Develop Reference

Regex extracting multiple matches for string [duplicate]

I'm trying to obtain all possible matches from a string using regex with javascript. It appears that my method of doing this is not matching parts of the string that have already been matched.
Variables:
var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';
var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;
Code:
var match = string.match(reg);
All matched results I get:
A1B1Y:A1B2Y
A1B5Y:A1B6Y
A1B9Y:A1B10Y
Matched results I want:
A1B1Y:A1B2Y
A1B2Y:A1B3Y
A1B5Y:A1B6Y
A1B6Y:A1B7Y
A1B9Y:A1B10Y
A1B10Y:A1B11Y
In my head, I want A1B1Y:A1B2Y to be a match along with A1B2Y:A1B3Y, even though A1B2Y in the string will need to be part of two matches.

Without modifying your regex, you can set it to start matching at the beginning of the second half of the match after each match using .exec and manipulating the regex object's lastIndex property.
var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';
var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;
var matches = [], found;
while (found = reg.exec(string)) {
matches.push(found[0]);
reg.lastIndex -= found[0].split(':')[1].length;
}
console.log(matches);
//["A1B1Y:A1B2Y", "A1B2Y:A1B3Y", "A1B5Y:A1B6Y", "A1B6Y:A1B7Y", "A1B9Y:A1B10Y", "A1B10Y:A1B11Y"]
Demo
As per Bergi's comment, you can also get the index of the last match and increment it by 1 so it instead of starting to match from the second half of the match onwards, it will start attempting to match from the second character of each match onwards:
reg.lastIndex = found.index+1;
Demo
The final outcome is the same. Though, Bergi's update has a little less code and performs slightly faster. =]

You cannot get the direct result from match, but it is possible to produce the result via RegExp.exec and with some modification to the regex:
var regex = /A[0-9]+B[0-9]+Y(?=(:A[0-9]+B[0-9]+Y))/g;
var input = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y'
var arr;
var results = [];
while ((arr = regex.exec(input)) !== null) {
results.push(arr[0] + arr[1]);
}
I used zero-width positive look-ahead (?=pattern) in order not to consume the text, so that the overlapping portion can be rematched.
Actually, it is possible to abuse replace method to do achieve the same result:
var input = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y'
var results = [];
input.replace(/A[0-9]+B[0-9]+Y(?=(:A[0-9]+B[0-9]+Y))/g, function ($0, $1) {
results.push($0 + $1);
return '';
});
However, since it is replace, it does extra useless replacement work.

Unfortunately, it's not quite as simple as a single string.match.
The reason is that you want overlapping matches, which the /g flag doesn't give you.
You could use lookahead:
var re = /A\d+B\d+Y(?=:A\d+B\d+Y)/g;
But now you get:
string.match(re); // ["A1B1Y", "A1B2Y", "A1B5Y", "A1B6Y", "A1B9Y", "A1B10Y"]
The reason is that lookahead is zero-width, meaning that it just says whether the pattern comes after what you're trying to match or not; it doesn't include it in the match.
You could use exec to try and grab what you want. If a regex has the /g flag, you can run exec repeatedly to get all the matches:
// using re from above to get the overlapping matches
var m;
var matches = [];
var re2 = /A\d+B\d+Y:A\d+B\d+Y/g; // make another regex to get what we need
while ((m = re.exec(string)) !== null) {
// m is a match object, which has the index of the current match
matches.push(string.substring(m.index).match(re2)[0]);
}
matches == [
"A1B1Y:A1B2Y",
"A1B2Y:A1B3Y",
"A1B5Y:A1B6Y",
"A1B6Y:A1B7Y",
"A1B9Y:A1B10Y",
"A1B10Y:A1B11Y"
];
Here's a fiddle of this in action. Open up the console to see the results
Alternatively, you could split the original string on :, then loop through the resulting array, pulling out the the ones that match when array[i] and array[i+1] both match like you want.

How to remove the last matched regex pattern in javascript

I have a text which goes like this...
var string = '~a=123~b=234~c=345~b=456'
I need to extract the string such that it splits into
['~a=123~b=234~c=345','']
That is, I need to split the string with /b=.*/ pattern but it should match the last found pattern. How to achieve this using RegEx?
Note: The numbers present after the equal is randomly generated.
Edit:
The above one was just an example. I did not make the question clear I guess.
Generalized String being...
<word1>=<random_alphanumeric_word>~<word2>=<random_alphanumeric_word>..~..~..<word2>=<random_alphanumeric_word>
All have random length and all wordi are alphabets, the whole string length is not fixed. the only text known would be <word2>. Hence I needed RegEx for it and pattern being /<word2>=.*/

This doesn't sound like a job for regexen considering that you want to extract a specific piece. Instead, you can just use lastIndexOf to split the string in two:
var lio = str.lastIndexOf('b=');
var arr = [];
var arr[0] = str.substr(0, lio);
var arr[1] = str.substr(lio);
http://jsfiddle.net/NJn6j/

I don't think I'd personally use a regex for this type of problem, but you can extract the last option pair with a regex like this:
var str = '~a=123~b=234~c=345~b=456';
var matches = str.match(/^(.*)~([^=]+=[^=]+)$/);
// matches[1] = "~a=123~b=234~c=345"
// matches[2] = "b=456"
Demo: http://jsfiddle.net/jfriend00/SGMRC/

Assuming the format is (~, alphanumeric name, =, and numbers) repeated arbitrary number of times. The most important assumption here is that ~ appear once for each name-value pair, and it doesn't appear in the name.
You can remove the last token by a simple replacement:
str.replace(/(.*)~.*/, '$1')
This works by using the greedy property of * to force it to match the last ~ in the input.
This can also be achieved with lastIndexOf, since you only need to know the index of the last ~:
str.substring(0, (str.lastIndexOf('~') + 1 || str.length() + 1) - 1)
(Well, I don't know if the code above is good JS or not... I would rather write in a few lines. The above is just for showing one-liner solution).

A RegExp that will give a result that you may could use is:
string.match(/[a-z]*?=(.*?((?=~)|$))/gi);
// ["a=123", "b=234", "c=345", "b=456"]
But in your case the simplest solution is to split the string before extract the content:
var results = string.split('~'); // ["", "a=123", "b=234", "c=345", "b=456"]
Now will be easy to extract the key and result to add to an object:
var myObj = {};
results.forEach(function (item) {
if(item) {
var r = item.split('=');
if (!myObj[r[0]]) {
myObj[r[0]] = [r[1]];
} else {
myObj[r[0]].push(r[1]);
}
}
});
console.log(myObj);
Object:
a: ["123"]
b: ["234", "456"]
c: ["345"]

(?=.*(~b=[^~]*))\1
will get it done in one match, but if there are duplicate entries it will go to the first. Performance also isn't great and if you string.replace it will destroy all duplicates. It would pass your example, but against '~a=123~b=234~c=345~b=234' it would go to the first 'b=234'.
.*(~b=[^~]*)
will run a lot faster, but it requires another step because the match comes out in a group:
var re = /.*(~b=[^~]*)/.exec(string);
var result = re[1]; //~b=234
var array = string.split(re[1]);
This method will also have the with exact duplicates. Another option is:
var regex = /.*(~b=[^~]*)/g;
var re = regex.exec(string);
var result = re[1];
// if you want an array from either side of the string:
var array = [string.slice(0, regex.lastIndex - re[1].length - 1), string.slice(regex.lastIndex, string.length)];
This actually finds the exact location of the last match and removes it regex.lastIndex - re[1].length - 1 is my guess for the index to remove the ellipsis from the leading side, but I didn't test it so it might be off by 1.

How can I split this string in JavaScript?

I have strings like this:
ab
rx'
wq''
pok'''
oyu,
mi,,,,
Basically, I want to split the string into two parts. The first part should have the alphabetical characters intact, the second part should have the non-alphabetical characters.
The alphabetical part is guaranteed to be 2-3 lowercase characters between a and z; the non-alphabetical part can be any length, and is gauranteed to only be the characters , or ', but not both in the one string (e.g. eex,', will never occur).
So the result should be:
[ab][]
[rx][']
[wq]['']
[pok][''']
[oyu][,]
[mi][,,,,]
How can I do this? I'm guessing a regular expression but I'm not particularly adept at coming up with them.

Regular expressions have is a nice special called "word boundary" (\b). You can use it, well, to detect the boundary of a word, which is a sequence of alpha-numerical characters.
So all you have to do is
foo.split(/\b/)
For example,
"pok'''".split(/\b/) // ["pok", "'''"]

If you can 100% guarantee that:
Letter-strings are 2 or 3 characters
There are always one or more primes/commas
There is never any empty space before, after or in-between the letters and the marks
(aside from line-break)
You can use:
/^([a-zA-Z]{2,3})('+|,+)$/gm
var arr = /^([a-zA-Z]{2,3})('+|,+)$/gm.exec("pok'''");
arr === ["pok'''", "pok", "'''"];
var arr = /^([a-zA-Z]{2,3})('+|,+)$/gm.exec("baf,,,");
arr === ["baf,,,", "baf", ",,,"];
Of course, save yourself some sanity, and save that RegEx as a var.
And as a warning, if you haven't dealt with RegEx like this:
If a match isn't found -- if you try to match foo','' by mixing marks, or you have 0-1 or 4+ letters, or 0 marks... ...then instead of getting an array back, you'll get null.
So you can do this:
var reg = /^([a-zA-Z]{2,3})('+|,+)$/gm,
string = "foobar'',,''",
result_array = reg.exec(string) || [string];
In this case, the result of the exec is null; by putting the || (or) there, we can return an array that has the original string in it, as index-0.
Why?
Because the result of a successful exec will have 3 slots; [*string*, *letters*, *marks*].
You might be tempted to just read the letters like result_array[1].
But if the match failed and result_array === null, then JavaScript will scream at you for trying null[1].
So returning the array at the end of a failed exec will allow you to get result_array[1] === undefined (ie: there was no match to the pattern, so there are no letters in index-1), rather than a JS error.

You could try something like that:
function splitString(string){
var match1 = null;
var match2 = null;
var stringArray = new Array();
match1 = string.indexOf(',');
match2 = string.indexOf('`');
if(match1 != 0){
stringArray = [string.slice(0,match1-1),string.slice(match1,string.length-1];
}
else if(match2 != 0){
stringArray = [string.slice(0,match2-1),string.slice(match2,string.length-1];
}
else{
stringArray = [string];
}
}

var str = "mi,,,,";
var idx = str.search(/\W/);
if(idx) {
var list = [str.slice(0, idx), str.slice(idx)]
}
You'll have the parts in list[0] and list[1].
P.S. There might be some better ways than this.

yourStr.match(/(\w{2,3})([,']*)/)

if (match = string.match(/^([a-z]{2,3})(,+?$|'+?$)/)) {
match = match.slice(1);
}

Javascript split only once and ignore the rest

I am parsing some key value pairs that are separated by colons. The problem I am having is that in the value section there are colons that I want to ignore but the split function is picking them up anyway.
sample:
Name: my name
description: this string is not escaped: i hate these colons
date: a date
On the individual lines I tried this line.split(/:/, 1) but it only matched the value part of the data. Next I tried line.split(/:/, 2) but that gave me ['description', 'this string is not escaped'] and I need the whole string.
Thanks for the help!

a = line.split(/:/);
key = a.shift();
val = a.join(':');

Use the greedy operator (?) to only split the first instance.
line.split(/: (.+)?/, 2);

If you prefer an alternative to regexp consider this:
var split = line.split(':');
var key = split[0];
var val = split.slice(1).join(":");
Reference: split, slice, join.

Slightly more elegant:
a = line.match(/(.*?):(.*)/);
key = a[1];
val = a[2];

May be this approach will be the best for such purpose:
var a = line.match(/([^:\s]+)\s*:\s*(.*)/);
var key = a[1];
var val = a[2];
So, you can use tabulations in your config/data files of such structure and also not worry about spaces before or after your name-value delimiter ':'.
Or you can use primitive and fast string functions indexOf and substr to reach your goal in, I think, the fastest way (by CPU and RAM)
for ( ... line ... ) {
var delimPos = line.indexOf(':');
if (delimPos <= 0) {
continue; // Something wrong with this "line"
}
var key = line.substr(0, delimPos).trim();
var val = line.substr(delimPos + 1).trim();
// Do all you need with this key: val
}

Split string in two at first occurrence
To split a string with multiple i.e. columns : only at the first column occurrence
use Positive Lookbehind (?<=)
const a = "Description: this: is: nice";
const b = "Name: My Name";
console.log(a.split(/(?<=^[^:]*):/)); // ["Description", " this: is: nice"]
console.log(b.split(/(?<=^[^:]*):/)); // ["Name", " My Name"]
it basically consumes from Start of string ^ everything that is not a column [^:] zero or more times *. Once the positive lookbehind is done, finally matches the column :.
If you additionally want to remove one or more whitespaces following the column,
use /(?<=^[^:]*): */
Explanation on Regex101.com

function splitOnce(str, sep) {
const idx = str.indexOf(sep);
return [str.slice(0, idx), str.slice(idx+1)];
}
splitOnce("description: this string is not escaped: i hate these colons", ":")

split string only on first instance of specified character

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)