Develop Reference

Javascipt regex to get string between two characters except escaped without lookbehind

I am looking for a specific javascript regex without the new lookahead/lookbehind features of Javascript 2018 that allows me to select text between two asterisk signs but ignores escaped characters.
In the following example only the text "test" and the included escaped characters are supposed to be selected according the rules above:
\*jdjdjdfdf*test*dfsdf\*adfasdasdasd*test**test\**sd* (Selected: "test", "test", "test\*")
During my research I found this solution Regex, everything between two characters except escaped characters /(?<!\\)(%.*?(?<!\\)%)/ but it uses negative lookbehinds which is supported in javascript 2018 but I need to support IE11 as well, so this solution doesn't work for me.
Then i found another approach which is almost getting there for me here: Javascript: negative lookbehind equivalent?. I altered the answer of Kamil Szot to fit my needs: ((?!([\\])).|^)(\*.*?((?!([\\])).|^)\*) Unfortuantely it doesn't work when two asterisks ** are in a row.
I have already invested a lot of hours and can't seem to get it right, any help is appreciated!
An example with what i have so far is here: https://www.regexpal.com/?fam=117350
I need to use the regexp in a string.replace call (str.replace(regexp|substr, newSubStr|function); so that I can wrap the found strings with a span element of a specific class.

You can use this regular expression:
(?:\\.|[^*])*\*((?:\\.|[^*])*)\*
Your code should then only take the (only) capture group of each match.
Like this:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /(?:\\.|[^*])*\*((?:\\.|[^*])*)\*/g
var match;
while (match = regex.exec(str)) {
console.log(match[1]);
}
If you need to replace the matches, for instance to wrap the matches in a span tag while also dropping the asterisks, then use two capture groups:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /((?:\\.|[^*])*)\*((?:\\.|[^*])*)\*/g
var result = str.replace(regex, "$1<span>$2</span>");
console.log(result);
One thing to be careful with: when you use string literals in JavaScript tests, escape the backslash (with another backslash). If you don't do that, the string actually will not have a backslash! To really get the backslash in the in-memory string, you need to escape the backslash.

const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr.match(/\*(\\.)*t(\\.)*e(\\.)*s(\\.)*t(\\.)*\*/g).map(m => m.substr(1, m.length-2));
console.log(m);
More generic code:
const prepareRegExp = (word, delimiter = '\\*') => {
const escaped = '(\\\\.)*';
return new RegExp([
delimiter,
escaped,
[...word].join(escaped),
escaped,
delimiter
].join``, 'g');
};
const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr
.match(prepareRegExp('test'))
.map(m => m.substr(1, m.length-2));
console.log(m);
https://instacode.dev/#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

remove last part of string following '&&&' with JavaScript Regex

I'm trying to use a regex in JS to remove the last part of a string. This substring starts with &&&, is followed by something not &&&, and ends with .pdf.
So, for example, the final regex should take a string like:
parent&&&child&&&grandchild.pdf
and match
parent&&&child
I'm not that great with regex's, so my best effort has been something like:
.*?(?:&&&.*\.pdf)
Which matches the whole string. Can anyone help me out?

You may use this greedy regex either in replace or in match:
var s = 'parent&&&child&&&grandchild.pdf';
// using replace
var r = s.replace(/(.*)&&&.*\.pdf$/, '$1');
console.log(r);
//=> parent&&&child
// using match
var m = s.match(/(.*)&&&.*\.pdf$/)
if (m) {
console.log(m[1]);
//=> parent&&&child
}
By using greedy pattern .* before &&& we make sure to match **last instance of &&& in input.

You want to remove the last portion, so replace it
var str = "parent&&&child&&&grandchild.pdf"
var result = str.replace(/&&&[^&]+\.pdf$/, '')
console.log(result)

JavaScript String test with array of RegEx

I have some doubts regarding RegEx in JavaScript as I am not good in RegEx.
I have a String and I want to compare it against some array of RegEx expressions.
First I tried for one RegEx and it's not working. I want to fix that also.
function check(str){
var regEx = new RegEx("(users)\/[\w|\W]*");
var result = regEx.test(str);
if(result){
//do something
}
}
It is not working properly.
If I pass users, it doesn't match. If I pass users/ or users/somestring, it is matching.
If I change the RegEx to (usersGroupList)[/\w|\W]*, then it is matching for any string that contains the string users
fdgdsfgguserslist/data
I want to match like if string is either users or it should contain users/something or users/
And also I want the string to compare it with similar regex array.
I want to compare the string str with users, users/something, list, list/something, anothermatch, anothermatch/something. If if it matches any of these expression i want to do something.
How can I do that?
Thanks

Then, you'll have to make the last group optional. You do that by capturing the /something part in a group and following it with ? which makes the previous token, here the captured group, optional.
var regEx = new RegExp("(users)(\/[\w|\W]*)?");

What about making:
the last group optional
starting from beginning of the string
Like this:
var regEx = new RegExp("^(users)(\/[\w|\W]*)?");
Same applies for all the others cases, e.g. for list:
var regEx = new RegExp("^(list)(\/[\w|\W]*)?");
All in One Approach
var regEx = new RegExp("^(users|list|anothermatch)(\/[\w|\W]*)?");
Even More Generic
var keyw = ["users", "list", "anothermatch"];
var keyws = keyw.join("|");
var regEx = new RegExp("^("+keyws+")(\/[\w|\W]*)?");

You haven't made the / optional. Try this instead
(users)\/?[\w|\W]*

using a lookahead to get the last occurrence of a pattern in javascript

I was able to build a regex to extract a part of a pattern:
var regex = /\w+\[(\w+)_attributes\]\[\d+\]\[own_property\]/g;
var match = regex.exec( "client_profile[foreclosure_defenses_attributes][0][own_property]" );
match[1] // "foreclosure_defenses"
However, I also have a situation where there will be a repetitive pattern like so:
"client_profile[lead_profile_attributes][foreclosure_defenses_attributes][0][own_property]"
In that case, I want to ignore [lead_profile_attributes] and just extract the portion of the last occurence as I did in the first example. In other words, I still want to match "foreclosure_defenses" in this case.
Since all patterns will be like [(\w+)_attributes], I tried to do a lookahead, but it is not working:
var regex = /\w+\[(\w+)_attributes\](?!\[(\w+)_attributes\])\[\d+\]\[own_property\]/g;
var match = regex.exec("client_profile[lead_profile_attributes][foreclosure_defenses_attributes][0][own_property]");
match // null
match returns null meaning that my regex isn't working as expected. I added the following:
\[(\w+)_attributes\](?!\[(\w+)_attributes\])
Because I want to match only the last occurrence of the following pattern:
[lead_profile_attributes][foreclosure_defenses_attributes]
I just want to grab the foreclosure_defenses, not the lead_profile.
What might I be doing wrong?

I think I got it working without positive lookahead:
regex = /(\[(\w+)_attributes\])+/
/(\[(\w+)_attributes\])+/
match = regex.exec(str);
["[a_attributes][b_attributes][c_attributes]", "[c_attributes]", "c"]

I was able to also achieve it through noncapturing groups. Output from chrome console:
var regex = /(?:\w+(\[\w+\]\[\d+\])+)(\[\w+\])/;
undefined
regex
/(?:\w+(\[\w+\]\[\d+\])+)(\[\w+\])/
str = "profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]";
"profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]"
match = regex.exec(str);
["profile[foreclosure_defenses_attributes][0][properties_attributes][0][other_stuff]", "[properties_attributes][0]", "[other_stuff]"]

JavaScript escape stars on regular expression

I am trying to get a serial number from a zigbee packet (i.e get from 702442500 *13*32*702442500#9).
So far, I've tried this:
test = "*#*0##*13*32*702442500#9##";
test.match("\*#\*0##\*13\*32\*(.*)#9##");
And this:
test.match("*#*0##*13*32*(.*)#9##");
With no luck. How do I get a valid regular expression that does what I want?

The below regex matches the number which has atleast three digits,
/([0-9][0-9][0-9]+)/
DEMO

If you want to extract the big number, you can use:
/\*#\*0##\*13\*32\*([^#]+)#9##/
Note that I use delimiters / that are needed to write a pattern in Javascript (without the regexp object syntax). When you use this syntax, (double)? quotes are not needed. I use [^#]+ instead of .* because it is more clear and more efficent for the regex engine.

The easiest way to grab that portion of the string would be to use
var regex = /(\*\d{3,}#)/g,
test = "*13*32*702442500#9";
var match = test.match(regex).slice(1,-1);
This captures a * followed by 3 or more \d (numbers) until it reaches an octothorpe. Using the global (/g) modifier will cause it to return an array of matches.
For example, if
var test = "*13*32*702442500#9
*#*0##*13*32*702442500#9##";
then, test.match(regex) will return ["*702442500#", "*702442500#"]. You can then slice the elements of this array:
var results = [],
test = "... above ... ",
regex = /(\*\d{3,}#)/g,
matches = test.match(regex);
matches.forEach(function (d) {
results.push(d.slice(1,-1));
})
// results : `["702442500", "702442500"]`