Javascript regex for matching valid characters without repetition of a particular character

Question

let rx = /(\/MxML\/[a-z0-9\[\]@/]*)/gi;
let s =
    'If (/MxML/trades[1]/value== 1 then /MxML/trades[type=2]/value must be /MxML/stream/pre/reference@href';

let m;
let res = [];

while ((m = rx.exec(s))) {
    res.push(m[1]);
}
console.log(res);

Pattern should start with /MxML/
Keep including acceptable characters [a-z0-9[]@/]
A single = is ok. But == or === are not ok.

Desired output

[
    '/MxML/trades[1]/value',
    '/MxML/trades[type=2]/value',
    '/MxML/stream/pre/reference@href',
];

My current (incorrect) output

[
    '/MxML/trades[1]/value',
    '/MxML/trades[type',
    '/MxML/stream/pre/reference@href',
];

I can think of a lookaround, but can't figure out how to use it here. (?<!\=)\=(?!\=)

I've been going through several other examples on Stackoverflow, but they are about avoiding all character repetition and not about avoiding a particular character(=) repetition in conjunction with other valid characters.

MonkeyZeus · Accepted Answer · 2022-03-09 17:21:30Z

1

Here is a regex which should be compatible in all browsers:

/(\/MxML\/(?:[a-z0-9\[\]@/]|=(?!=))*)/gi

/(\/MxML\/ - your regex
(?: - start non-capture group; we need this to implement an "or" operator and not mess up your use of m[1]
- [a-z0-9\[\]@/] - your regex of whitelisted chars; notice the asterisk was removed
- | - or
- =(?!=) - allow for an equal sign not followed by an equal sign
)* - close non-capturing group and capture between zero and infinity instances of what's inside
)/gi - your regex

let rx = /(\/MxML\/(?:[a-z0-9\[\]@/]|=(?!=))*)/gi;
let s =
    'If (/MxML/trades[1]/value== 1 then /MxML/trades[type=2]/value must be /MxML/stream/pre/reference@href';

let m;
let res = [];

while ((m = rx.exec(s))) {
    res.push(m[1]);
}
console.log(res);

edited Mar 9, 2022 at 17:21

answered Mar 9, 2022 at 17:14

MonkeyZeus

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3 Comments

Kaya Toast Over a year ago

Thanks. Can you add a line of explanation please, including what is : for.

MonkeyZeus Over a year ago

@KayaToast It's a non-capturing group. I didn't want to mess up your use of m[1]

MonkeyZeus Over a year ago

@KayaToast I edited my answer to explain the regex a bit more

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