Javascript Regex I'm struggling with - javascript

I am admittedly NOT a regex person but usually I can figure my way around something. This one has me stumped...
I need to match and replace a double greater than str (>>) that has an optional leading space. I know this doesn't work but something along the lines of...
/\s\s+[>>]/
But that's obviously no good.
Would appreciate any help. This site has been an amazing resource for me over the years and I can't believe I'm only getting around to posting something now, so it goes to show even a knucklehead like me has been able to benefit without bothering people... until now:) Thanks in advance.


For >> both inside a string and with leading whitespace, try:
/(\s*)(>>){1}/


If you want the space to be optional, then you can simply do this :
/>>/
And you may use it as a replacement pattern with the g modifier :
str = str.replace(/>>/g, 'something')
If you want to check that a string is >> with maybe some space before, then use
/^\s?>>$/


Breaking down your example:
\s will match any white-space character
\s+ will match one or more white-space characters
[>>] will match one > (see more below on this)
So your expression will match a > preceeded by at least two white-space characters.
If you want to match zero-or-more you will have to use *; fex \s*.
Square brackets are used to denote sets of characters, and will match any of the characters in the set; fex [abc] will match a, b or c, but only one character at time.
Single characters in a regular expression will match that character; fex > will match one greater-than sign.
Putting it together we get the following regular expression for your case:
/\s*>>/


This should work with the optional space.
/\s{0,}>>/g
Visit this link to test the matches.


If you want it to match an unlimited amount of leading space characters:
/ *>>/
If you want it to match 0 or 1 leading space character:
/ ?>>/


use this :
str.replace(/\s+>>/g, 'whatever');


This regex should work.
/\s*[>]{2}/
This is cleaner
/\s*>>/
Tested:
var pattern = /\s*>>/;
s1 = " >>";
s2 = ">>";
s3 = ">> surrounding text";
s4 = "surrounding >> text";
s5 = "surrounding>>text";
s1.match(pattern);
[" >>"]
s2.match(pattern);
[">>"]
s3.match(pattern);
[">>"]
s4.match(pattern);
[" >>"]
s5.match(pattern);
[">>"]
Replacement example
var pattern = /\s*>>/;
var s6 = " >> surrounding text";
s6.replace(pattern, ">");
"> surrounding text"

Related

RegEx for exact word search starting with Umlaut [duplicate]

I am building search and I am going to use javascript autocomplete with it. I am from Finland (finnish language) so I have to deal with some special characters like ä, ö and å
When user types text in to the search input field I try to match the text to data.
Here is simple example that is not working correctly if user types for example "ää". Same thing with "äl"
var title = "this is simple string with finnish word tämä on ääkköstesti älkää ihmetelkö";
// Does not work
var searchterm = "äl";
// does not work
//var searchterm = "ää";
// Works
//var searchterm = "wi";
if ( new RegExp("\\b"+searchterm, "gi").test(title) ) {
$("#result").html("Match: ("+searchterm+"): "+title);
} else {
$("#result").html("nothing found with term: "+searchterm);
}
http://jsfiddle.net/7TsxB/
So how can I get those ä,ö and å characters to work with javascript regex?
I think I should use unicode codes but how should I do that? Codes for those characters are:
[\u00C4,\u00E4,\u00C5,\u00E5,\u00D6,\u00F6]
=> äÄåÅöÖ

There appears to be a problem with Regex and the word boundary \b matching the beginning of a string with a starting character out of the normal 256 byte range.
Instead of using \b, try using (?:^|\\s)
var title = "this is simple string with finnish word tämä on ääkköstesti älkää ihmetelkö";
// Does not work
var searchterm = "äl";
// does not work
//var searchterm = "ää";
// Works
//var searchterm = "wi";
if ( new RegExp("(?:^|\\s)"+searchterm, "gi").test(title) ) {
$("#result").html("Match: ("+searchterm+"): "+title);
} else {
$("#result").html("nothing found with term: "+searchterm);
}
Breakdown:
(?: parenthesis () form a capture group in Regex. Parenthesis started with a question mark and colon ?: form a non-capturing group. They just group the terms together
^ the caret symbol matches the beginning of a string
| the bar is the "or" operator.
\s matches whitespace (appears as \\s in the string because we have to escape the backslash)
) closes the group
So instead of using \b, which matches word boundaries and doesn't work for unicode characters, we use a non-capturing group which matches the beginning of a string OR whitespace.

The \b character class in JavaScript RegEx is really only useful with simple ASCII encoding. \b is a shortcut code for the boundary between \w and \W sets or \w and the beginning or end of the string. These character sets only take into account ASCII "word" characters, where \w is equal to [a-zA-Z0-9_] and \W is the negation of that class.
This makes the RegEx character classes largely useless for dealing with any real language.
\s should work for what you want to do, provided that search terms are only delimited by whitespace.

this question is old, but I think I found a better solution for boundary in regular expressions with unicode letters.
Using XRegExp library you can implement a valid \b boundary expanding this
XRegExp('(?=^|$|[^\\p{L}])')
the result is a 4000+ char long, but it seems to work quite performing.
Some explanation: (?= ) is a zero-length lookahead that looks for a begin or end boundary or a non-letter unicode character. The most important think is the lookahead, because the \b doesn't capture anything: it is simply true or false.

\b is a shortcut for the transition between a letter and a non-letter character, or vice-versa.
Updating and improving on max_masseti's answer:
With the introduction of the /u modifier for RegExs in ES2018, you can now use \p{L} to represent any unicode letter, and \P{L} (notice the uppercase P) to represent anything but.
EDIT: Previous version was incomplete.
As such:
const text = 'A Fé, o Império, e as terras viciosas';
text.split(/(?<=\p{L})(?=\P{L})|(?<=\P{L})(?=\p{L})/);
// ['A', ' Fé', ',', ' o', ' Império', ',', ' e', ' as', ' terras', ' viciosas']
We're using a lookbehind (?<=...) to find a letter and a lookahead (?=...) to find a non-letter, or vice versa.

I would recommend you to use XRegExp when you have to work with a specific set of characters from Unicode, the author of this library mapped all kind of regional sets of characters making the work with different languages easier.

Despite the fact the issue seems to be 8 years old, I run into a similar problem (I had to match Cyrillic letters) not so far ago. I spend a whole day on this and could not find any appropriate answer here on StackOverflow. So, to avoid others making lots of effort, I'd like to share my solution.
Yes, \b word boundary works only with Latin letters (Word boundary: \b):
Word boundary \b doesn’t work for non-Latin alphabets
The word boundary test \b checks that there should be \w on the one side from the position and "not \w" – on the other side.
But \w means a Latin letter a-z (or a digit or an underscore), so the test doesn’t work for other characters, e.g. Cyrillic letters or hieroglyphs.
Yes, JavaScript RegExp implementation hardly supports UTF-8 encoding.
So, I tried implementing own word boundary feature with the support of non-Latin characters. To make word boundary work just with Cyrillic characters I created such regular expression:
new RegExp(`(?<![\u0400-\u04ff])${cyrillicSearchValue}(?![\u0400-\u04ff])`,'gi')
Where \u0400-\u04ff is a range of Cyrillic characters provided in the table of codes. It is not an ideal solution, however, it works properly in most cases.
To make it work in your case, you just have to pick up an appropriate range of codes from the list of Unicode characters.
To try out my example run the code snippet below.
function getMatchExpression(cyrillicSearchValue) {
return new RegExp(
`(?<![\u0400-\u04ff])${cyrillicSearchValue}(?![\u0400-\u04ff])`,
'gi',
);
}
const sentence = 'Будь-який текст кирилицею, де необхідно знайти слово з контексту';
console.log(sentence.match(getMatchExpression('текст')));
// expected output: ["текст"]
console.log(sentence.match(getMatchExpression('но')));
// expected output: null

I noticed something really weird with \b when using Unicode:
/\bo/.test("pop"); // false (obviously)
/\bä/.test("päp"); // true (what..?)
/\Bo/.test("pop"); // true
/\Bä/.test("päp"); // false (what..?)
It appears that meaning of \b and \B are reversed, but only when used with non-ASCII Unicode? There might be something deeper going on here, but I'm not sure what it is.
In any case, it seems that the word boundary is the issue, not the Unicode characters themselves. Perhaps you should just replace \b with (^|[\s\\/-_&]), as that seems to work correctly. (Make your list of symbols more comprehensive than mine, though.)

My idea is to search with codes representing the Finnish letters
new RegExp("\\b"+asciiOnly(searchterm), "gi").test(asciiOnly(title))
My original idea was to use plain encodeURI but the % sign seemed to interfere with the regexp.
http://jsfiddle.net/7TsxB/5/
I wrote a crude function using encodeURI to encode every character with code over 128 but removing its % and adding 'QQ' in the beginning. It is not the best marker but I couldn't get non alphanumeric to work.

What you are looking for is the Unicode word boundaries standard:
http://unicode.org/reports/tr29/tr29-9.html#Word_Boundaries
There is a JavaScript implementation here (unciodejs.wordbreak.js)
https://github.com/wikimedia/unicodejs

I had a similar problem, where I was trying to replace all of a particular unicode word with a different unicode word, and I cannot use lookbehind because it's not supported in the JS engine this code will be used in. I ultimately resolved it like this:
const needle = "КАРТОПЛЯ";
const replace = "БАРАБОЛЯ";
const regex = new RegExp(
String.raw`(^|[^\n\p{L}])`
+ needle
+ String.raw`(?=$|\P{L})`,
"gimu",
);
const result = (
'КАРТОПЛЯ сдффКАРТОПЛЯдадф КАРТОПЛЯ КАРТОПЛЯ КАРТОПЛЯ??? !!!КАРТОПЛЯ ;!;!КАРТОПЛЯ/#?#?'
+ '\n\nКАРТОПЛЯ КАРТОПЛЯ - - -КАРТОПЛЯ--'
)
.replace(regex, function (match, ...args) {
return args[0] + replace;
});
console.log(result)
output:
БАРАБОЛЯ сдффКАРТОПЛЯдадф БАРАБОЛЯ БАРАБОЛЯ БАРАБОЛЯ??? !!!БАРАБОЛЯ ;!;!БАРАБОЛЯ/#?#?
БАРАБОЛЯ БАРАБОЛЯ - - -БАРАБОЛЯ--
Breaking it apart
The first regex: (^|[^\n\p{L}])
^| = Start of the line or
[^\n\p{L}] = Any character which is not a letter or a newline
The second regex: (?=$|\P{L})
?= = Lookahead
$| = End of the line or
\P{L} = Any character which is not a letter
The first regex captures the group and is then used via args[0] to put it back into the string during replacement, thereby avoiding a lookbehind. The second regex utilized lookahead.
Note that the second one MUST be a lookahead because if we capture it then overlapping regex matches will not trigger (e.g. КАРТОПЛЯ КАРТОПЛЯ КАРТОПЛЯ would only match on the 1st and 3rd ones).

Trying to find text "myTest":
/(?<![\p{L}\p{N}_])myTest(?![\p{L}\p{N}_])/gu
Similar to NetBeans or Notepad++ form. Trying to find the expression without any letter or number or underscore (like \w characters of word boundary \b) in any unicode characters of letter and number before or after the expression.

I have had a similar problem, but I had to replace an array of terms. All solutions, which I have found did not worked, if two terms were in the text next to each other (because their boundaries overlaped). So I had to use a little modified approach:
var text = "Ještě. že; \"už\" à. Fürs, 'anlässlich' že že že.";
var terms = ["à","anlässlich","Fürs","už","Ještě", "že"];
var replaced = [];
var order = 0;
for (i = 0; i < terms.length; i++) {
terms[i] = "(^\|[ \n\r\t.,;'\"\+!?-])(" + terms[i] + ")([ \n\r\t.,;'\"\+!?-]+\|$)";
}
var re = new RegExp(terms.join("|"), "");
while (true) {
var replacedString = "";
text = text.replace(re, function replacer(match){
var beginning = match.match("^[ \n\r\t.,;'\"\+!?-]+");
if (beginning == null) beginning = "";
var ending = match.match("[ \n\r\t.,;'\"\+!?-]+$");
if (ending == null) ending = "";
replacedString = match.replace(beginning,"");
replacedString = replacedString.replace(ending,"");
replaced.push(replacedString);
return beginning+"{{"+order+"}}"+ending;
});
if (replacedString == "") break;
order += 1;
}
See the code in a fiddle: http://jsfiddle.net/antoninslejska/bvbLpdos/1/
The regular expression is inspired by: http://breakthebit.org/post/3446894238/word-boundaries-in-javascripts-regular
I can't say, that I find the solution elegant...

The correct answer to the question is given by andrefs.
I will only rewrite it more clearly, after putting all required things together.
For ASCII text, you can use \b for matching a word boundary both at the start and the end of a pattern. When using Unicode text, you need to use 2 different patterns for doing the same:
Use (?<=^|\P{L}) for matching the start or a word boundary before the main pattern.
Use (?=\P{L}|$) for matching the end or a word boundary after the main pattern.
Additionally, use (?i) in the beginning of everything, to make all those matchings case-insensitive.
So the resulting answer is: (?i)(?<=^|\P{L})xxx(?=\P{L}|$), where xxx is your main pattern. This would be the equivalent of (?i)\bxxx\b for ASCII text.
For your code to work, you now need to do the following:
Assign to your variable "searchterm", the pattern or words you want to find.
Escape the variable's contents. For example, replace '\' with '\\' and also do the same for any reserved special character of regex, like '\^', '\$', '\/', etc. Check here for a question on how to do this.
Insert the variable's contents to the pattern above, in the place of "xxx", by simply using the string.replace() method.

bad but working:
var text = " аб аб АБ абвг ";
var ttt = "(аб)"
var p = "(^|$|[^A-Za-zА-Я-а-я0-9()])"; // add other word boundary symbols here
var exp = new RegExp(p+ttt+p,"gi");
text = text.replace(exp, "$1($2)$3").replace(exp, "$1($2)$3");
const t1 = performance.now();
console.log(text);
result (without qutes):
" (аб) (аб) (АБ) абвг "

I struggled hard on this. Working with French accented characters, and I managed to find this solution :
const myString = "MyString";
const regex = new RegExp(
"(?:[^À-ú]|^)\\b(" + myString + ")\\b(?:[^À-ú]|$)",
"ig"
);
What id does :
It keeps checking word-boundaries with \b before and after "MyString".
In addition to that, (?:[^À-ú]|^) and (?:[^À-ú]|$) will check if MyString is not surrounded by any accented characters
It will not work with cyrillic but it may be possible to find the range of cirillic charactes and edit [^À-ú] in consequence.
Warning, it captures only the group (MyString) but the total match contains previous and next characters
See example : https://regex101.com/r/5P0ZIe/1
Match examples :
MyString
match : "MyString"
group 1 : "MyString"
Lorem ipsum. MyString dolor sit amet
match : " MyString "
group 1 : "MyString"
(MyString)
match : "(MyString)"
group 1 : "MyString"
BetweenCharactersMyStringIsNotFound
match : Nothing
group 1 : Nothing
éMyStringé
match : Nothing
group 1 : Nothing
ùMyString
match : Nothing
group 1 : Nothing
MyStringÖ
match : Nothing
group 1 : Nothing

Regex treating accentuated letters as word boundary [duplicate]

I am building search and I am going to use javascript autocomplete with it. I am from Finland (finnish language) so I have to deal with some special characters like ä, ö and å
When user types text in to the search input field I try to match the text to data.
Here is simple example that is not working correctly if user types for example "ää". Same thing with "äl"
var title = "this is simple string with finnish word tämä on ääkköstesti älkää ihmetelkö";
// Does not work
var searchterm = "äl";
// does not work
//var searchterm = "ää";
// Works
//var searchterm = "wi";
if ( new RegExp("\\b"+searchterm, "gi").test(title) ) {
$("#result").html("Match: ("+searchterm+"): "+title);
} else {
$("#result").html("nothing found with term: "+searchterm);
}
http://jsfiddle.net/7TsxB/
So how can I get those ä,ö and å characters to work with javascript regex?
I think I should use unicode codes but how should I do that? Codes for those characters are:
[\u00C4,\u00E4,\u00C5,\u00E5,\u00D6,\u00F6]
=> äÄåÅöÖ

There appears to be a problem with Regex and the word boundary \b matching the beginning of a string with a starting character out of the normal 256 byte range.
Instead of using \b, try using (?:^|\\s)
var title = "this is simple string with finnish word tämä on ääkköstesti älkää ihmetelkö";
// Does not work
var searchterm = "äl";
// does not work
//var searchterm = "ää";
// Works
//var searchterm = "wi";
if ( new RegExp("(?:^|\\s)"+searchterm, "gi").test(title) ) {
$("#result").html("Match: ("+searchterm+"): "+title);
} else {
$("#result").html("nothing found with term: "+searchterm);
}
Breakdown:
(?: parenthesis () form a capture group in Regex. Parenthesis started with a question mark and colon ?: form a non-capturing group. They just group the terms together
^ the caret symbol matches the beginning of a string
| the bar is the "or" operator.
\s matches whitespace (appears as \\s in the string because we have to escape the backslash)
) closes the group
So instead of using \b, which matches word boundaries and doesn't work for unicode characters, we use a non-capturing group which matches the beginning of a string OR whitespace.

The \b character class in JavaScript RegEx is really only useful with simple ASCII encoding. \b is a shortcut code for the boundary between \w and \W sets or \w and the beginning or end of the string. These character sets only take into account ASCII "word" characters, where \w is equal to [a-zA-Z0-9_] and \W is the negation of that class.
This makes the RegEx character classes largely useless for dealing with any real language.
\s should work for what you want to do, provided that search terms are only delimited by whitespace.

this question is old, but I think I found a better solution for boundary in regular expressions with unicode letters.
Using XRegExp library you can implement a valid \b boundary expanding this
XRegExp('(?=^|$|[^\\p{L}])')
the result is a 4000+ char long, but it seems to work quite performing.
Some explanation: (?= ) is a zero-length lookahead that looks for a begin or end boundary or a non-letter unicode character. The most important think is the lookahead, because the \b doesn't capture anything: it is simply true or false.

\b is a shortcut for the transition between a letter and a non-letter character, or vice-versa.
Updating and improving on max_masseti's answer:
With the introduction of the /u modifier for RegExs in ES2018, you can now use \p{L} to represent any unicode letter, and \P{L} (notice the uppercase P) to represent anything but.
EDIT: Previous version was incomplete.
As such:
const text = 'A Fé, o Império, e as terras viciosas';
text.split(/(?<=\p{L})(?=\P{L})|(?<=\P{L})(?=\p{L})/);
// ['A', ' Fé', ',', ' o', ' Império', ',', ' e', ' as', ' terras', ' viciosas']
We're using a lookbehind (?<=...) to find a letter and a lookahead (?=...) to find a non-letter, or vice versa.

I would recommend you to use XRegExp when you have to work with a specific set of characters from Unicode, the author of this library mapped all kind of regional sets of characters making the work with different languages easier.

Despite the fact the issue seems to be 8 years old, I run into a similar problem (I had to match Cyrillic letters) not so far ago. I spend a whole day on this and could not find any appropriate answer here on StackOverflow. So, to avoid others making lots of effort, I'd like to share my solution.
Yes, \b word boundary works only with Latin letters (Word boundary: \b):
Word boundary \b doesn’t work for non-Latin alphabets
The word boundary test \b checks that there should be \w on the one side from the position and "not \w" – on the other side.
But \w means a Latin letter a-z (or a digit or an underscore), so the test doesn’t work for other characters, e.g. Cyrillic letters or hieroglyphs.
Yes, JavaScript RegExp implementation hardly supports UTF-8 encoding.
So, I tried implementing own word boundary feature with the support of non-Latin characters. To make word boundary work just with Cyrillic characters I created such regular expression:
new RegExp(`(?<![\u0400-\u04ff])${cyrillicSearchValue}(?![\u0400-\u04ff])`,'gi')
Where \u0400-\u04ff is a range of Cyrillic characters provided in the table of codes. It is not an ideal solution, however, it works properly in most cases.
To make it work in your case, you just have to pick up an appropriate range of codes from the list of Unicode characters.
To try out my example run the code snippet below.
function getMatchExpression(cyrillicSearchValue) {
return new RegExp(
`(?<![\u0400-\u04ff])${cyrillicSearchValue}(?![\u0400-\u04ff])`,
'gi',
);
}
const sentence = 'Будь-який текст кирилицею, де необхідно знайти слово з контексту';
console.log(sentence.match(getMatchExpression('текст')));
// expected output: ["текст"]
console.log(sentence.match(getMatchExpression('но')));
// expected output: null

I noticed something really weird with \b when using Unicode:
/\bo/.test("pop"); // false (obviously)
/\bä/.test("päp"); // true (what..?)
/\Bo/.test("pop"); // true
/\Bä/.test("päp"); // false (what..?)
It appears that meaning of \b and \B are reversed, but only when used with non-ASCII Unicode? There might be something deeper going on here, but I'm not sure what it is.
In any case, it seems that the word boundary is the issue, not the Unicode characters themselves. Perhaps you should just replace \b with (^|[\s\\/-_&]), as that seems to work correctly. (Make your list of symbols more comprehensive than mine, though.)

My idea is to search with codes representing the Finnish letters
new RegExp("\\b"+asciiOnly(searchterm), "gi").test(asciiOnly(title))
My original idea was to use plain encodeURI but the % sign seemed to interfere with the regexp.
http://jsfiddle.net/7TsxB/5/
I wrote a crude function using encodeURI to encode every character with code over 128 but removing its % and adding 'QQ' in the beginning. It is not the best marker but I couldn't get non alphanumeric to work.

What you are looking for is the Unicode word boundaries standard:
http://unicode.org/reports/tr29/tr29-9.html#Word_Boundaries
There is a JavaScript implementation here (unciodejs.wordbreak.js)
https://github.com/wikimedia/unicodejs

I had a similar problem, where I was trying to replace all of a particular unicode word with a different unicode word, and I cannot use lookbehind because it's not supported in the JS engine this code will be used in. I ultimately resolved it like this:
const needle = "КАРТОПЛЯ";
const replace = "БАРАБОЛЯ";
const regex = new RegExp(
String.raw`(^|[^\n\p{L}])`
+ needle
+ String.raw`(?=$|\P{L})`,
"gimu",
);
const result = (
'КАРТОПЛЯ сдффКАРТОПЛЯдадф КАРТОПЛЯ КАРТОПЛЯ КАРТОПЛЯ??? !!!КАРТОПЛЯ ;!;!КАРТОПЛЯ/#?#?'
+ '\n\nКАРТОПЛЯ КАРТОПЛЯ - - -КАРТОПЛЯ--'
)
.replace(regex, function (match, ...args) {
return args[0] + replace;
});
console.log(result)
output:
БАРАБОЛЯ сдффКАРТОПЛЯдадф БАРАБОЛЯ БАРАБОЛЯ БАРАБОЛЯ??? !!!БАРАБОЛЯ ;!;!БАРАБОЛЯ/#?#?
БАРАБОЛЯ БАРАБОЛЯ - - -БАРАБОЛЯ--
Breaking it apart
The first regex: (^|[^\n\p{L}])
^| = Start of the line or
[^\n\p{L}] = Any character which is not a letter or a newline
The second regex: (?=$|\P{L})
?= = Lookahead
$| = End of the line or
\P{L} = Any character which is not a letter
The first regex captures the group and is then used via args[0] to put it back into the string during replacement, thereby avoiding a lookbehind. The second regex utilized lookahead.
Note that the second one MUST be a lookahead because if we capture it then overlapping regex matches will not trigger (e.g. КАРТОПЛЯ КАРТОПЛЯ КАРТОПЛЯ would only match on the 1st and 3rd ones).

Trying to find text "myTest":
/(?<![\p{L}\p{N}_])myTest(?![\p{L}\p{N}_])/gu
Similar to NetBeans or Notepad++ form. Trying to find the expression without any letter or number or underscore (like \w characters of word boundary \b) in any unicode characters of letter and number before or after the expression.

I have had a similar problem, but I had to replace an array of terms. All solutions, which I have found did not worked, if two terms were in the text next to each other (because their boundaries overlaped). So I had to use a little modified approach:
var text = "Ještě. že; \"už\" à. Fürs, 'anlässlich' že že že.";
var terms = ["à","anlässlich","Fürs","už","Ještě", "že"];
var replaced = [];
var order = 0;
for (i = 0; i < terms.length; i++) {
terms[i] = "(^\|[ \n\r\t.,;'\"\+!?-])(" + terms[i] + ")([ \n\r\t.,;'\"\+!?-]+\|$)";
}
var re = new RegExp(terms.join("|"), "");
while (true) {
var replacedString = "";
text = text.replace(re, function replacer(match){
var beginning = match.match("^[ \n\r\t.,;'\"\+!?-]+");
if (beginning == null) beginning = "";
var ending = match.match("[ \n\r\t.,;'\"\+!?-]+$");
if (ending == null) ending = "";
replacedString = match.replace(beginning,"");
replacedString = replacedString.replace(ending,"");
replaced.push(replacedString);
return beginning+"{{"+order+"}}"+ending;
});
if (replacedString == "") break;
order += 1;
}
See the code in a fiddle: http://jsfiddle.net/antoninslejska/bvbLpdos/1/
The regular expression is inspired by: http://breakthebit.org/post/3446894238/word-boundaries-in-javascripts-regular
I can't say, that I find the solution elegant...

The correct answer to the question is given by andrefs.
I will only rewrite it more clearly, after putting all required things together.
For ASCII text, you can use \b for matching a word boundary both at the start and the end of a pattern. When using Unicode text, you need to use 2 different patterns for doing the same:
Use (?<=^|\P{L}) for matching the start or a word boundary before the main pattern.
Use (?=\P{L}|$) for matching the end or a word boundary after the main pattern.
Additionally, use (?i) in the beginning of everything, to make all those matchings case-insensitive.
So the resulting answer is: (?i)(?<=^|\P{L})xxx(?=\P{L}|$), where xxx is your main pattern. This would be the equivalent of (?i)\bxxx\b for ASCII text.
For your code to work, you now need to do the following:
Assign to your variable "searchterm", the pattern or words you want to find.
Escape the variable's contents. For example, replace '\' with '\\' and also do the same for any reserved special character of regex, like '\^', '\$', '\/', etc. Check here for a question on how to do this.
Insert the variable's contents to the pattern above, in the place of "xxx", by simply using the string.replace() method.

bad but working:
var text = " аб аб АБ абвг ";
var ttt = "(аб)"
var p = "(^|$|[^A-Za-zА-Я-а-я0-9()])"; // add other word boundary symbols here
var exp = new RegExp(p+ttt+p,"gi");
text = text.replace(exp, "$1($2)$3").replace(exp, "$1($2)$3");
const t1 = performance.now();
console.log(text);
result (without qutes):
" (аб) (аб) (АБ) абвг "

I struggled hard on this. Working with French accented characters, and I managed to find this solution :
const myString = "MyString";
const regex = new RegExp(
"(?:[^À-ú]|^)\\b(" + myString + ")\\b(?:[^À-ú]|$)",
"ig"
);
What id does :
It keeps checking word-boundaries with \b before and after "MyString".
In addition to that, (?:[^À-ú]|^) and (?:[^À-ú]|$) will check if MyString is not surrounded by any accented characters
It will not work with cyrillic but it may be possible to find the range of cirillic charactes and edit [^À-ú] in consequence.
Warning, it captures only the group (MyString) but the total match contains previous and next characters
See example : https://regex101.com/r/5P0ZIe/1
Match examples :
MyString
match : "MyString"
group 1 : "MyString"
Lorem ipsum. MyString dolor sit amet
match : " MyString "
group 1 : "MyString"
(MyString)
match : "(MyString)"
group 1 : "MyString"
BetweenCharactersMyStringIsNotFound
match : Nothing
group 1 : Nothing
éMyStringé
match : Nothing
group 1 : Nothing
ùMyString
match : Nothing
group 1 : Nothing
MyStringÖ
match : Nothing
group 1 : Nothing

JQuery match with RegEx not working

I have a filename that will be something along the lines of this:
Annual-GDS-Valuation-30th-Dec-2016-082564K.docx
It will contain 5 numbers followed by a single letter, but it may be in a different position in the file name. The leading zero may or may not be there, but it is not required.
This is the code I come up with after checking examples, however SelectedFileClientID is always null
var SelectedFileClientID = files.match(/^d{5}\[a-zA-Z]{1}$/);
I'm not sure what is it I am doing wrong.
Edit:
The 0 has nothing to do with the code I am trying to extract. It may or may not be there, and it could even be a completely different character, or more than one, but has nothing to do with it at all. The client has decided they want to put additional characters there.

There are at least 3 issues with your regex: 1) the pattern is enclosed with anchors, and thus requires a full string match, 2) the d matches a letter d, not a digit, you need \d to match a digit, 3) a \[ matches a literal [, so the character class is ruined.
Use
/\d{5}[a-zA-Z]/
Details:
\d{5} - 5 digits
[a-zA-Z] - an ASCII letter
JS demo:
var s = 'Annual-GDS-Valuation-30th-Dec-2016-082564K.docx';
var m = s.match(/\d{5}[a-zA-Z]/);
console.log(m[0]);

All right, there are a few things wrong...
var matches = files.match(/\-0?(\d{5}[a-zA-Z])\.[a-z]{3,}$/);
var SelectedFileClientID = matches ? matches[1] : '';
So:
First, I get the matches on your string -- .match()
Then, your file name will not start with the digits - so drop the ^
You had forgotten the backslash for digits: \d
Do not backslash your square bracket - it's here used as a regular expression token
no need for the {1} for your letters: the square bracket content is enough as it will match one, and only one letter.
Hope this helps!

Try this pattern , \d{5}[a-zA-Z]

Try - 0?\d{5}[azA-Z]
As you mentioned 0 may or may not be there. so 0? will take that into account.
Alternatively it can be done like this. which can match any random character.
(\w+|\W+|\d+)?\d{5}[azA-Z]

RegEx to find all occurrences of a character not between two other characters

I'm looking for a regex (that I can implement in Javascript, so no lookbehinds) which will match all occurrences of a character, as long as it doesn't appear between two other characters. For example, I want to match all hyphens as long as they are not between plus signs.
----- // should match.
+---+ // should not match
---+---+--- // should only match the first 3 and last 3 characters.
I've tried adapting the method used in this post like so:
[-]+(?![^+]*\+)+
But it is not matching as desired. Any help is greatly appreciated.
P.S. Looking specifically for a REGEX solution. I realize this may not be the optimal solution but I'm specifically trying to improve my knowledge of regex.

If the delimiters (+ in your case) always come in pairs, you could use this:
var str = 'a-b-+-c-+-d-e'; // delimter is +
matches = str.match(/-(?=([^+]*\+[^+]*\+)*[^+]*$)/g, '');
console.log(matches); // 4 matches of `-`
This matches every hyphen that has an even number of pluses following it (could be zero).

Please try the following:
(?:^-+|-+$)
Regex101 demo: https://regex101.com/r/524iej/1

You can use use .replace to remove all text between 2 +s:
str = str.replace(/\+-*\+/g, '');
Example:
var str = '---+---+---';
str = str.replace(/\+-*\+/g, '');
//=> ------

Creating javascript regex tp replace characters using whitelist

I'm trying to create a regex which will replace all the characters which are not in the specified white list (letters,digits,whitespaces, brackets, question mark and explanation mark)
This is the code :
var regEx = /^[^(\s|\w|\d|()|?|!|<br>)]*?$/;
qstr += tempStr.replace(regEx, '');
What is wrong with it ?
Thank you

The anchors are wrong - they only allow the regex to match the entire string
The lazy quantifier is wrong - you wouldn't want the regex to match 0 characters (if you have removed the anchors)
The parentheses and pipe characters are wrong - you don't need them in a character class.
The <br> is wrong - you can't match specific substrings in a character class.
The \d is superfluous since it's already contained in \w (thanks Alex K.!)
You're missing the global modifier to make sure you can do more than one replace.
You should be using + instead of * in order not to replace lots of empty strings with themselves.
Try
var regEx = /[^\s\w()?!]+/g;
and handle the <br>s independently (before that regex is applied, or the brackets will be removed).

You'll want to use the g (global) modifier:
var regEx = /^[^(\s|\w|\d|()|?|!|<br>)]*?$/g; // <-- `g` goes there
qstr += tempStr.replace(regEx, '');
This allows your expression to match multiple times.

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