Same regex pattern, different results

Asked Oct 28 '21 at 18:40

Active Oct 28 '21 at 18:40

Viewed 28 times

I assume these two are equivalent:

let pattern1 = /\((.?)*\)/g;
let pattern2 = new RegExp('\((.?)*\)', 'g');

but they produce different results when applied on the following string:

let str = 'abc(a, b);'
let result1 = str.match(pattern1);
let result2 = str.match(pattern2);

Why?

asked Oct 28 '21 at 18:40

2

Well, these are not the same. The second one is actually `'((.?)*)'`. You need to escape the `\\` otherwise it detects a escape sequence. – MinusFour Oct 28 '21 at 18:42
@MinusFour: Do you mean the escape character is extra? – Arnold Zahrneinder Oct 28 '21 at 18:43
@MinusFour: I am escaping the parenthesis – Arnold Zahrneinder Oct 28 '21 at 18:44
1

String literal notation will interpret the backslash for its own use, and so it does not result in a literal backslash that the regular expression needs. It can be confusing, but both string literals and regular expressions use backslash as an escape character, so you need one for each phase: a double backslash – trincot Oct 28 '21 at 18:45
@MinusFour: That isn't what I want, you are wrong! – Arnold Zahrneinder Oct 28 '21 at 18:45
No, Arnold, you misunderstand their comment. – trincot Oct 28 '21 at 18:46
@trincot: That was a silly mistake of mine! you are correct and thanks for point out. – Arnold Zahrneinder Oct 28 '21 at 18:47
1

Basically, you need to double escape it. You escape the \ in the string so you actually get a \ character. Which is then used to escape the `(` in the regular expression. – MinusFour Oct 28 '21 at 18:47
@MinusFour: Yes, that's correct, I was distracted by the regex pattern itself. silly mistake – Arnold Zahrneinder Oct 28 '21 at 18:48

0 Answers0