使用正则表达式捕获第一部分，跳过中间部分，并捕获最后一部分

Question

I've some lines like the example below, I want to capture everything in the first bracket, then everything in the some texts part.

{Foo} - {Bar} - some texts

Expected result: {Foo} - some texts

All the texts and the ones in the brackets will be different, so I will need to based it off on the {} and - to capture what I need.

What I've came up with gets everything in the first bracket, \\{(.*?)\\} . However I'm not too sure how I can skip the middle part and just grab the some texts . I'm looking at using lookaround assertions, is that the best way to approach this?

Thanks.

Answer 1

Use 2 Capture Groups to Exclude the Middle

 /(\\{.+?\\})\\s-\\s\\{.+?\\}\\s-\\s(.+?)\\b/gi;

^{☝ ☝
Replace from here to ..........................there with \\s-\\s}

This pattern will match multiples on the same line and/or multiple lines.

---

Demo

 var str = `{Foo} - {Bar} - some text {Foo} - {Bar} - some different text {Foo} - {Bar} - this text is always matched correctly because the word boundary at the end {Foo} - {Bar} - so this pattern is ok sharing the same line and/or {Foo} - {Bar} - multiple lines`; var rgx = /(\\{.+?\\})\\s-\\s\\{.+?\\}\\s-\\s(.+?)\\b/gi; var res = str.replace(rgx, '$1 - $2'); console.log(res);

Answer 2

This works in perl, it is skipping everything up to and including .+\\} - then matching whatever is left:

$ cat x.pl
use strict;

my $str = '{Foo} - {Bar} - some texts';

if ($str =~ /\{(.*?)\}.+\} - (.+)/) {
    print "$1, $2\n";
}


$ perl x.pl
Foo, some texts

Answer 3

以下正则表达式应该作为/({[^}]+})(?:.+})?(.+)/与包含{Foo} 的capture group 1和capture group 2有一些文本