've also written this to handle certain failures (like if null gets passed or an empty string, etc.). In those cases, the following function will return false.
Use
splitAtSearch('Version 12.34.56', '.') // Returns ['Version 12', '.34.56']
Function
/**
* Splits string based on first result in search
* #param {string} string - String to split
* #param {string} search - Characters to split at
* #return {array|false} - Strings, split at search
* False on blank string or invalid type
*/
function splitAtSearch( string, search ) {
let isValid = string !== '' // Disallow Empty
&& typeof string === 'string' // Allow strings
|| typeof string === 'number' // Allow numbers
if (!isValid) { return false } // Failed
else { string += '' } // Ensure string type
// Search
let searchIndex = string.indexOf(search)
let isBlank = (''+search) === ''
let isFound = searchIndex !== -1
let noSplit = searchIndex === 0
let parts = []
// Remains whole
if (!isFound || noSplit || isBlank) {
parts[0] = string
}
// Requires splitting
else {
parts[0] = string.substring(0, searchIndex)
parts[1] = string.substring(searchIndex)
}
return parts
}
Examples
splitAtSearch('') // false
splitAtSearch(true) // false
splitAtSearch(false) // false
splitAtSearch(null) // false
splitAtSearch(undefined) // false
splitAtSearch(NaN) // ['NaN']
splitAtSearch('foobar', 'ba') // ['foo', 'bar']
splitAtSearch('foobar', '') // ['foobar']
splitAtSearch('foobar', 'z') // ['foobar']
splitAtSearch('foobar', 'foo') // ['foobar'] not ['', 'foobar']
splitAtSearch('blah bleh bluh', 'bl') // ['blah bleh bluh']
splitAtSearch('blah bleh bluh', 'ble') // ['blah ', 'bleh bluh']
splitAtSearch('$10.99', '.') // ['$10', '.99']
splitAtSearch(3.14159, '.') // ['3', '.14159']
For those trying to get everything after the first occurrence:
Something like "Nic K Cage" to "K Cage".
You can use slice to get everything from a certain character. In this case from the first space:
const delim = " "
const name = "Nic K Cage"
const result = name.split(delim).slice(1).join(delim) // prints: "K Cage"
Or if OP's string had two hyphens:
const text = "sometext-20202-03"
// Option 1
const opt1 = text.slice(text.indexOf('-')).slice(1) // prints: 20202-03
// Option 2
const opt2 = text.split('-').slice(1).join("-") // prints: 20202-03
Efficient, compact and works in the general case:
s='sometext-20202'
s.slice(s.lastIndexOf('-')+1)
Use a regular expression of the form: \w-\d+ where a \w represents a word and \d represents a digit. They won't work out of the box, so play around. Try this.
You can use split method for it. And if you should take string from specific pattern you can use split with req. exp.:
var string = "sometext-20202";
console.log(string.split(/-(.*)/)[1])
Everyone else has posted some perfectly reasonable answers. I took a different direction. Without using split, substring, or indexOf. Works great on i.e. and firefox. Probably works on Netscape too.
Just a loop and two ifs.
function getAfterDash(str) {
var dashed = false;
var result = "";
for (var i = 0, len = str.length; i < len; i++) {
if (dashed) {
result = result + str[i];
}
if (str[i] === '-') {
dashed = true;
}
}
return result;
};
console.log(getAfterDash("adfjkl-o812347"));
My solution is performant and handles edge cases.
The point of the above code was to procrastinate work, please don't actually use it.
To use any delimiter and get first or second part
//To divide string using deimeter - here #
//str: full string that is to be splitted
//delimeter: like '-'
//part number: 0 - for string befor delimiter , 1 - string after delimiter
getPartString(str, delimter, partNumber) {
return str.split(delimter)[partNumber];
}
Related
disclaimer - absolutely new to regexes....
I have a string like this:
subject=something||x-access-token=something
For this I need to extract two values. Subject and x-access-token.
As a starting point, I wanted to collect two strings: subject= and x-access-token=. For this here is what I did:
/[a-z,-]+=/g.exec(mystring)
It returns only one element subject=. I expected both of them. Where i am doing wrong?
The g modifier does not affect exec, because exec only returns the first match by specification. What you want is the match method:
mystring.match(/[a-z,-]+=/g)
No regex necessary. Write a tiny parser, it's easy.
function parseValues(str) {
var result = {};
str.split("||").forEach(function (item) {
var parts = item.split("=");
result[ parts[0] /* key */ ] = parts[1]; /* value */
});
return result;
}
usage
var obj = parseValues("subject=something||x-access-token=something-else");
// -> {subject: "something", x-access-token: "something-else"}
var subj = obj.subject;
// -> "something"
var token = obj["x-access-token"];
// -> "something-else"
Additional complications my arise when there is an escaping schema involved that allows you to have || inside a value, or when a value can contain an =.
You will hit these complications with regex approach as well, but with a parser-based approach they will be much easier to solve.
You have to execute exec twice to get 2 extracted strings.
According to MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec
If your regular expression uses the "g" flag, you can use the exec() method multiple times to find successive matches in the same string.
Usually, people extract all strings matching the pattern one by one with a while loop. Please execute following code in browser console to see how it works.
var regex = /[a-z,-]+=/g;
var string = "subject=something||x-access-token=something";
while(matched = regex.exec(string)) console.log(matched);
You can convert the string into a valid JSON string, then parse it to retrieve an object containing the expected data.
var str = 'subject=something||x-access-token=something';
var obj = JSON.parse('{"' + str.replace(/=/g, '":"').replace(/\|\|/g, '","') + '"}');
console.log(obj);
I don't think you need regexp here, just use the javascript builtin function "split".
var s = "subject=something1||x-access-token=something2";
var r = s.split('||'); // r now is an array: ["subject=something1", "x-access-token=something2"]
var i;
for(i=0; i<r.length; i++){
// for each array's item, split again
r[i] = r[i].split('=');
}
At the end you have a matrix like the following:
y x 0 1
0 subject something1
1 x-access-token something2
And you can access the elements using x and y:
"subject" == r[0][0]
"x-access-token" == r[1][0]
"something2" == r[1][1]
If you really want to do it with a pure regexp:
var input = 'subject=something1||x-access-token=something2'
var m = /subject=(.*)\|\|x-access-token=(.*)/.exec(input)
var subject = m[1]
var xAccessToken = m[2]
console.log(subject);
console.log(xAccessToken);
However, it would probably be cleaner to split it instead:
console.log('subject=something||x-access-token=something'
.split(/\|\|/)
.map(function(a) {
a = a.split(/=/);
return { key: a[0], val: a[1] }
}));
I would like to use Javascript Regex instead of split.
Here is the example string:
var str = "123:foo";
The current method calls:
str.split(":")[1]
This will return "foo", but it raises an Error when given a bad string that doesn't have a :.
So this would raise an error:
var str = "fooblah";
In the case of "fooblah" I'd like to just return an empty string.
This should be pretty simple, but went looking for it, and couldn't figure it out. Thank you in advance.
Remove the part up to and including the colon (or the end of the string, if there's no colon):
"123:foo".replace(/.*?(:|$)/, '') // "foo"
"foobar" .replace(/.*?(:|$)/, '') // ""
How this regexp works:
.* Grab everything
? non-greedily
( until we come to
: a colon
| or
$ the end of the string
)
A regex won't help you. Your error likely arises from trying to use undefined later. Instead, check the length of the split first.
var arr = str.split(':');
if (arr.length < 2) {
// Do something to handle a bad string
} else {
var match = arr[1];
...
}
Here's what I've always used, with different variations; this is just a simple version of it:
function split(str, d) {
var op = "";
if(str.indexOf(d) > 0) {
op = str.split(d);
}
return(op);
}
Fairly simple, either returns an array or an empty string.
var str1 = "123:foo", str2 = "fooblah";
var res = function (s) {
return /:/.test(s) && s.replace(/.*(?=:):/, "") || ""
};
console.log(res(str1), res(str2))
Here is a solution using a single regex, with the part you want in the capturing group:
^[^:]*:([^:]+)
I have a string that and I am trying to extract the characters before the quote.
Example is extract the 14 from 14' - €14.99
I am using the follwing code to acheive this.
$menuItem.text().match(/[^']*/)[0]
My problem is that if the string is something like €0.88 I wish to get an empty string returned. However I get back the full string of €0.88.
What I am I doing wrong with the match?
This is the what you should use to split:
string.slice(0, string.indexOf("'"));
And then to handle your non existant value edge case:
function split(str) {
var i = str.indexOf("'");
if(i > 0)
return str.slice(0, i);
else
return "";
}
Demo on JsFiddle
Nobody seems to have presented what seems to me as the safest and most obvious option that covers each of the cases the OP asked about so I thought I'd offer this:
function getCharsBefore(str, chr) {
var index = str.indexOf(chr);
if (index != -1) {
return(str.substring(0, index));
}
return("");
}
try this
str.substring(0,str.indexOf("'"));
Here is an underscore mixin in coffescript
_.mixin
substrBefore : ->
[char, str] = arguments
return "" unless char?
fn = (s)-> s.substr(0,s.indexOf(char)+1)
return fn(str) if str?
fn
or if you prefer raw javascript : http://jsfiddle.net/snrobot/XsuQd/
You can use this to build a partial like:
var beforeQuote = _.substrBefore("'");
var hasQuote = beforeQuote("14' - €0.88"); // hasQuote = "14'"
var noQoute = beforeQuote("14 €0.88"); // noQuote = ""
Or just call it directly with your string
var beforeQuote = _.substrBefore("'", "14' - €0.88"); // beforeQuote = "14'"
I purposely chose to leave the search character in the results to match its complement mixin substrAfter (here is a demo: http://jsfiddle.net/snrobot/SEAZr/ ). The later mixin was written as a utility to parse url queries. In some cases I am just using location.search which returns a string with the leading ?.
I use "split":
let string = "one-two-three";
let um = string.split('-')[0];
let dois = string.split('-')[1];
let tres = string.split('-')[2];
document.write(tres) //three
Given a string
'1.2.3.4.5'
I would like to get this output
'1.2345'
(In case there are no dots in the string, the string should be returned unchanged.)
I wrote this
function process( input ) {
var index = input.indexOf( '.' );
if ( index > -1 ) {
input = input.substr( 0, index + 1 ) +
input.slice( index ).replace( /\./g, '' );
}
return input;
}
Live demo: http://jsfiddle.net/EDTNK/1/
It works but I was hoping for a slightly more elegant solution...
There is a pretty short solution (assuming input is your string):
var output = input.split('.');
output = output.shift() + '.' + output.join('');
If input is "1.2.3.4", then output will be equal to "1.234".
See this jsfiddle for a proof. Of course you can enclose it in a function, if you find it necessary.
EDIT:
Taking into account your additional requirement (to not modify the output if there is no dot found), the solution could look like this:
var output = input.split('.');
output = output.shift() + (output.length ? '.' + output.join('') : '');
which will leave eg. "1234" (no dot found) unchanged. See this jsfiddle for updated code.
It would be a lot easier with reg exp if browsers supported look behinds.
One way with a regular expression:
function process( str ) {
return str.replace( /^([^.]*\.)(.*)$/, function ( a, b, c ) {
return b + c.replace( /\./g, '' );
});
}
You can try something like this:
str = str.replace(/\./,"#").replace(/\./g,"").replace(/#/,".");
But you have to be sure that the character # is not used in the string; or replace it accordingly.
Or this, without the above limitation:
str = str.replace(/^(.*?\.)(.*)$/, function($0, $1, $2) {
return $1 + $2.replace(/\./g,"");
});
You could also do something like this, i also don't know if this is "simpler", but it uses just indexOf, replace and substr.
var str = "7.8.9.2.3";
var strBak = str;
var firstDot = str.indexOf(".");
str = str.replace(/\./g,"");
str = str.substr(0,firstDot)+"."+str.substr(1,str.length-1);
document.write(str);
Shai.
Here is another approach:
function process(input) {
var n = 0;
return input.replace(/\./g, function() { return n++ > 0 ? '' : '.'; });
}
But one could say that this is based on side effects and therefore not really elegant.
This isn't necessarily more elegant, but it's another way to skin the cat:
var process = function (input) {
var output = input;
if (typeof input === 'string' && input !== '') {
input = input.split('.');
if (input.length > 1) {
output = [input.shift(), input.join('')].join('.');
}
}
return output;
};
Not sure what is supposed to happen if "." is the first character, I'd check for -1 in indexOf, also if you use substr once might as well use it twice.
if ( index != -1 ) {
input = input.substr( 0, index + 1 ) + input.substr(index + 1).replace( /\./g, '' );
}
var i = s.indexOf(".");
var result = s.substr(0, i+1) + s.substr(i+1).replace(/\./g, "");
Somewhat tricky. Works using the fact that indexOf returns -1 if the item is not found.
Trying to keep this as short and readable as possible, you can do the following:
JavaScript
var match = string.match(/^[^.]*\.|[^.]+/g);
string = match ? match.join('') : string;
Requires a second line of code, because if match() returns null, we'll get an exception trying to call join() on null. (Improvements welcome.)
Objective-J / Cappuccino (superset of JavaScript)
string = [string.match(/^[^.]*\.|[^.]+/g) componentsJoinedByString:''] || string;
Can do it in a single line, because its selectors (such as componentsJoinedByString:) simply return null when sent to a null value, rather than throwing an exception.
As for the regular expression, I'm matching all substrings consisting of either (a) the start of the string + any potential number of non-dot characters + a dot, or (b) any existing number of non-dot characters. When we join all matches back together, we have essentially removed any dot except the first.
var input = '14.1.2';
reversed = input.split("").reverse().join("");
reversed = reversed.replace(\.(?=.*\.), '' );
input = reversed.split("").reverse().join("");
Based on #Tadek's answer above. This function takes other locales into consideration.
For example, some locales will use a comma for the decimal separator and a period for the thousand separator (e.g. -451.161,432e-12).
First we convert anything other than 1) numbers; 2) negative sign; 3) exponent sign into a period ("-451.161.432e-12").
Next we split by period (["-451", "161", "432e-12"]) and pop out the right-most value ("432e-12"), then join with the rest ("-451161.432e-12")
(Note that I'm tossing out the thousand separators, but those could easily be added in the join step (.join(','))
var ensureDecimalSeparatorIsPeriod = function (value) {
var numericString = value.toString();
var splitByDecimal = numericString.replace(/[^\d.e-]/g, '.').split('.');
if (splitByDecimal.length < 2) {
return numericString;
}
var rightOfDecimalPlace = splitByDecimal.pop();
return splitByDecimal.join('') + '.' + rightOfDecimalPlace;
};
let str = "12.1223....1322311..";
let finStr = str.replace(/(\d*.)(.*)/, '$1') + str.replace(/(\d*.)(.*)/, '$2').replace(/\./g,'');
console.log(finStr)
const [integer, ...decimals] = '233.423.3.32.23.244.14...23'.split('.');
const result = [integer, decimals.join('')].join('.')
Same solution offered but using the spread operator.
It's a matter of opinion but I think it improves readability.
Say, I have a string
"hello is it me you're looking for"
I want to cut part of this string out and return the new string, something like
s = string.cut(0,3);
s would now be equal to:
"lo is it me you're looking for"
EDIT: It may not be from 0 to 3. It could be from 5 to 7.
s = string.cut(5,7);
would return
"hellos it me you're looking for"
You're almost there. What you want is:
http://www.w3schools.com/jsref/jsref_substr.asp
So, in your example:
Var string = "hello is it me you're looking for";
s = string.substr(3);
As only providing a start (the first arg) takes from that index to the end of the string.
Update, how about something like:
function cut(str, cutStart, cutEnd){
return str.substr(0,cutStart) + str.substr(cutEnd+1);
}
Use
substring
function
Returns a subset of a string between
one index and another, or through the
end of the string.
substring(indexA, [indexB]);
indexA
An integer between 0 and one less than the length of the string.
indexB
(optional) An integer between 0 and the length of the string.
substring extracts characters from indexA up to but not including indexB. In particular:
* If indexA equals indexB, substring returns an empty string.
* If indexB is omitted, substring extracts characters to the end
of the string.
* If either argument is less than 0 or is NaN, it is treated as if
it were 0.
* If either argument is greater than stringName.length, it is treated as
if it were stringName.length.
If indexA is larger than indexB, then the effect of substring is as if the two arguments were swapped; for example, str.substring(1, 0) == str.substring(0, 1).
Some other more modern alternatives are:
Split and join
function cutFromString(oldStr, fullStr) {
return fullStr.split(oldStr).join('');
}
cutFromString('there ', 'Hello there world!'); // "Hello world!"
Adapted from MDN example
String.replace(), which uses regex. This means it can be more flexible with case sensitivity.
function cutFromString(oldStrRegex, fullStr) {
return fullStr.replace(oldStrRegex, '');
}
cutFromString(/there /i , 'Hello THERE world!'); // "Hello world!"
s = string.cut(5,7);
I'd prefer to do it as a separate function, but if you really want to be able to call it directly on a String from the prototype:
String.prototype.cut= function(i0, i1) {
return this.substring(0, i0)+this.substring(i1);
}
string.substring() is what you want.
Just as a reference for anyone looking for similar function, I have a String.prototype.bisect implementation that splits a string 3-ways using a regex/string delimiter and returns the before,delimiter-match and after parts of the string....
/*
Splits a string 3-ways along delimiter.
Delimiter can be a regex or a string.
Returns an array with [before,delimiter,after]
*/
String.prototype.bisect = function( delimiter){
var i,m,l=1;
if(typeof delimiter == 'string') i = this.indexOf(delimiter);
if(delimiter.exec){
m = this.match(delimiter);
i = m.index;
l = m[0].length
}
if(!i) i = this.length/2;
var res=[],temp;
if(temp = this.substring(0,i)) res.push(temp);
if(temp = this.substr(i,l)) res.push(temp);
if(temp = this.substring(i+l)) res.push(temp);
if(res.length == 3) return res;
return null;
};
/* though one could achieve similar and more optimal results for above with: */
"my string to split and get the before after splitting on and once".split(/and(.+)/,2)
// outputs => ["my string to split ", " get the before after splitting on and once"]
As stated here: https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Objects/String/split
If separator is a regular expression that contains capturing parentheses, then each time separator is matched the results (including any undefined results) of the capturing parentheses are spliced into the output array. However, not all browsers support this capability.
You need to do something like the following:
var s = "I am a string";
var sSubstring = s.substring(2); // sSubstring now equals "am a string".
You have two options about how to go about it:
http://www.quirksmode.org/js/strings.html#substring
http://www.quirksmode.org/js/strings.html#substr
Try the following:
var str="hello is it me you're looking for";
document.write(str.substring(3)+"<br />");
You can check this link
this works well
function stringCutter(str,cutCount,caretPos){
let firstPart = str.substring(0,caretPos-cutCount);
let secondPart = str.substring(caretPos,str.length);
return firstPart + secondPart;
}