Javascript RegEx Lazy match

Question

I am trying to find for loop pattern in javascript code, and replace syntax (from : to in), using below regex way,

var str="for(var x in []) for(var y in [])";

str.replace( new RegExp( '(for\\s*\\(.+\\s+):(\\s+.+\\))', 'ig' ), "\$1in\$2" )

i.e.

for(var x : list)
{
 // something
}

with

for(var x in list)
{
 // something
}

However I am facing issues when there are multiple for loops in same line.

for(var x : list) { for(var y : list) {
     // something
 }
}

which is valid syntax, however due to Greedy regex approach it converts as below:

for(var x : list) { for(var y in list) {
         // something
 }
}

I tried to explore lazy regex syntax but couldn't make it work. How can I achieve this ?

Regex is no good for dealing with nested structures. I can make it work for current scenario but it may still fail in general when you have more nesting patterns. — Pushpesh Kumar Rajwanshi
– Pushpesh Kumar Rajwanshi, Commented May 3, 2019 at 8:04
Couldn't you just do it like str.split(':').join(' in ')? Or you only want to do it as part of for loops? — Icepickle
– Icepickle, Commented May 3, 2019 at 8:10
Icepickle, No, because ':' is used in various ways i.e. for objects ==> var objectA = {'A','Apple','B':'Banana'}; — Pratik
– Pratik, Commented May 3, 2019 at 8:11
@PushpeshKumarRajwanshiHow shall I handle this thn ? Using recursive match ? Or any other possible way ? — Pratik
– Pratik, Commented May 3, 2019 at 8:12

Nina Scholz · Accepted Answer · 2019-05-03 08:37:53Z

2

You could add some lazy quantifiers to all * and +. and take for as part of the replacement, because of matching.

var str = "for(var x : []) for(var y : [])";

console.log(str.replace(/for\s*?(\(.+?\s+?):(\s+?.+?\))/ig, "for $1in$2"));

A bit shorter and it includes for in the first group.

var str = "for(var x : []) for(var y : [])";

console.log(str.replace(/(for\s*?\(.+?):(.+?\))/ig, "$1in$2"));

edited May 3, 2019 at 8:37

answered May 3, 2019 at 8:18

Nina Scholz

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1 Comment

Pushpesh Kumar Rajwanshi Over a year ago

Writing .+?\s+? is redundant can be just written as .+? and same for \s+?.+?

Pushpesh Kumar Rajwanshi · Accepted Answer · 2019-05-03 09:20:04Z

Instead of using lazy quantifiers, you can use negated character set as they perform better and you can use this regex,

(for\s*\([^:]+):([^)]+\))

and replace it with,

$1 in $2

Also, you don't have to use .+\\s+ as this is redundant and instead you can just write .+? and even better to use negated character set for it to work faster and similarly after : you can write \\s+.+ as .+? but again negated character class is better choice like I mentioned in my answer.

Another point that could lead you into issues is, you should not use this \$1in\$2 for replacement, and instead use $1 in $2 firstly you don't need to escape $ as \$ and secondly because in case your for loop is like this, for(var x:list) i.e. without having space between colon and surrounding variables, then the output of replacement you may get is for(var xinlist) which would make it invalid. Which is why I suggested above in my answer to replace with $1 in $2 so in has space at both sides.

Regex Demo

JS codes,

const s = `for(var x : list)
{
 // something
}

for(var x : list) { for(var y : list) {
     // something
 }
}`

console.log(s.replace(/(for\s*\([^:]+):([^)]+\))/g, '$1 in $2'))

htho · Accepted Answer · 2019-05-03 08:15:35Z

0

The lazy behaviour can be achived with a ? after the quantifier.

const str = "for(var x : list) { for(var y : list) {"
str.replace( new RegExp( '(for\\s*?\\(.+?\\s+?):(\\s+.+\\))', 'ig' ), "\$1in\$2" )

btw. JavaScript RegEx literals are much easier to read:

str.replace( /(for\s*?\(.+?\s+?):(\s+.+\))/ig, "\$1in\$2" )

answered May 3, 2019 at 8:15

htho

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htho Over a year ago

"achieve using" what?

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