I was trying to use a regular expression to match the inner text between two characters, but I am getting the wrong text
I tried putting [A-z]* instead of .* for matching only the inner text and it worked. But I need to match non-letter characters too.
/\[?(,? ?\[(\[(.+)-(.+)\])\])\]?/g
This is my regular expression and i want to match the characters between the square brackets:
[[[hello-hello]],[[hi-hi]]]
The bold characters are the one matched.
I'd expect to match [[[hello-hello]],[[hi-hi]]] in match 1 and [[[hello-hello]],[[hi-hi]]] in match two.
I was trying to use a regular expression to match the inner text between two characters, but I am getting the wrong text
I tried putting [A-z]* instead of .* for matching only the inner text and it worked. But I need to match non-letter characters too.
/\[?(,? ?\[(\[(.+)-(.+)\])\])\]?/g
This is my regular expression and i want to match the characters between the square brackets:
[[[hello-hello]],[[hi-hi]]]
The bold characters are the one matched.
Share Improve this question edited May 26, 2019 at 14:54 Emma 27.8k11 gold badges48 silver badges71 bronze badges asked May 25, 2019 at 16:21 Pato05Pato05 3563 silver badges14 bronze badges 4I'd expect to match [[[hello-hello]],[[hi-hi]]] in match 1 and [[[hello-hello]],[[hi-hi]]] in match two.
- 1 Set the capture to non-greedy (.+?) Or use ([^\]]+) – Denis Giffeler Commented May 25, 2019 at 16:25
-
3
Also be aware that
[A-z]will consume square brackets. – bobble bubble Commented May 25, 2019 at 16:26 - Can you add example input and output? – hong4rc Commented May 25, 2019 at 16:27
- 1 @DenisGiffeler yes, it works! I can either set the U flag or the ?. Put your ment as an answer and I will mark it – Pato05 Commented May 25, 2019 at 16:35
5 Answers
Reset to default 3If everything in between the [] would be desired, then we might simplify our expression to maybe:
(?:\[+)(.+?)(?:\]+)
Here, we capture our likely desired substring in this capturing group:
(.+?)
Then, we add two boundaries on its left and right sides using two non-capturing groups:
(?:\[+)
(?:\]+)
Demo
const regex = /(?:\[+)(.+?)(?:\]+)/g;
const str = `[[[hello-hello]]
[[hi-hi]]]
[[hi hi]]]`;
const subst = `$1`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);RegEx
If this expression wasn't desired, it can be modified/changed in regex101..
RegEx Circuit
jex.im visualizes regular expressions:
I'd use something like this:
\[(?!\[)([^\]]*)\]
This will match a [ character, if not followed by a [ character. It will then match any amount of non ] characters capturing them in group 1. Followed by matching a ] character.
const text = "[[[hello-hello]],[[hi-hi]]]";
const regex = /\[(?!\[)([^\]]*)\]/g;
var match;
while (match = regex.exec(text)) {
console.log(match);
}Alternatively you can leave out the capturing group and drop the first and last character of every match.
const text = "[[[hello-hello]],[[hi-hi]]]";
const regex = /\[(?!\[)[^\]]*\]/g;
console.log(
text.match(regex)
.map(match => match.slice(1, -1))
);You could use 1 capturing group to capture your values.
The values before and after the hyphen could be matches using a negated character class \[([^][\n-]+ matching not an opening or closing bracket, a hyphen or a newline.
In your pattern you use a dot which will match any character except a newline so the negated character class contains a newline to prevent crossing lines.
\[([^\][\n-]+-[^\][\n-]+)]
Explanation
\[Match[(Start capturing group[^\][\n-]+Negated character class, match 1+ times not],[,-or a newline-Match-[^\][\n-]+Match 1+ times not],[,-or a newline
)Close capturing group]Match]char
Regex demo
const regex = /\[([^\][\n-]+-[^\][\n-]+)]/g;
const str = `[[[hello-hello]],[[hi-hi]]]`;
let m;
while ((m = regex.exec(str)) !== null) {
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
console.log(m[1]);
}Here is the regex I came up with:
\[+([a-z- A-Z]+)\]+
Demo
RegEx
(?<=\[)([a-z- A-Z]+)(?=\])
(?<=\\[): Start with a bracket, but does not contain a bracket.
(?=\\]): End with a bracket, but does not contain a bracket.
Detailed explanations can be found in this link.