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javascript, regex parse string content in curly brackets - Stack Overflow

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i am new to regex. I am trying to parse all contents inside curly brackets in a string. I looked up this post as a reference and did exactly as one of the answers suggest, however the result is unexpected.

Here is what i did

var abc = "test/abcd{string1}test{string2}test" //any string
var regex = /{(.+?)}/
regex.exec(abc) // i got ["{string1}", "string1"]
             //where i am expecting ["string1", "string2"]

i think i am missing something, what am i doing wrong?

update

i was able to get it with /g for a global search

var regex = /{(.*?)}/g
abc.match(regex) //gives ["{string1}", "{string2}"]

how can i get the string w/o brackets?

i am new to regex. I am trying to parse all contents inside curly brackets in a string. I looked up this post as a reference and did exactly as one of the answers suggest, however the result is unexpected.

Here is what i did

var abc = "test/abcd{string1}test{string2}test" //any string
var regex = /{(.+?)}/
regex.exec(abc) // i got ["{string1}", "string1"]
             //where i am expecting ["string1", "string2"]

i think i am missing something, what am i doing wrong?

update

i was able to get it with /g for a global search

var regex = /{(.*?)}/g
abc.match(regex) //gives ["{string1}", "{string2}"]

how can i get the string w/o brackets?

Share Improve this question edited May 23, 2017 at 12:00 CommunityBot 11 silver badge asked Mar 20, 2012 at 18:02 MengQi HanMengQi Han 3101 gold badge3 silver badges9 bronze badges
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6 Answers 6

Reset to default 32
"test/abcd{string1}test{string2}test".match(/[^{}]+(?=\})/g)

produces

["string1", "string2"]

It assumes that every } has a corresponding { before it and {...} sections do not nest. It will also not capture the content of empty {} sections.

var abc = "test/abcd{string1}test{string2}test" //any string
var regex = /{(.+?)}/g
var matches;

while(matches = regex.exec(abc))
    console.log(matches);

Try this:

var abc = "test/abcd{string1}test{string2}test" //any string
var regex = /{(.+?)}/g //g flag so the regex is global
abc.match(regex) //find every match

a good place to read about Regex in javascript is here, and a nice place to test is here

good luck!

Nothing wrong. But you'll need to look at your capturing groups (the second element in the array) to get the content you wanted (you can ignore the first). To get all occurences, it's not enough to run exec once, you'll need to loop over the results using match.

Edit: nevermind that, afaik you can't access capturing groups with match. A simpler solution would be using a positive lookahead, as Mike Samuel suggested.

This result:

["{string1}", "string1"]

is showing you that for the first match, the entire regex matched "{string1}" and the first capturing parentheses matched "string1".

If you want to get all matches and see all capturing parens of each match, you can use the "g" flag and loop through, calling exec() multiple times like this:

var abc = "test/abcd{string1}test{string2}test"; //any string
var regex = /{(.+?)}/g;
var match, results = [];
while (match = regex.exec(abc)) {
    results.push(match[1]);   // save first captured parens sub-match into results array
}

// results == ["string1", "string2"]

You can see it work here: http://jsfiddle.net/jfriend00/sapfm/

try this for file

const fs = require('fs');

fs.readFile('logs.txt', function(err, data) {
if(err) throw err;
const paragraph = "'" + data + "'";
const regex = /\d+\<;>\S+\<;>(\d+)\<;/g;
const found = paragraph.match(regex);

console.log(found);
})
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