Develop Reference

Regex match cookie value and remove hyphens

I'm trying to extract out a group of words from a larger string/cookie that are separated by hyphens. I would like to replace the hyphens with a space and set to a variable. Javascript or jQuery.
As an example, the larger string has a name and value like this within it:
facility=34222%7CConner-Department-Store;
(notice the leading "C")
So first, I need to match()/find facility=34222%7CConner-Department-Store; with regex. Then break it down to "Conner Department Store"
var cookie = document.cookie;
var facilityValue = cookie.match( REGEX ); ??

var test = "store=874635%7Csomethingelse;facility=34222%7CConner-Department-Store;store=874635%7Csomethingelse;";
var test2 = test.replace(/^(.*)facility=([^;]+)(.*)$/, function(matchedString, match1, match2, match3){
return decodeURIComponent(match2);
});
console.log( test2 );
console.log( test2.split('|')[1].replace(/[-]/g, ' ') );

If I understood it correctly, you want to make a phrase by getting all the words between hyphens and disallowing two successive Uppercase letters in a word, so I'd prefer using Regex in that case.
This is a Regex solution, that works dynamically with any cookies in the same format and extract the wanted sentence from it:
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Demo:
var str = "facility=34222%7CConner-Department-Store;";
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Explanation:
Use this Regex (/([A-Z][a-z]+)-?/g to match the words between -.
Replace any - occurence in the matched words.
Then just join these matches array with white space.

Ok,
first, you should decode this string as follows:
var str = "facility=34222%7CConner-Department-Store;"
var decoded = decodeURIComponent(str);
// decoded = "facility=34222|Conner-Department-Store;"
Then you have multiple possibilities to split up this string.
The easiest way is to use substring()
var solution1 = decoded.substring(decoded.indexOf('|') + 1, decoded.length)
// solution1 = "Conner-Department-Store;"
solution1 = solution1.replace('-', ' ');
// solution1 = "Conner Department Store;"
As you can see, substring(arg1, arg2) returns the string, starting at index arg1 and ending at index arg2. See Full Documentation here
If you want to cut the last ; just set decoded.length - 1 as arg2 in the snippet above.
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1)
//returns "Conner-Department-Store"
or all above in just one line:
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1).replace('-', ' ')
If you want still to use a regular Expression to retrieve (perhaps more) data out of the string, you could use something similar to this snippet:
var solution2 = "";
var regEx= /([A-Za-z]*)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/;
if (regEx.test(decoded)) {
solution2 = decoded.match(regEx);
/* returns
[0:"facility=34222|Conner-Department-Store",
1:"facility",
2:"34222",
3:"Conner-Department-Store",
index:0,
input:"facility=34222|Conner-Department-Store;"
length:4] */
solution2 = solution2[3].replace('-', ' ');
// "Conner Department Store"
}
I have applied some rules for the regex to work, feel free to modify them according your needs.
facility can be any Word built with alphabetical characters lower and uppercase (no other chars) at any length
= needs to be the char =
34222 can be any number but no other characters
| needs to be the char |
Conner-Department-Store can be any characters except one of the following (reserved delimiters): :/?#[]#;,'
Hope this helps :)
edit: to find only the part
facility=34222%7CConner-Department-Store; just modify the regex to
match facility= instead of ([A-z]*)=:
/(facility)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/

You can use cookies.js, a mini framework from MDN (Mozilla Developer Network).
Simply include the cookies.js file in your application, and write:
docCookies.getItem("Connor Department Store");

Regular expression to test exact word with a trailing dot

I am having a dynamic variable which I need to match against a source.
Source: 'text clientLogin padding float';
search: '.clientLog'
The search text has a leading dot('.') which should be omitted while comparing.
For the above example I should:
Get the part of search after '.'(dot)
Check the source for the search text i.e clientLog & return true if whole word matches.(in this example it should return false as source has clientLogin).
I am trying to use RegEx to achieve this
var regEx = new RegExp(k); // k is the search text
if(regEx.test(classNames)){....
Above code is inside jQuery.each,k is the key of the object which is being iterated.I did not figure out how to omit the '.' but read somewhere to implement Word Boundries for the exact match.
Please suggest.
thanks

Try this:
var
source = 'text clientLogin padding float',
search = '.clientLog',
pattern = '\\b'+search.replace(/^\./, '')+'\\b',
result = new RegExp(pattern).test(source);
Notes:
We strip off the leading '.' from the search string while building the pattern
We use word boundary markers (\b). This helps ensure that "login" is not considered a valid match for "log", for example, like in your case.
The double-escaping (\\b) is necessary as we're building our pattern as a string - this is necessary for dynamic patterns fed to the RegExp constructor.

Stripping text
In JavaScript, you can strip text with the substring() method like this:
var str = "Hello World!";
var sub_str = str.substring(1, str.length);
In substring(x, y), x is the starting index of the new string, y is the ending index. The indecies in JavaScript start at 0, so we have to use the next index to omit the first character in the string.
You can also read it up here on W3Schools.
Regular Expressions
You can search RegEx patterns in strings like this:
var str = "Hello World!";
var pos = str.search(/World/); // Attention: No quotes here!
pos equals the index of the first match of the given expression. If your expression did not match your string, then pos will equal -1.
Note, that str.search(/World/); is basicly the same as str.search(new RegExp("World"));
You can also read it up here on W3Schools.
To check, if your string contains that classname, you could do this:
var str = "classname1 classname2 classname3";
var search_str = ".classname2";
if(str.search(new RegExp("\\b(" + search_str.substring(1, search_str.length) + ")\\b")) > -1){
// classname found
} else {
//classname not found
}

What's the exact regex to match the proper string?

My string has [1212,1212],[1212,11212],...
I'd like to extract each value into an array for example I'd want 1212,1212 as one pair and evaluate a series of steps.
Tried /[[0-9],[0-9]]/ but It wasn't doing the task as I wanted. Basically I'm a noob in Regex, could someone please help.
Thanks in advance

You need some modifications for your regular expression for it to work correctly:
/\[[0-9]+,[0-9]+\]/g
You need to escape square brackets [ because they have special meaning.
[0-9] matches only one digits, you need the + quantifier to match one or more digits and thus [0-9]+.
Use the global modifier g to extract all matches.
Then you can extract all the values into an array like this:
var input = "[1212,1212],[1212,11212]";
var pattern = /\[[0-9]+,[0-9]+\]/g;
var result = [];
var currentMatch;
while((currentMatch = pattern.exec(input)) != null) {
result.push(currentMatch.toString());
}
result;
Or if you don't need to find the matches successively one at a time, then you can use String.match() as #Matthew Mcveigh did:
var input = "[1212,1212],[1212,11212]";
var result = input.match(/\[[0-9]+,[0-9]+\]/g);

It seems like you just need to match one or more digits before and after a comma, so you could do the following:
"[1212,1212],[1212,11212]".match(/\d+,\d+/g)
Which will give you the array: ["1212,1212", "1212,11212"]
To extract the pairs:
var result = "[1212,1212],[1212,11212]".match(/\d+,\d+/g);
for (var i = 0; i < result.length; i++) {
var pair = result[i].match(/\d+/g),
left = pair[0], right = pair[1];
alert("left: " + left + ", right: " + right);
}

You need to escape the literal brackets that you want to match. You can also use \d to match "any digit", which makes it tidier. Also, you're only matching one digit. You need to match "one or more" (+ quantifier)
/\[\d+,\d+\]/g
That g modifier finds all matches in the string, otherwise only the first one is found.

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.

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)