Change char in a string in JavaScript

Question

What regular expression should I use with the 'replace()' function in JavaScript to change every occurrence of char '_' in 0, but stop working as long as finding the char '.'?

Example:

_____323.____ ---> 00323._

____032.0____ --> 0032.0_

Are there ways more efficient than to use 'replace()'?

I am working with numbers. In particular, they can be both integer that float, so my string could never have two dots like in __32.12.32 or __31.34.45. At maximum just one dot.

What can I add in this:

/_(?=[\d_.])/g

to also find '_' followed by nothing?

Example: 0__ or 2323.43_

This does not work:

/_(?=[\d_.$])/g

You can use replace() along with reg-ex, what is your choice? — Dhaval Marthak
– Dhaval Marthak, Commented Sep 3, 2013 at 13:35
@dystroy sorry, i dropped my comment before seeing yours and your answer. — mvw
– mvw, Commented Sep 3, 2013 at 13:38
@JasonMcCreary I understand "they can be both integer that float [...] At maximum just one dot" as a NO. — Denys Séguret
– Denys Séguret, Commented Sep 3, 2013 at 13:56
Given that _032.0_ is possible, why isn't _032._0 possible ? — user557597
– user557597, Commented Sep 3, 2013 at 16:12

Denys Séguret · Accepted Answer · 2013-09-03 13:47:21Z

8

You could use

str = str.replace(/[^.]*/,function(a){ return a.replace(/_/g,'0') })

Reference

edited Sep 3, 2013 at 13:47

answered Sep 3, 2013 at 13:35

Denys Séguret

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5 Comments

Umur Kontacı Over a year ago

Can you explain how replace works when 2nd parameter is a function?

Jongware Over a year ago

@Umur: Look it up here: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Umur Kontacı Over a year ago

@Jongware I know how that works actually, I wanted it to make it a better answer since it is a smart one.

Denys Séguret Over a year ago

@UmurKontacı I added the reference.

Royi Namir Over a year ago

very nice usage of the global flag ( when is not present) +1

Aegis · Accepted Answer · 2013-09-03 14:06:16Z

1

Without replace/regex:

var foo = function (str) {
  var result = "", char, i, l;
  for (i = 0, l = str.length; i < l; i++) {
    char = str[i];
    if (char == '.') {
      break;
    } else if (char == '_') {
      result += '0';
    } else {
      result += str[i];
    }
    char = str[i];
  }
  return result + str.slice(i);
}

With regex: dystroy

Benchmark for the various answers in this post: http://jsperf.com/regex-vs-no-regex-replace

edited Sep 3, 2013 at 14:06

answered Sep 3, 2013 at 13:44

Aegis

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9 Comments

MDEV Over a year ago

Non-regex needn't be that large an answer: var str = "__323.__345._"; var newstr = str.substr(0,str.indexOf('.')).replace(/_/g,'0')+str.substr(str.indexOf('.'));

Denys Séguret Over a year ago

@SmokeyPHP There's a regex in your comment.

MDEV Over a year ago

@dystroy Funny.. yes of course - I mean of course a solution that takes most of the effort away from the regex (my tests prove up to 2x performance increase)

MDEV Over a year ago

@Aegis That's fab, thanks for getting involved and all, but I meant against the regex solution provided by dystroy.

Aegis Over a year ago

@SmokeyPHP I know, I'll add that to the test. My point is just that if you want performance then avoiding regex is a good idea, even just one :p

|

score 1 · Accepted Answer · 2013-09-04 18:35:09Z

1

Unless you have some other obscure condition -

find:

 _(?=[\d_.])

replace:

Or "To find also _ followed by nothing, example: 0__ or 2323.43_"

_(?=[\d_.]|$)

edited Sep 4, 2013 at 18:35

answered Sep 3, 2013 at 15:48

user557597

Comments

Denys Séguret · Accepted Answer · 2013-09-03 13:48:53Z

0

You could use lookahead assertion in regex...

"__3_23._45_4".replace(/_(?![^.]*$)/g,'0')

Result: 003023._45_4

Explanation:

/          # start regex
_          # match `_`
(?!        # negative lookahead assertion
[^.]*$     # zero or more (`*`) not dots (`[^.]`) followed by the end of the string
)          # end negative lookahead assertion
/g         # end regex with global flag set

edited Sep 3, 2013 at 13:48

Denys Séguret

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answered Sep 3, 2013 at 13:42

Billy Moon

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3 Comments

MDEV Over a year ago

Note: fails on __323.__345._ (if that situation were to crop up)

rook Over a year ago

As for me, it stops only before the last . character, ignoring all previous (replacing undescores).

rook Over a year ago

That don't work even in Python: re.sub(r'_(?![^.]*$)', "0", "__3_23._45_4._5") result: '003023.04504._5'

Billy Moon · Accepted Answer · 2013-09-03 14:10:22Z

0

var str = "___345_345.4w3__w45.234__34_";

var dot = false;
str.replace(/[._]/g, function(a){
  if(a=='.' || dot){
    dot = true;
    return a
  } else {
    return 0
  }
})

answered Sep 3, 2013 at 14:10

Billy Moon

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1 Comment

Denys Séguret Over a year ago

About your str : "At maximum just one dot".

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