Develop Reference

How to get the string value after the colon?

I really need your help,
How can I check a string to see if it has a ":" and then if it does, to get the string value after the ":"
ie.
var x = "1. Search - File Number: XAL-2017-463288"
var y = "XAL-2017-463288"

//check for the colon
if (x.indexOf(':') !== -1) {
//split and get
var y = x.split(':')[1];
}

Split on the colon, and grab the second member of the result. This assumes you want the first colon found.
var y = x.split(":")[1];
If the string didn't have a :, then y will be undefined, so a separate check isn't needed.
Or you could use .indexOf(). This assumes there's definitely a colon. Otherwise you'll get the whole string.
var y = x.slice(x.indexOf(":") + 1);
If you wanted to check for the colon, then save the result of .indexOf() to a variable first, and only do the .slice() if the index was not -1.

how to retrieve a string between to same charecter

I know how to use substring() but here I have a problem, I'd like to retrieve a number between two "_" from a unknown string length. here is my string for example.
7_28_li
and I want to get the 28. How can I proceed to do so ?
Thanks.

Regex
'7_28_li'.match(/_(\d+)_/)[1]
The slashes inside match make it's contents regex.
_s are taken literally
( and ) are for retrieving the contents (the target number) later
\d is a digit character
+ is "one or more".
The [1] on the end is accesses what got matched from the first set of parens, the one or more (+) digits (\d).
Loop
var str = '7_28_li';
var state = 0; //How many underscores have gone by
var num = '';
for (var i = 0; i < str.length; i++) {
if (str[i] == '_') state++;
else if (state == 1) num += str[i];
};
num = parseInt(num);
Probably more efficient, but kind of long and ugly.
Split
'7_28_li'.split('_')[1]
Split it into an array, then get the second element.
IndexOf
var str = "7_28_li";
var num = str.substring(str.indexOf('_') + 1, str.indexOf('_', 2));
Get the start and end point. Uses the little-known second parameter of indexOf. This works better than lastIndexOf because it is guaranteed to give the first number between _s, even when there are more than 2 underscores.

First find the index of _, and then find the next position of _. Then get the substring between them.
var data = "7_28_li";
var idx = data.indexOf("_");
console.log(data.substring(idx + 1, data.indexOf("_", idx + 1)));
# 28
You can understand that better, like this
var data = "7_28_li";
var first = data.indexOf("_");
var next = data.indexOf("_", first + 1);
console.log(data.substring(first + 1, next));
# 28
Note: The second argument to indexOf is to specify where to start looking from.

Probably the easiest way to do it is to call split on your string, with your delimiter ("_" in this case) as the argument. It'll return an array with 7, 28, and li as elements, so you can select the middle one.
"7_28_li".split("_")[1]
This will work if it'll always be 3 elements. If it's more, divide the length property by 2 and floor it to get the right element.
var splitstring = "7_28_li".split("_")
console.log(splitstring[Math.floor(splitstring.length/2)]);
I'm not sure how you want to handle even length strings, but all you have to do is set up an if statement and then do whatever you want.

If you know there would be 2 underscore, you can use this
var str = "7_28_li";
var res = str.substring(str.indexOf("_") +1, str.lastIndexOf("_"));
If you want to find the string between first 2 underscores
var str = "7_28_li";
var firstIndex = str.indexOf("_");
var secondIndex = str.indexOf("_", firstIndex+1);
var res = str.substring(firstIndex+1, secondIndex);

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.

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", ":")

How to extract a string using JavaScript Regex?

I'm trying to extract a substring from a file with JavaScript Regex. Here is a slice from the file :
DATE:20091201T220000
SUMMARY:Dad's birthday
the field I want to extract is "Summary". Here is the approach:
extractSummary : function(iCalContent) {
/*
input : iCal file content
return : Event summary
*/
var arr = iCalContent.match(/^SUMMARY\:(.)*$/g);
return(arr);
}

function extractSummary(iCalContent) {
var rx = /\nSUMMARY:(.*)\n/g;
var arr = rx.exec(iCalContent);
return arr[1];
}
You need these changes:
Put the * inside the parenthesis as
suggested above. Otherwise your matching
group will contain only one
character.
Get rid of the ^ and $. With the global option they match on start and end of the full string, rather than on start and end of lines. Match on explicit newlines instead.
I suppose you want the matching group (what's
inside the parenthesis) rather than
the full array? arr[0] is
the full match ("\nSUMMARY:...") and
the next indexes contain the group
matches.
String.match(regexp) is
supposed to return an array with the
matches. In my browser it doesn't (Safari on Mac returns only the full
match, not the groups), but
Regexp.exec(string) works.

You need to use the m flag:
multiline; treat beginning and end characters (^ and $) as working
over multiple lines (i.e., match the beginning or end of each line
(delimited by \n or \r), not only the very beginning or end of the
whole input string)
Also put the * in the right place:
"DATE:20091201T220000\r\nSUMMARY:Dad's birthday".match(/^SUMMARY\:(.*)$/gm);
//------------------------------------------------------------------^ ^
//-----------------------------------------------------------------------|

Your regular expression most likely wants to be
/\nSUMMARY:(.*)$/g
A helpful little trick I like to use is to default assign on match with an array.
var arr = iCalContent.match(/\nSUMMARY:(.*)$/g) || [""]; //could also use null for empty value
return arr[0];
This way you don't get annoying type errors when you go to use arr

This code works:
let str = "governance[string_i_want]";
let res = str.match(/[^governance\[](.*)[^\]]/g);
console.log(res);
res will equal "string_i_want". However, in this example res is still an array, so do not treat res like a string.
By grouping the characters I do not want, using [^string], and matching on what is between the brackets, the code extracts the string I want!
You can try it out here: https://www.w3schools.com/jsref/tryit.asp?filename=tryjsref_match_regexp
Good luck.

(.*) instead of (.)* would be a start. The latter will only capture the last character on the line.
Also, no need to escape the :.

You should use this :
var arr = iCalContent.match(/^SUMMARY\:(.)*$/g);
return(arr[0]);

this is how you can parse iCal files with javascript
function calParse(str) {
function parse() {
var obj = {};
while(str.length) {
var p = str.shift().split(":");
var k = p.shift(), p = p.join();
switch(k) {
case "BEGIN":
obj[p] = parse();
break;
case "END":
return obj;
default:
obj[k] = p;
}
}
return obj;
}
str = str.replace(/\n /g, " ").split("\n");
return parse().VCALENDAR;
}
example =
'BEGIN:VCALENDAR\n'+
'VERSION:2.0\n'+
'PRODID:-//hacksw/handcal//NONSGML v1.0//EN\n'+
'BEGIN:VEVENT\n'+
'DTSTART:19970714T170000Z\n'+
'DTEND:19970715T035959Z\n'+
'SUMMARY:Bastille Day Party\n'+
'END:VEVENT\n'+
'END:VCALENDAR\n'
cal = calParse(example);
alert(cal.VEVENT.SUMMARY);