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Having hard time with jQuery and replace string value

Im currently developing a posting [like What's on your mind] feature where im using twemoji plugin for emojis.
For some security reasons, i have to convert the emoji into its alt code/image filename before it stores to the database.
And convert it back to image when its being displayed on the feeds.
In my case I use [emoji=filename.png]
for example i have this string:
var string = "[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9-fe0f.png]";
string.replace(/-fe0f.png/g, '.png')
.replace(/\[emoji=(.*?)\]/g,'<img src="https://example.net/images/$1">');
the snippet above is working fine, but the only problem is it removes All -fe0f.png in the filename which causes some broken image.
What I want to achive is to remove the -fe0f.png part only when the filename length is <= 14. or maybe if the file name is consist of something like this: (char)-fe0f.png , but if it has more than (char) like (char)-(char)-(char)-fe0f.png, it should still remain the same..
the result should be:
from
[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9-fe0f.png]
to
[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9.png]
UPDATE:
I just noticed now that there are filenames like this 30-fe0f-20e3.png
but it needs to remove -fe0f in the middle.
so instead of [emoji=30-fe0f-20e3.png],
i need to have [emoji=30-20e3.png]

The file name length limit is equal to fourteen. Thus, there should be "nine" characters before the "-fe0f"
[^=] means all characters except "="
<![^=])a means there must not "=" before the "a"
<![^=]{9})a means it must not has a "=" character during the nine characters before the letter "a".
(?<![^=]{9})-fe0f.png means it must not has a "=" character during the nine characters before the "-fe0f.png".
So your new code should be like the below:
var string = "[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9-fe0f.png]";
string.replace(/(?<![^=]{9})-fe0f.png/g, '.png')
.replace(/\[emoji=(.*?)\]/g,'<img src="https://example.net/images/$1">');

Replacing the data in the example string:
const regex = /(\[emoji=[^\s\]\[]{0,13})-fe0f(\.png)/g;
let string = "[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9-fe0f.png]";
string = string.replace(regex, '$1$2');
console.log(string);
You can do the replacement in one replace call with a match and a capture group, matching 0-13 characters after emoji=
\[emoji=([^\s\]\[]{0,13})-fe0f\.png]
The pattern matches:
\[emoji= Match [emoji=
( Capture group 1
[^\s\]\[]{0,13} Match 0-13 times a non whitespace char except for [ and ]
) Close group 1
-fe0f\.png] Match literally (note to escape the dot)
regex demo
const regex = /\[emoji=([^\s\]\[]{0,13})-fe0f\.png]/g;
let string = "[emoji=1f938.png] [emoji=1f938-200d-2642-fe0f.png] [emoji=26f9-fe0f.png]";
string = string.replace(regex, '<img src="https://example.net/images/$1.png">');
console.log(string);

This should do it if you are just trying to not replace for greater than 14 chars.
if (string.length > 14) {
// do your replace here
}
Now, not sure if you are suggesting that if there's more than one "-" that you don't want to replace either.

Get a substring from a string for a regular expression in JavaScript

I have a string of the following form:
data-translate='view-7631b26ea80b1b601c313b15cc4e2ab03faedf30'>Avatar data
It can be in different languages, but in any case I need to get a string which is between the characters ' '
That is, in the example above, I need to get the following string:
view-7631b26ea80b1b601c313b15cc4e2ab03faedf30
Can I do this using the method string.replace(regexp, str) ?
I've highlighted the desired line using the following regular expression:
/'\b(.*)\b'/gm
Now, using the method string.replace I need to delete everything except that...
Got any suggestions?

Use match method.
var data = "data-translate='view-7631b26ea80b1b601c313b15cc4e2ab03faedf30'>Avatar data";
data = data.match(/'\b(.*)\b'/gm)

You have good solid anchor text in either side, so:
var match = /data-translate='([^']+)'/.exec(str);
var substr = match && match[1];
Live Example:
var str = "data-translate='view-7631b26ea80b1b601c313b15cc4e2ab03faedf30'>Avatar data";
var match = /data-translate='([^']+)'/.exec(str);
var substr = match && match[1];
document.body.innerHTML =
"<pre>Got: [" + substr + "]</pre>";
But again, as I said in a comment, using a simple regular expression to extract information from HTML is usually doomed to fail. For instance, you probably don't want to match this:
<p>The string is data-translate='view-7631b26ea80b1b601c313b15cc4e2ab03faedf30'</p>
...and yet, a simple regex solution will do exactly that. To properly handle HTML, you must use a proper HTML parser.

You can also try this one:
/\'([^\']+)\'/gm

Javascript Regexp Duplicate Line Matching not working correctly

I am writing a Javascript code to parse some grammar files, it is quite some code but I will post relevant information here. I am using Javascript Regexp in order to match a duplicate line held within a string. The string contains, for example (assume the string name is lines):
if
else
;
print
{
}
test1
test1
=
+
-
*
/
(
)
num
string
comment
id
test2
test2
What should happen, is a match found on 'test1' and 'test2'. It should then delete the duplicate, leaving 1 instance of test1 and test2. What is happening is no match at all. I am confident in my regex but javascript may be doing something I am not expecting. Here is the code doing the work on the string given above:
var rex = new RegExp("(.*)(\r?\n\1)+","g");
var re = '/(.*)(\r?\n\1)+/g';
rex.lastIndex = 0;
var m = rex.exec(lines);
if (m) {
alert("Found Duplicate");
var linenum = lines.search(re); //Get line number of error
alert("Error: Symbol Defined twice\n");
alert("Error occured on line: " + linenum);
lines = lines.replace(rex,""); //Gets rid of the duplicate
}
It never gets into the if(m) statement. Therefore no match is found. I tested the regex here: http://regexpal.com/ using the regex in my code as well as the example text provided. It matches just fine, so I am at kind of a loss. If anyone can help, it would be great.
Thank you.
Edit:
Forgot to add, I am testing this in firefox, and it only has to work in firefox. Not sure if that matters.

First error: \ in a JS string is also an escape character.
var rex = new RegExp("(.*)(\r?\n\1)+","g");
should be written
var rex = new RegExp("(.*)(\\r?\\n\\1)+","g");
// or, shorter:
var rex = /(.*)(\r?\n\1)+/g;
if you want to make it work. In the case of the RegExp constructor, you’re passing the pattern as a string to the constructor function. This means you need to escape each \ backslash that occurs in the pattern. If you use a regexp literal, you don’t need to escape them, since they’re not in a string, but retain their ‘normal’ properties in the regexp pattern.
Second error, your expression
var re = '/(.*)(\r?\n\1)+/g';
is wrong. What you’re doing here is assigning a string literal to a variable. I’m assuming you meant to assign a regular expression literal, which should be written like this:
var re = /(.*)(\r?\n\1)+/g;
Third error: the last line
lines = lines.replace(rex,""); //Gets rid of the duplicate
removes both instances of all duplicate lines! If you want to keep the first instance of each duplicate, you should use
lines = lines.replace(rex, "$1");
And finally, this method only detects two consecutive identical lines. Is that what you want, or do you need to detect any duplicates, wherever they may be?

var str = 'if\nelse\n;\nprint\n{\n}\ntest1\ntest1\n=\n+\n-\n*\n/\n(\n)\nnum\nstring\ncomment\nid\ntest2\ntest2\ntest2\ntest2\ntest2';
console.log(str);
str = str.replace(/\r\n?/g,'');
// I prefer replacing all the newline characters with \n's here
str = str.replace(/(^|\n)([^\n]*)(\n\2)+/g,function(m0,m1,m2,m3,ind) {
var line = str.substr(0,ind).split(/\n/).length + 1;
var msg = '[Found duplicate]';
msg += '\nFollowing symbol defined more than once';
msg += '\n\tsymbol: ' + m2;
msg += '\n\ton line ' + line;
console.log(msg);
return m1 + m2;
});
console.log(str);
Otherwise you can skip the first line and change the pattern into
/(^|\r\n?|\n)([^\r\n]*)((?:\r\n?|\n)\2)+/g
Note that [^\n]* will also catch multiple empty lines. If you want to make sure it matches (and replaces) non-empty lines then you might want to use [^\n]+.
[EDIT]
For the record, each m represents each arguments object, so m0 is the whole match, m1 is the 1st subgroup ((^|\n)), m2 is the 2nd subgroup (([^\n]*)) and m3 is the last subgroup ((\n\2)). I could have used arguments[n] instead but these are shorter.
As with the return value, due to lack of lookbehind in the regex flavor used by Javascript, this pattern is catching a possible preceding newline (unless it is the first line) so it needs to return the match and that preceding newline if any. That's why it shouldn't be returning m2 only.

String manipulation - getting value after the last position of a char

How I can get the value after last char(. ; + _ etc.):
e.g.
string.name+org.com
I want to get "com".
Is there any function in jQuery?

Use lastIndexOf and substr to find the character and get the part of the string after it:
var extension = name.substr(name.lastIndexOf(".") + 1);
Demo: http://jsfiddle.net/Guffa/K3BWn/

A simple and readable approch to get the substring after the last occurrence of a character from a defined set is to split the string with a regular expression containing a character class and then use pop() to get the last element of the resulting array:
The pop() method removes the last element from an array and returns that element.
See a JS demo below:
var s = 'string.name+org.com';
var result = s.split(/[.;+_]/).pop();
console.log(result);
to split at all non-overlapping occurrences of the regex by default.
NOTE: If you need to match ^, ], \ or -, you may escape them and use anywhere inside the character class (e.g. /[\^\-\]\\]/). It is possible to avoid escaping ^ (if you do not put it right after the opening [), - (if it is right after the opening [, right before the closing ], after a valid range, or between a shorthand character class and another symbol): /[-^\]\\]/.
Also, if you need to split with a single char, no regex is necessary:
// Get the substring after the last dot
var result = 'string.name+org.com'.split('.').pop();
console.log(result);

Not jQuery, just JavaScript: lastIndexOf and substring would do it (not since the update indicating multiple characters). As would a regular expression with a capture group containing a character class followed by an end-of-string anchor, e.g. /([^.;+_]+)$/ used with RegExp#exec or String#match.
E.g. (live copy | source):
var match = /([^.;+_]+)$/.exec(theStringToTest),
result = match && match[1];

var s = "string.name+org.com",
lw = s.replace(/^.+[\W]/, '');
console.log(lw) /* com */
this will also work for
string.name+org/com
string.name+org.info

You can use RegExp Object.
Try this code:
"http://stackoverflow.com".replace(/.*\./,"");

I'll throw in a crazy (i.e. no RegExp) one:
var s = 'string.name+org.com';
var a = s.split('.'); //puts all sub-Strings delimited by . into an Array
var result = a[a.length-1]; //gets the last element of that Array
alert(result);​
EDIT: Since the update of the question is demanding mutiple delimiters to work this is probably not the way to go. Too crazy.....

use javascript function like
url.substr(url.length - 3);

maybe this is too late to consider, this codes works fine for me using jquery
var afterDot = value.substr(value.lastIndexOf('_') + 1);
You could just replate '_' to '.'
var myString = 'asd/f/df/xc/asd/test.jpg'
var parts = myString.split('/');
var answer = parts[parts.length - 1];
console.log(answer);

Regex in javascript complex

string str contains somewhere within it http://www.example.com/ followed by 2 digits and 7 random characters (upper or lower case). One possibility is http://www.example.com/45kaFkeLd or http://www.example.com/64kAleoFr. So the only certain aspect is that it always starts with 2 digits.
I want to retrieve "64kAleoFr".
var url = str.match([regex here]);

The regex you’re looking for is /[0-9]{2}[a-zA-Z]{7}/.
var string = 'http://www.example.com/64kAleoFr',
match = (string.match(/[0-9]{2}[a-zA-Z]{7}/) || [''])[0];
console.log(match); // '64kAleoFr'
Note that on the second line, I use the good old .match() trick to make sure no TypeError is thrown when no match is found. Once this snippet has executed, match will either be the empty string ('') or the value you were after.

you could use
var url = str.match(/\d{2}.{7}$/)[0];
where:
\d{2} //two digits
.{7} //seven characters
$ //end of the string
if you don't know if it will be at the end you could use
var url = str.match(/\/\d{2}.{7}$/)[0].slice(1); //grab the "/" at the begining and slice it out

what about using split ?
alert("http://www.example.com/64kAleoFr".split("/")[3]);

var url = "http://www.example.com/",
re = new RegExp(url.replace(/\./g,"\\.") + "(\\d{2}[A-Za-z]{7})");
str = "This is a string with a url: http://www.example.com/45kaFkeLd in the middle.";
var code = str.match(re);
if (code != null) {
// we have a match
alert(code[1]); // "45kaFkeLd"
}​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​
The url needs to be part of the regex if you want to avoid matching other strings of characters elsewhere in the input. The above assumes that the url should be configurable, so it constructs a regex from the url variable (noting that "." has special meaning in a regex so it needs to be escaped). The bit with the two numbers and seven letter is then in parentheses so it can be captured.
Demo: http://jsfiddle.net/nnnnnn/NzELc/

http://www\\.example\\.com/([0-9]{2}\\w{7}) this is your pattern. You'll get your 2 digits and 7 random characters in group 1.

If you notice your example strings, both strings have few digits and a random string after a slash (/) and if the pattern is fixed then i would rather suggest you to split your string with slash and get the last element of the array which was the result of the split function.
Here is how:
var string = "http://www.example.com/64kAleoFr"
ar = string.split("/");
ar[ar.length - 1];
Hope it helps