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Removing commas unless inside quotes, not using regexp

I am trying to remove commas in a string unless they appear inside quotes.
var mystring = "this, is, a, test, example, \"i, dont know\", jumps" ;
var newchar = '';
mystring = mystring.split(',').join(newchar);// working correctly
document.write(mystring);
Output I have is
this is a test example "i dont know" jumps
Expected output
this is a test example "i, dont know" jumps
A couple of questions. How can I find the index of string so that inside the quotation it will include comma but outside of quotation " it will not include comma ,. I know I have to use indexOf and substring but I don't know how to format it? (No regex please as I'm new to JavaScript and I'm just focusing on the basics.)

Loop through the string, remembering whether or not you are inside a set of quotation marks, and building a new string to which the commas inside quotes are not added:
var inQuotes = false; // Are we inside quotes?
var result = ''; // New string we will build.
for (var i = 0; i < str.length; i++) { // Loop through string.
var chr = str[i]; // Extract character.
var isComma = chr === ','; // Is this a comma?
var isQuote = chr === '"'; // Is this a quote?
if (inQuotes || !isComma) { // Include this character?
if (isQuote) inQuotes = !inQuotes; // If quote, reverse quote status.
result += chr; // Add character to result.
}
}
This solution has the advantage compared to the accepted one that it will work properly even if the input has multiple quotes strings inside it.

This will work, but it's not ideal for all cases. Example: It will not work for a string with more than 2 quotation marks.
var mystring = "this, is, a, test, example, \"i, dont know\", jumps" ;
var newchar = '';
var firstIndex = mystring.indexOf("\"");
var lastIndex = mystring.lastIndexOf("\"");
var substring1 = mystring.substring(0,firstIndex).split(',').join(newchar);
var substring2 = mystring.substring(lastIndex).split(',').join(newchar);
mystring = substring1 + mystring.substring(firstIndex, lastIndex) + substring2;
document.write(mystring);

Some day you need to start using regexp, than regexr.com is your friend. The regexp solution is simple:
var mystring = "this, is, a, test, example, \"i, dont know\", jumps" ;
var newchar = '_';
mystring = mystring.match(/(".*?"|[^",\s]+)(?=\s*,|\s*$)/g).join(newchar);// working correctly
document.write(mystring);

Regex match quotes inside bracket regex

I'm working on a regex that must match only the text inside quotes but not in a comment, my macthes must only the strings in bold
<"love";>
>/*"love"*/<
<>'love'<>
"lo
more love
ve"
I'm stunck on this:
/(?:((\"|\')(.|\n)*?(\"|\')))(?=(?:\/\**\*\/))/gm
The first one (?:((\"|\')(.|\n)*?(\"|\'))) match all the strings
the second one (?=(?:\/\**\*\/)) doesn't match text inside quotes inside /* "mystring" */
bit my logic is cleary wrong
Any suggestion?
Thanks

Maybe you just need to use a negative lookahead to check for the comment end */?
But first, I'd split the string into separate lines
var arrayOfLines = input_str.split(/\r?\n/);
or, without empty lines:
var arrayOfLines = input_str.match(/[^\r\n]+/g);
and then use this regex:
["']([^'"]+)["'](?!.*\*\/)
Sample code:
var rebuilt_string = ''
var re = /["']([^'"]+)["'](?!.*\*\/)/g;
var subst = '<b>$1</b>';
for (i = 0; i < arrayOfLines.length; i++)
{
rebuilt_string = rebuilt_string + arrayOfLines[i].replace(re, subst) + "\r\n";
}

The way to avoid commented parts is to match them before. The global pattern looks like this:
/(capture parts to avoid)|target/
Then use a callback function for the replacement (when the capture group exists, return the match without change, otherwise, replace the match with what you want.
Example:
var result = text.replace(/(\/\*[^*]*(?:\*+(?!\/)[^*]*)*\*\/)|"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*'/g,
function (m, g1) {
if (g1) return g1;
return '<b>' + m + '</b>';
});

Find string in array and remove

I want to find a string in an array sql and remove the string. The string would be like:
" specimen.snop_code = ''"
There will be 4 digits between the single qoutes, which could be anything. I was thinking of using regex to find the string.
Tried just using pop() but I need to target the string to be removed from the array. Note that I need to remove all instances of the string. So something like:
disease_filter = new RegExp(" specimen.snop_code = ''", 'g');
for (var i=sql.length-1; i>=0; i--) {
if (sql[i] === disease_filter) {
array.splice(i, 1);
}
}
So how can I make " specimen.snop_code = '*'" into a regular expression with a wildcard as shown between the single quotes?

You can use .replace with a regex as the first parameter:
var input = " specimen.snop_code = 'something'";
var disease_filter = input.replace(/'(.*)'/gi, "'other stuff'");
// disease_filter is now "specimen.snop_code = 'other stuff'"
edit: removed unneccesary escaping as commented.

Use dynamic (variable) string as regex pattern in JavaScript

I want to add a (variable) tag to values with regex, the pattern works fine with PHP but I have troubles implementing it into JavaScript.
The pattern is (value is the variable):
/(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/is
I escaped the backslashes:
var str = $("#div").html();
var regex = "/(?!(?:[^<]+>|[^>]+<\\/a>))\\b(" + value + ")\\b/is";
$("#div").html(str.replace(regex, "" + value + ""));
But this seem not to be right, I logged the pattern and its exactly what it should be.
Any ideas?

To create the regex from a string, you have to use JavaScript's RegExp object.
If you also want to match/replace more than one time, then you must add the g (global match) flag. Here's an example:
var stringToGoIntoTheRegex = "abc";
var regex = new RegExp("#" + stringToGoIntoTheRegex + "#", "g");
// at this point, the line above is the same as: var regex = /#abc#/g;
var input = "Hello this is #abc# some #abc# stuff.";
var output = input.replace(regex, "!!");
alert(output); // Hello this is !! some !! stuff.
JSFiddle demo here.
In the general case, escape the string before using as regex:
Not every string is a valid regex, though: there are some speciall characters, like ( or [. To work around this issue, simply escape the string before turning it into a regex. A utility function for that goes in the sample below:
function escapeRegExp(stringToGoIntoTheRegex) {
return stringToGoIntoTheRegex.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
}
var stringToGoIntoTheRegex = escapeRegExp("abc"); // this is the only change from above
var regex = new RegExp("#" + stringToGoIntoTheRegex + "#", "g");
// at this point, the line above is the same as: var regex = /#abc#/g;
var input = "Hello this is #abc# some #abc# stuff.";
var output = input.replace(regex, "!!");
alert(output); // Hello this is !! some !! stuff.
JSFiddle demo here.
Note: the regex in the question uses the s modifier, which didn't exist at the time of the question, but does exist -- a s (dotall) flag/modifier in JavaScript -- today.

If you are trying to use a variable value in the expression, you must use the RegExp "constructor".
var regex = "(?!(?:[^<]+>|[^>]+<\/a>))\b(" + value + ")\b";
new RegExp(regex, "is")

I found I had to double slash the \b to get it working. For example to remove "1x" words from a string using a variable, I needed to use:
str = "1x";
var regex = new RegExp("\\b"+str+"\\b","g"); // same as inv.replace(/\b1x\b/g, "")
inv=inv.replace(regex, "");

You don't need the " to define a regular expression so just:
var regex = /(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/is; // this is valid syntax
If value is a variable and you want a dynamic regular expression then you can't use this notation; use the alternative notation.
String.replace also accepts strings as input, so you can do "fox".replace("fox", "bear");
Alternative:
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/", "is");
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(" + value + ")\b/", "is");
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(.*?)\b/", "is");
Keep in mind that if value contains regular expressions characters like (, [ and ? you will need to escape them.

I found this thread useful - so I thought I would add the answer to my own problem.
I wanted to edit a database configuration file (datastax cassandra) from a node application in javascript and for one of the settings in the file I needed to match on a string and then replace the line following it.
This was my solution.
dse_cassandra_yaml='/etc/dse/cassandra/cassandra.yaml'
// a) find the searchString and grab all text on the following line to it
// b) replace all next line text with a newString supplied to function
// note - leaves searchString text untouched
function replaceStringNextLine(file, searchString, newString) {
fs.readFile(file, 'utf-8', function(err, data){
if (err) throw err;
// need to use double escape '\\' when putting regex in strings !
var re = "\\s+(\\-\\s(.*)?)(?:\\s|$)";
var myRegExp = new RegExp(searchString + re, "g");
var match = myRegExp.exec(data);
var replaceThis = match[1];
var writeString = data.replace(replaceThis, newString);
fs.writeFile(file, writeString, 'utf-8', function (err) {
if (err) throw err;
console.log(file + ' updated');
});
});
}
searchString = "data_file_directories:"
newString = "- /mnt/cassandra/data"
replaceStringNextLine(dse_cassandra_yaml, searchString, newString );
After running, it will change the existing data directory setting to the new one:
config file before:
data_file_directories:
- /var/lib/cassandra/data
config file after:
data_file_directories:
- /mnt/cassandra/data

Much easier way: use template literals.
var variable = 'foo'
var expression = `.*${variable}.*`
var re = new RegExp(expression, 'g')
re.test('fdjklsffoodjkslfd') // true
re.test('fdjklsfdjkslfd') // false

Using string variable(s) content as part of a more complex composed regex expression (es6|ts)
This example will replace all urls using my-domain.com to my-other-domain (both are variables).
You can do dynamic regexs by combining string values and other regex expressions within a raw string template. Using String.raw will prevent javascript from escaping any character within your string values.
// Strings with some data
const domainStr = 'my-domain.com'
const newDomain = 'my-other-domain.com'
// Make sure your string is regex friendly
// This will replace dots for '\'.
const regexUrl = /\./gm;
const substr = `\\\.`;
const domain = domainStr.replace(regexUrl, substr);
// domain is a regex friendly string: 'my-domain\.com'
console.log('Regex expresion for domain', domain)
// HERE!!! You can 'assemble a complex regex using string pieces.
const re = new RegExp( String.raw `([\'|\"]https:\/\/)(${domain})(\S+[\'|\"])`, 'gm');
// now I'll use the regex expression groups to replace the domain
const domainSubst = `$1${newDomain}$3`;
// const page contains all the html text
const result = page.replace(re, domainSubst);
note: Don't forget to use regex101.com to create, test and export REGEX code.

var string = "Hi welcome to stack overflow"
var toSearch = "stack"
//case insensitive search
var result = string.search(new RegExp(toSearch, "i")) > 0 ? 'Matched' : 'notMatched'
https://jsfiddle.net/9f0mb6Lz/
Hope this helps

How can I concatenate regex literals in JavaScript?

Is it possible to do something like this?
var pattern = /some regex segment/ + /* comment here */
/another segment/;
Or do I have to use new RegExp() syntax and concatenate a string? I'd prefer to use the literal as the code is both more self-evident and concise.

Here is how to create a regular expression without using the regular expression literal syntax. This lets you do arbitary string manipulation before it becomes a regular expression object:
var segment_part = "some bit of the regexp";
var pattern = new RegExp("some regex segment" + /*comment here */
segment_part + /* that was defined just now */
"another segment");
If you have two regular expression literals, you can in fact concatenate them using this technique:
var regex1 = /foo/g;
var regex2 = /bar/y;
var flags = (regex1.flags + regex2.flags).split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
var regex3 = new RegExp(expression_one.source + expression_two.source, flags);
// regex3 is now /foobar/gy
It's just more wordy than just having expression one and two being literal strings instead of literal regular expressions.

Just randomly concatenating regular expressions objects can have some adverse side effects. Use the RegExp.source instead:
var r1 = /abc/g;
var r2 = /def/;
var r3 = new RegExp(r1.source + r2.source,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '') +
(r1.multiline ? 'm' : ''));
console.log(r3);
var m = 'test that abcdef and abcdef has a match?'.match(r3);
console.log(m);
// m should contain 2 matches
This will also give you the ability to retain the regular expression flags from a previous RegExp using the standard RegExp flags.
jsFiddle

I don't quite agree with the "eval" option.
var xxx = /abcd/;
var yyy = /efgh/;
var zzz = new RegExp(eval(xxx)+eval(yyy));
will give "//abcd//efgh//" which is not the intended result.
Using source like
var zzz = new RegExp(xxx.source+yyy.source);
will give "/abcdefgh/" and that is correct.
Logicaly there is no need to EVALUATE, you know your EXPRESSION. You just need its SOURCE or how it is written not necessarely its value. As for the flags, you just need to use the optional argument of RegExp.
In my situation, I do run in the issue of ^ and $ being used in several expression I am trying to concatenate together! Those expressions are grammar filters used accross the program. Now I wan't to use some of them together to handle the case of PREPOSITIONS.
I may have to "slice" the sources to remove the starting and ending ^( and/or )$ :)
Cheers, Alex.

Problem If the regexp contains back-matching groups like \1.
var r = /(a|b)\1/ // Matches aa, bb but nothing else.
var p = /(c|d)\1/ // Matches cc, dd but nothing else.
Then just contatenating the sources will not work. Indeed, the combination of the two is:
var rp = /(a|b)\1(c|d)\1/
rp.test("aadd") // Returns false
The solution:
First we count the number of matching groups in the first regex, Then for each back-matching token in the second, we increment it by the number of matching groups.
function concatenate(r1, r2) {
var count = function(r, str) {
return str.match(r).length;
}
var numberGroups = /([^\\]|^)(?=\((?!\?:))/g; // Home-made regexp to count groups.
var offset = count(numberGroups, r1.source);
var escapedMatch = /[\\](?:(\d+)|.)/g; // Home-made regexp for escaped literals, greedy on numbers.
var r2newSource = r2.source.replace(escapedMatch, function(match, number) { return number?"\\"+(number-0+offset):match; });
return new RegExp(r1.source+r2newSource,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '')
+ (r1.multiline ? 'm' : ''));
}
Test:
var rp = concatenate(r, p) // returns /(a|b)\1(c|d)\2/
rp.test("aadd") // Returns true

Providing that:
you know what you do in your regexp;
you have many regex pieces to form a pattern and they will use same flag;
you find it more readable to separate your small pattern chunks into an array;
you also want to be able to comment each part for next dev or yourself later;
you prefer to visually simplify your regex like /this/g rather than new RegExp('this', 'g');
it's ok for you to assemble the regex in an extra step rather than having it in one piece from the start;
Then you may like to write this way:
var regexParts =
[
/\b(\d+|null)\b/,// Some comments.
/\b(true|false)\b/,
/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|length|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/,
/(\$|jQuery)/,
/many more patterns/
],
regexString = regexParts.map(function(x){return x.source}).join('|'),
regexPattern = new RegExp(regexString, 'g');
you can then do something like:
string.replace(regexPattern, function()
{
var m = arguments,
Class = '';
switch(true)
{
// Numbers and 'null'.
case (Boolean)(m[1]):
m = m[1];
Class = 'number';
break;
// True or False.
case (Boolean)(m[2]):
m = m[2];
Class = 'bool';
break;
// True or False.
case (Boolean)(m[3]):
m = m[3];
Class = 'keyword';
break;
// $ or 'jQuery'.
case (Boolean)(m[4]):
m = m[4];
Class = 'dollar';
break;
// More cases...
}
return '<span class="' + Class + '">' + m + '</span>';
})
In my particular case (a code-mirror-like editor), it is much easier to perform one big regex, rather than a lot of replaces like following as each time I replace with a html tag to wrap an expression, the next pattern will be harder to target without affecting the html tag itself (and without the good lookbehind that is unfortunately not supported in javascript):
.replace(/(\b\d+|null\b)/g, '<span class="number">$1</span>')
.replace(/(\btrue|false\b)/g, '<span class="bool">$1</span>')
.replace(/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/g, '<span class="keyword">$1</span>')
.replace(/\$/g, '<span class="dollar">$</span>')
.replace(/([\[\](){}.:;,+\-?=])/g, '<span class="ponctuation">$1</span>')

It would be preferable to use the literal syntax as often as possible. It's shorter, more legible, and you do not need escape quotes or double-escape backlashes. From "Javascript Patterns", Stoyan Stefanov 2010.
But using New may be the only way to concatenate.
I would avoid eval. Its not safe.

You could do something like:
function concatRegex(...segments) {
return new RegExp(segments.join(''));
}
The segments would be strings (rather than regex literals) passed in as separate arguments.

You can concat regex source from both the literal and RegExp class:
var xxx = new RegExp(/abcd/);
var zzz = new RegExp(xxx.source + /efgh/.source);

Use the constructor with 2 params and avoid the problem with trailing '/':
var re_final = new RegExp("\\" + ".", "g"); // constructor can have 2 params!
console.log("...finally".replace(re_final, "!") + "\n" + re_final +
" works as expected..."); // !!!finally works as expected
// meanwhile
re_final = new RegExp("\\" + "." + "g"); // appends final '/'
console.log("... finally".replace(re_final, "!")); // ...finally
console.log(re_final, "does not work!"); // does not work

No, the literal way is not supported. You'll have to use RegExp.

the easier way to me would be concatenate the sources, ex.:
a = /\d+/
b = /\w+/
c = new RegExp(a.source + b.source)
the c value will result in:
/\d+\w+/

I prefer to use eval('your expression') because it does not add the /on each end/ that ='new RegExp' does.