排除正则表达式中的字符串

Question

For example, I have some expression like that

expression1
expression2 expression3

I want to match "expression2 expression3" in a regular expression if "expression1" is not an unwanted string (Let me call it unwanted.). So, it should be like that:

unwanted
expression2 expression3 // Not Matched...

string
expression2 expression3 // Matched...

How can I do this? I have tried something like that:

(?!unwanted\n)(expression2)[ ]+(expression3)

But it doesn't work. What can be the problem?

Thanks in advance...

Answer 1

Capture the unwanted string into Group 1 and check if it is undefined . If it is, there is no unwanted text and you may grab that match, else, discard it:

 var regex = /(unwanted\\n)?(expression\\d+)\\s+(expression\\d+)/g; var str = "unwanted\\nexpression2 expression3\\n\\nstring\\nexpression4 expression5"; var res = [], m; while ((m = regex.exec(str)) !== null) { if (m[1] === undefined) res.push(m[0]); } console.log(res); 

Answer 2

Maybe you could work on from this:

(?!unwanted.*)(?:^.{1,8}).*\s*(expression2 expression3)

Basically it makes sure the expressions are preceded by at least the number of characters in unwanted , that isn't unwanted .

If you only want the expressions you'll get them in capture group 1.

See it here at regex101 .

Edit: Looking at this one's I realize it's probably full of caveats, but it might get you started. (can't see them now though ;)

Edit 2: Alternative (better I believe) requiring something , but not allowing unwanted anywhere, on the line before:

^(?!.*unwanted.*).+[\r\n](expression2 expression3)

Here it is at regex101 .