Develop Reference

Split a String till a word in javascript

I have to split a string till a word in JavaScript.
var urlCurrent = "www.example.com/web/americas/home";
var siteNames = ["americas","international","europe","asia-pacific","africa-middle-east","russia","india"];
var siteNamesJoin = siteNames.join('|');
var siteUrlCurrent = urlCurrent.split(siteNamesJoin);
Here, I have to split the urlCurrent string using the words in the array. so that at the end I have to get www.example.com/web/americas. I am not getting the regex for that.

You may use
new RegExp("^.*?(?:" + siteNamesJoin + "|$)")
The pattern will look like
^.*?(?:americas|international|europe|asia-pacific|africa-middle-east|russia|india|$)
See the regex graph:
Details
^ - start of string
.*? - any 0 or more chars other than line break chars, as few as possible
(?:americas|international|europe|asia-pacific|africa-middle-east|russia|india|$) - any of the values in between pipes or end of string.
See JS demo:
var urlCurrent = "www.example.com/web/americas/home";
var siteNames = ["americas","international","europe","asia-pacific","africa-middle-east","russia","india"];
var siteNamesJoin = siteNames.join('|');
var match = urlCurrent.match(new RegExp("^.*?(?:" + siteNamesJoin + "|$)"));
var siteUrlCurrent = match ? match[0] : "";
console.log(siteUrlCurrent);
NOTE: if the siteNames "words" may contains special regex metacharacters, you will need to escape the siteNames items:
var siteNamesJoin = siteNames.map(function (x) { return x.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&') }).join('|');
Also, if those words must match in between / or end of string, you may adjust the pattern:
var match = urlCurrent.match(new RegExp("^(.*?)/(?:(?:" + siteNamesJoin + ")(?![^/])|$)"));
See another demo.
var urlCurrent = "www.example.com/web/americas/home";
var siteNames = ["americas","international","europe","asia-pacific","africa-middle-east","russia","india"];
var siteNamesJoin = siteNames.map(function (x) { return x.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&') }).join('|');
var match = urlCurrent.match(new RegExp("^.*?/(?:(?:" + siteNamesJoin + ")(?![^/])|$)"));
var siteUrlCurrent = match ? match[0] : "";
console.log(siteUrlCurrent);

using regex & match
var urlCurrent = "www.example.com/web/americas/home";
var siteNames = ["americas","international","europe","asia-pacific","africa-middle-east","russia","india"];
var regex = new RegExp(siteNames.join('|'));
var match = urlCurrent.match(regex);
if(match) {
var url = urlCurrent.substr(0, match[0].length + match.index)
console.log(url);
} else {
//invalid url
}

Maybe this helps you, i guess you don´t need a regex:
var urlCurrent = "www.example.com/web/americas/home";
var siteNames = ["americas","international","europe","asia-pacific","africa-middle-east","russia","india"];
var siteNamesJoin = siteNames.join('|');
var siteUrlCurrent = urlCurrent.split(siteNamesJoin);
const pos = urlCurrent.split("/")[2];
let url;
for(const index in siteNames) {
if(pos === siteNames[index]) {
url = urlCurrent.substring(0,urlCurrent.lastIndexOf("/"));
}
}
console.log(url);

I have to make a split function in JavaScript

Edit
sorry if the question wasn't clear
here is the question..
create your version of javascript split function,
you may use indexOf and substring to help.
so if i give you a string "heellloolllloolllo" and i want to remove "llll" the function should return "heellloooolllo"
This what I did so far:
function split() {
var entered_string = document.forms["form1"]["str"].value;
var deleted_char = document.forms["form1"]["char"].value;
var index = entered_string.indexOf(deleted_char);
var i = deleted_char.length;
var result;
var x ;
for (x = 0; x< entered_string.length; x++ )
{
if (index < 0) {
result = entered_string;
} else {
result = entered_string.substring(0, index) +entered_string.substring(index+i);
}
}
alert(result)
}

Use the replace() function with the g at the end of your regular expression. It's called a "global modifier".
var string = 'heellloolllloolllo';
var res = string.replace(/llll/g, '');
console.log(res)
If your substring is a variable then you need to construct a new Regex object and set the g as the second parameter.
var string = 'heellloolllloolllo';
var find = 'llll';
var regex = new RegExp(find,'g');
var res = string.replace(regex, '');
console.log(res)
There are other useful modifiers you can use:
g - Global replace. Replace all instances of the matched string in the provided text.
i - Case insensitive replace. Replace all instances of the matched string, ignoring differences in case.
m - Multi-line replace. The regular expression should be tested for matches over multiple lines.
See this post for more information, credit to #codejoe.

Using String#replace and RegExp (the clean way)
var str = 'llllheellloolllloolllollll';
var matchStr = 'llll';
function removeSubString(str, matchStr) {
var re = new RegExp(matchStr, 'g');
return str.replace(re,"");
}
console.log(removeSubString(str, matchStr));
Using String#indexOf and String#substring
var str = 'llllheellloolllloolllollll';
var matchStr = 'llll';
function removeSubString(str, matchStr) {
var index = str.indexOf(matchStr);
while(index != -1) {
var firstSubStr = str.substring(0, index);
var lastSubStr = str.substring(index + matchStr.length);
str = firstSubStr + lastSubStr;
index = str.indexOf(matchStr);
}
return str;
}
console.log(removeSubString(str,matchStr))

Trouble with Regular Expression and special charters

I am using a regular Expression for a glosary function on a website, but it can not "handle" special charters as æ, ø and å. The regEx is as follows:
var re = new RegExp("\\b" + pat + "\\b", "g");
How can i modify the RegEx above to handle special charters?

You may use the solution from here:
function getWholeWords(input, word) {
var pL = "a-zA-Z\\xAA\\xB5\\xBA\\xC0-\\xD6\\xD8-\\xF6\\xF8-\\u02C1\\u02C6-\\u02D1\\u02E0-\\u02E4\\u02EC\\u02EE\\u0370-\\u0374\\u0376\\u0377\\u037A-\\u037D\\u037F\\u0386\\u0388-\\u038A\\u038C\\u038E-\\u03A1\\u03A3-\\u03F5\\u03F7-\\u0481\\u048A-\\u052F\\u0531-\\u0556\\u0559\\u0561-\\u0587\\u05D0-\\u05EA\\u05F0-\\u05F2\\u0620-\\u064A\\u066E\\u066F\\u0671-\\u06D3\\u06D5\\u06E5\\u06E6\\u06EE\\u06EF\\u06FA-\\u06FC\\u06FF\\u0710\\u0712-\\u072F\\u074D-\\u07A5\\u07B1\\u07CA-\\u07EA\\u07F4\\u07F5\\u07FA\\u0800-\\u0815\\u081A\\u0824\\u0828\\u0840-\\u0858\\u08A0-\\u08B2\\u0904-\\u0939\\u093D\\u0950\\u0958-\\u0961\\u0971-\\u0980\\u0985-\\u098C\\u098F\\u0990\\u0993-\\u09A8\\u09AA-\\u09B0\\u09B2\\u09B6-\\u09B9\\u09BD\\u09CE\\u09DC\\u09DD\\u09DF-\\u09E1\\u09F0\\u09F1\\u0A05-\\u0A0A\\u0A0F\\u0A10\\u0A13-\\u0A28\\u0A2A-\\u0A30\\u0A32\\u0A33\\u0A35\\u0A36\\u0A38\\u0A39\\u0A59-\\u0A5C\\u0A5E\\u0A72-\\u0A74\\u0A85-\\u0A8D\\u0A8F-\\u0A91\\u0A93-\\u0AA8\\u0AAA-\\u0AB0\\u0AB2\\u0AB3\\u0AB5-\\u0AB9\\u0ABD\\u0AD0\\u0AE0\\u0AE1\\u0B05-\\u0B0C\\u0B0F\\u0B10\\u0B13-\\u0B28\\u0B2A-\\u0B30\\u0B32\\u0B33\\u0B35-\\u0B39\\u0B3D\\u0B5C\\u0B5D\\u0B5F-\\u0B61\\u0B71\\u0B83\\u0B85-\\u0B8A\\u0B8E-\\u0B90\\u0B92-\\u0B95\\u0B99\\u0B9A\\u0B9C\\u0B9E\\u0B9F\\u0BA3\\u0BA4\\u0BA8-\\u0BAA\\u0BAE-\\u0BB9\\u0BD0\\u0C05-\\u0C0C\\u0C0E-\\u0C10\\u0C12-\\u0C28\\u0C2A-\\u0C39\\u0C3D\\u0C58\\u0C59\\u0C60\\u0C61\\u0C85-\\u0C8C\\u0C8E-\\u0C90\\u0C92-\\u0CA8\\u0CAA-\\u0CB3\\u0CB5-\\u0CB9\\u0CBD\\u0CDE\\u0CE0\\u0CE1\\u0CF1\\u0CF2\\u0D05-\\u0D0C\\u0D0E-\\u0D10\\u0D12-\\u0D3A\\u0D3D\\u0D4E\\u0D60\\u0D61\\u0D7A-\\u0D7F\\u0D85-\\u0D96\\u0D9A-\\u0DB1\\u0DB3-\\u0DBB\\u0DBD\\u0DC0-\\u0DC6\\u0E01-\\u0E30\\u0E32\\u0E33\\u0E40-\\u0E46\\u0E81\\u0E82\\u0E84\\u0E87\\u0E88\\u0E8A\\u0E8D\\u0E94-\\u0E97\\u0E99-\\u0E9F\\u0EA1-\\u0EA3\\u0EA5\\u0EA7\\u0EAA\\u0EAB\\u0EAD-\\u0EB0\\u0EB2\\u0EB3\\u0EBD\\u0EC0-\\u0EC4\\u0EC6\\u0EDC-\\u0EDF\\u0F00\\u0F40-\\u0F47\\u0F49-\\u0F6C\\u0F88-\\u0F8C\\u1000-\\u102A\\u103F\\u1050-\\u1055\\u105A-\\u105D\\u1061\\u1065\\u1066\\u106E-\\u1070\\u1075-\\u1081\\u108E\\u10A0-\\u10C5\\u10C7\\u10CD\\u10D0-\\u10FA\\u10FC-\\u1248\\u124A-\\u124D\\u1250-\\u1256\\u1258\\u125A-\\u125D\\u1260-\\u1288\\u128A-\\u128D\\u1290-\\u12B0\\u12B2-\\u12B5\\u12B8-\\u12BE\\u12C0\\u12C2-\\u12C5\\u12C8-\\u12D6\\u12D8-\\u1310\\u1312-\\u1315\\u1318-\\u135A\\u1380-\\u138F\\u13A0-\\u13F4\\u1401-\\u166C\\u166F-\\u167F\\u1681-\\u169A\\u16A0-\\u16EA\\u16F1-\\u16F8\\u1700-\\u170C\\u170E-\\u1711\\u1720-\\u1731\\u1740-\\u1751\\u1760-\\u176C\\u176E-\\u1770\\u1780-\\u17B3\\u17D7\\u17DC\\u1820-\\u1877\\u1880-\\u18A8\\u18AA\\u18B0-\\u18F5\\u1900-\\u191E\\u1950-\\u196D\\u1970-\\u1974\\u1980-\\u19AB\\u19C1-\\u19C7\\u1A00-\\u1A16\\u1A20-\\u1A54\\u1AA7\\u1B05-\\u1B33\\u1B45-\\u1B4B\\u1B83-\\u1BA0\\u1BAE\\u1BAF\\u1BBA-\\u1BE5\\u1C00-\\u1C23\\u1C4D-\\u1C4F\\u1C5A-\\u1C7D\\u1CE9-\\u1CEC\\u1CEE-\\u1CF1\\u1CF5\\u1CF6\\u1D00-\\u1DBF\\u1E00-\\u1F15\\u1F18-\\u1F1D\\u1F20-\\u1F45\\u1F48-\\u1F4D\\u1F50-\\u1F57\\u1F59\\u1F5B\\u1F5D\\u1F5F-\\u1F7D\\u1F80-\\u1FB4\\u1FB6-\\u1FBC\\u1FBE\\u1FC2-\\u1FC4\\u1FC6-\\u1FCC\\u1FD0-\\u1FD3\\u1FD6-\\u1FDB\\u1FE0-\\u1FEC\\u1FF2-\\u1FF4\\u1FF6-\\u1FFC\\u2071\\u207F\\u2090-\\u209C\\u2102\\u2107\\u210A-\\u2113\\u2115\\u2119-\\u211D\\u2124\\u2126\\u2128\\u212A-\\u212D\\u212F-\\u2139\\u213C-\\u213F\\u2145-\\u2149\\u214E\\u2183\\u2184\\u2C00-\\u2C2E\\u2C30-\\u2C5E\\u2C60-\\u2CE4\\u2CEB-\\u2CEE\\u2CF2\\u2CF3\\u2D00-\\u2D25\\u2D27\\u2D2D\\u2D30-\\u2D67\\u2D6F\\u2D80-\\u2D96\\u2DA0-\\u2DA6\\u2DA8-\\u2DAE\\u2DB0-\\u2DB6\\u2DB8-\\u2DBE\\u2DC0-\\u2DC6\\u2DC8-\\u2DCE\\u2DD0-\\u2DD6\\u2DD8-\\u2DDE\\u2E2F\\u3005\\u3006\\u3031-\\u3035\\u303B\\u303C\\u3041-\\u3096\\u309D-\\u309F\\u30A1-\\u30FA\\u30FC-\\u30FF\\u3105-\\u312D\\u3131-\\u318E\\u31A0-\\u31BA\\u31F0-\\u31FF\\u3400-\\u4DB5\\u4E00-\\u9FCC\\uA000-\\uA48C\\uA4D0-\\uA4FD\\uA500-\\uA60C\\uA610-\\uA61F\\uA62A\\uA62B\\uA640-\\uA66E\\uA67F-\\uA69D\\uA6A0-\\uA6E5\\uA717-\\uA71F\\uA722-\\uA788\\uA78B-\\uA78E\\uA790-\\uA7AD\\uA7B0\\uA7B1\\uA7F7-\\uA801\\uA803-\\uA805\\uA807-\\uA80A\\uA80C-\\uA822\\uA840-\\uA873\\uA882-\\uA8B3\\uA8F2-\\uA8F7\\uA8FB\\uA90A-\\uA925\\uA930-\\uA946\\uA960-\\uA97C\\uA984-\\uA9B2\\uA9CF\\uA9E0-\\uA9E4\\uA9E6-\\uA9EF\\uA9FA-\\uA9FE\\uAA00-\\uAA28\\uAA40-\\uAA42\\uAA44-\\uAA4B\\uAA60-\\uAA76\\uAA7A\\uAA7E-\\uAAAF\\uAAB1\\uAAB5\\uAAB6\\uAAB9-\\uAABD\\uAAC0\\uAAC2\\uAADB-\\uAADD\\uAAE0-\\uAAEA\\uAAF2-\\uAAF4\\uAB01-\\uAB06\\uAB09-\\uAB0E\\uAB11-\\uAB16\\uAB20-\\uAB26\\uAB28-\\uAB2E\\uAB30-\\uAB5A\\uAB5C-\\uAB5F\\uAB64\\uAB65\\uABC0-\\uABE2\\uAC00-\\uD7A3\\uD7B0-\\uD7C6\\uD7CB-\\uD7FB\\uF900-\\uFA6D\\uFA70-\\uFAD9\\uFB00-\\uFB06\\uFB13-\\uFB17\\uFB1D\\uFB1F-\\uFB28\\uFB2A-\\uFB36\\uFB38-\\uFB3C\\uFB3E\\uFB40\\uFB41\\uFB43\\uFB44\\uFB46-\\uFBB1\\uFBD3-\\uFD3D\\uFD50-\\uFD8F\\uFD92-\\uFDC7\\uFDF0-\\uFDFB\\uFE70-\\uFE74\\uFE76-\\uFEFC\\uFF21-\\uFF3A\\uFF41-\\uFF5A\\uFF66-\\uFFBE\\uFFC2-\\uFFC7\\uFFCA-\\uFFCF\\uFFD2-\\uFFD7\\uFFDA-\\uFFDC";
var rx = RegExp("(?:^|[^_0-9" + pL + "])(" + word + ")(?![_0-9" + pL + "])", "ig"); // Build the regex (might be moved out from the function)
var words = [];
while ((m = rx.exec(input)) !== null) {
words.push(m[1]); // Add an occurrence
}
return words;
}
var word = "æøå";
var input = "æøå, AæøåZ, BæøåY, and æøå!";
document.body.innerHTML = "<pre>" + JSON.stringify(getWholeWords(input, word), 0, 4) + "</pre>";
Or a regex that will look for a word only if it is enclosed with whitespace/start/end of the string:
var re = new RegExp("(?:^|\\s)(" + pat.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&') + ")(?!\\S)", "g");
and grab Group 1 value.

Replace capture group content in JavaScript

In my JavaScript code I have a regular expression with capture groups (that is configured by library user) and a source string which matches this regular expression. The regular expression matches whole string (i.e. it has ^ and $ characters at its start and end).
A silly example:
var regex = /^([a-zA-Z]{2})-([0-9]{3})_.*$/;
var sourceStr = "ab-123_foo";
I want to reassemble the source string, replacing values in the capture groups and leaving the rest of the string intact. Note that, while this example has most of the "rest of the string" at its end, it actually may be anywhere else.
For example:
var replacements = [ "ZX", "321" ];
var expectedString = "ZX-321_foo";
Is there a way to do this in JavaScript?
NB: The regular expression is configured by the library user via the legacy API. I can not ask user to provide a second regular expression to solve this problem.

Without changing the regex the best I can think of is a callback that replaces the matches
sourceStr = sourceStr.replace(regex, function(match, $1, $2, offset, str) {
return str.replace($1, replacements[0]).replace($2, replacements[1]);
});
That's not a very good solution, as it would fail on something like
var sourceStr = "ab_ab-123_foo";
as it would replace the first ab instead of the matched one etc. but works for the given example and any string that doesn't repeat the matched characters
var regex = /^([a-zA-Z]{2})-([0-9]{3})_.*$/;
var sourceStr = "ab-123_foo";
var replacements = [ "ZX", "321" ];
sourceStr = sourceStr.replace(regex, function(match, $1, $2, offset, str) {
return str.replace($1, replacements[0]).replace($2, replacements[1]);
});
document.body.innerHTML = sourceStr;

I think this is close. It satisfies the two test cases but I'm unsure about leading and trailing groupings.
function replacer (regex, sourceStr, replacements) {
// Make a new regex that adds groups to ungrouped items.
var groupAll = "";
var lastIndex = 0;
var src = regex.source;
var reGroup=/\(.*?\)/g;
var match;
while(match = reGroup.exec(src)){
groupAll += "(" + src.substring(lastIndex, match.index) + ")";
groupAll += match[0];
lastIndex = match.index + match[0].length;
}
var reGroupAll = new RegExp(groupAll);
// Replace the original groupings with the replacements
// and append what was previously ungrouped.
var rep = sourceStr.replace(reGroupAll, function(){
// (match, $1, $2, ..., index, source)
var len = arguments.length - 2;
var ret = "";
for (var i = 1,j=0; i < len; i+=2,j++) {
ret += arguments[i];
ret += replacements[j];
}
return ret;
});
return rep;
}
var regex = /^([a-zA-Z]{2})-([0-9]{3})_.*$/;
var sourceStr = "ab-123_foo";
var replacements = [ "ZX", "321" ];
var expectedString = "ZX-321_foo";
var replaced = replacer(regex, sourceStr, replacements);
console.log(replaced);
console.log(replaced === expectedString);
regex = /^.*_([a-zA-Z]{2})-([0-9]{3})$/;
sourceStr = "ab_ab-123";
expectedString = "ab_ZX-321";
var replaced = replacer(regex, sourceStr, replacements);
console.log(replaced);
console.log(replaced === expectedString);
Output:
ZX-321_foo
true
ab_ZX-321
true

Remove all occurrences except last?

I want to remove all occurrences of substring = . in a string except the last one.
E.G:
1.2.3.4
should become:
123.4

You can use regex with positive look ahead,
"1.2.3.4".replace(/[.](?=.*[.])/g, "");

2-liner:
function removeAllButLast(string, token) {
/* Requires STRING not contain TOKEN */
var parts = string.split(token);
return parts.slice(0,-1).join('') + token + parts.slice(-1)
}
Alternative version without the requirement on the string argument:
function removeAllButLast(string, token) {
var parts = string.split(token);
if (parts[1]===undefined)
return string;
else
return parts.slice(0,-1).join('') + token + parts.slice(-1)
}
Demo:
> removeAllButLast('a.b.c.d', '.')
"abc.d"
The following one-liner is a regular expression that takes advantage of the fact that the * character is greedy, and that replace will leave the string alone if no match is found. It works by matching [longest string including dots][dot] and leaving [rest of string], and if a match is found it strips all '.'s from it:
'a.b.c.d'.replace(/(.*)\./, x => x.replace(/\./g,'')+'.')
(If your string contains newlines, you will have to use [.\n] rather than naked .s)

You can do something like this:
var str = '1.2.3.4';
var last = str.lastIndexOf('.');
var butLast = str.substring(0, last).replace(/\./g, '');
var res = butLast + str.substring(last);
Live example:
http://jsfiddle.net/qwjaW/

You could take a positive lookahead (for keeping the last dot, if any) and replace the first coming dots.
var string = '1.2.3.4';
console.log(string.replace(/\.(?=.*\.)/g, ''));

A replaceAllButLast function is more useful than a removeAllButLast function. When you want to remove just replace with an empty string:
function replaceAllButLast(str, pOld, pNew) {
var parts = str.split(pOld)
if (parts.length === 1) return str
return parts.slice(0, -1).join(pNew) + pOld + parts.slice(-1)
}
var test = 'hello there hello there hello there'
test = replaceAllButLast(test, ' there', '')
console.log(test) // hello hello hello there

Found a much better way of doing this. Here is replaceAllButLast and appendAllButLast as they should be done. The latter does a replace whilst preserving the original match. To remove, just replace with an empty string.
var str = "hello there hello there hello there"
function replaceAllButLast(str, regex, replace) {
var reg = new RegExp(regex, 'g')
return str.replace(reg, function(match, offset, str) {
var follow = str.slice(offset);
var isLast = follow.match(reg).length == 1;
return (isLast) ? match : replace
})
}
function appendAllButLast(str, regex, append) {
var reg = new RegExp(regex, 'g')
return str.replace(reg, function(match, offset, str) {
var follow = str.slice(offset);
var isLast = follow.match(reg).length == 1;
return (isLast) ? match : match + append
})
}
var replaced = replaceAllButLast(str, / there/, ' world')
console.log(replaced)
var appended = appendAllButLast(str, / there/, ' fred')
console.log(appended)
Thanks to #leaf for these masterpieces which he gave here.

You could reverse the string, remove all occurrences of substring except the first, and reverse it again to get what you want.

function formatString() {
var arr = ('1.2.3.4').split('.');
var arrLen = arr.length-1;
var outputString = '.' + arr[arrLen];
for (var i=arr.length-2; i >= 0; i--) {
outputString = arr[i]+outputString;
}
alert(outputString);
}
See it in action here: http://jsbin.com/izebay

var s='1.2.3.4';
s=s.split('.');
s.splice(s.length-1,0,'.');
s.join('');
123.4