Firefox Javascript正则表达式获取方括号内但不包括方括号的数字数组

Question

Sample input

var abc = "blah blah [abc] , [123] [ab12], [456] [cdef] 789 ghi000 "

Expected output

["123", "456"]

I am trying to write regex that should match only the numbers within the square brackets and return an array of those numbers (excluding the []).

The regex that I tried used lookbehind so it worked from Chrome but failed for firefox. I found later that lookbehind is not yet supported in firefox (Visit https://bugzilla.mozilla.org/show_bug.cgi?id=1225665 ).

abc.match(/(?<=\[)(\d+)/g);

I get the expected output on Chrome 76:

["123", "456"]

but I get error on Firefox 68:

SyntaxError: invalid regexp group

How can I write a regex that works on both and generate expected result.

Answer 1

You can use match and map

 var abc = "blah blah [abc] , [123] [ab12], [456] [cdef] 789 ghi000 " let output = abc.match(/\\[\\d+\\]/g).map(m=>m.replace(/\\[(\\d+)\\]/g, "$1")) console.log(output) 

Or you can use exec

 var regex1 = /\\[(\\d+)\\]/g var str1 = "blah blah [abc] , [123] [ab12], [456] [cdef] 789 ghi000 " var array1; while ((array1 = regex1.exec(str1)) !== null) { console.log(`Found ${array1[1]}`); } 

Answer 2

Use capturing groups, eg:

var abc = "blah blah [abc] , [123] [ab12], [456] [cdef] 789 ghi000 ";
Array.from(abc.matchAll(/\[(\d+)\]/g)).map(m => m[1])

matchAll (note the browser compatibility) finds every occurrence of [NNN] and captures the digits inside the square brackets as match[1] .

Array.from() converts the iterator returned from matchAll to an array, which can then be processed to extract the capture.