Regex: match text between two strings not including my string

Question

I have a string and want to match:

await Listing.find({
    Test: 1
});

But don’t want to match it if it ends with .lean();

await Listing.find({
    Test: 1
}).lean();

I have this regex but it’s not working: (?<=await)([\S\s]*?)(?!.+\.lean())(;)

Poul Bak · Accepted Answer · 2022-06-21 05:58:03Z

1

You can use this modified regex:

(?<=await)([\S\s]*?)(?<!.+\.lean\(\))(;)

All I have changed is:

Making \.lean a negative look BEHIND

ESCAPING the parentheses.

answered Jun 21, 2022 at 5:58

Poul Bak

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Artyom Vancyan · Accepted Answer · 2022-06-21 07:29:45Z

1

^await(?:(?!\.lean\(\)).)*;$

Short Explanation

^await String starts with await
(?:(?!\.lean\(\)).)*; Contains anything except .lean() till the last ; colon

Also, see the regex demo

JavaScript Example

let regex = /^await(?:(?!\.lean\(\)).)*;$/s;

console.log(regex.test(`await Listing.find({
    Test: 1
});`));

console.log(regex.test(`await Listing.find({
    Test: 1
}).lean();`));

edited Jun 21, 2022 at 7:29

answered Jun 21, 2022 at 7:13

Artyom Vancyan

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The fourth bird · Accepted Answer · 2022-06-21 09:43:29Z

1

If you want to stay between a single opening and closing parenthesis, you don't need any assertion:

\bawait\s[^()]*\([^()]*\);

Regex demo

answered Jun 21, 2022 at 9:43

The fourth bird

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Dev01 · Accepted Answer · 2022-06-21 20:09:04Z

0

I ended up using this in vscode to find all mongoose queries that don't use lean so that I can increase the speed of queries and reduce memory footprint:

await (.*)find(([^;]|\n)*?)(?<!.\.lean\(\))(;)

answered Jun 21, 2022 at 20:09

Dev01

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