Javascript regex difficult replace

Question

I am new with regex, could you help me to solve this?

I want to replace

Some text and XXblaaaaa# and text again XXbl!abla# or XXblalala# or XXbabla#

with

Some text and OOTEXT** and text again OOTEXT** or OOTEXT**

I made a jsfiddle to test t.

My regex for test is: /((XX).+?(?=(#)))/g;, i.e., find XX untill #.

But how can I set the following condition: untill # if string between XX and # does not contain !?

Ah, now I understand. For each ! you want to ignore the next #, instead of skipping the entire XX..!..# thing. — Martin Ender
– Martin Ender, Commented Aug 22, 2013 at 13:02
Sorry, a open new question stackoverflow.com/questions/18382328/… — Dmitry
– Dmitry, Commented Aug 22, 2013 at 13:48

ColBeseder · Accepted Answer · 2013-08-22 12:57:15Z

1

/XX[^!]*?#/g

This will replace everything from XX to #, as long as it doesn't include an !

Used like this:

"Some text and XXblaaaaa# and text again XXbl!abla# or XXblalala# or XXbabla#".replace(/XX[^!]*?#/g, "OOTEXT");

EDIT:

A more efficient version would be like this:

    /XX[^!#]*#/g

And if you don't want to allow whitespace, then add \s in the character class

/XX[^!#\s]*#/g

edited Aug 22, 2013 at 12:57

answered Aug 22, 2013 at 12:40

ColBeseder

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11 Comments

Martin Ender Over a year ago

Why not put the # inside the character class, too, to get rid of the ungreediness?

ColBeseder Over a year ago

I think this way is easier to understand because it looks more like the OP's description. "From XX to # without including !"

Martin Ender Over a year ago

Sure, but that doesn't mean we shouldn't advocate best practices ;).

Brad Christie Over a year ago

@m.buettner: Sometimes readability over a [nominal] optimization is better. ;-)

ColBeseder Over a year ago

I agree. But readability is also a best practice. I'll add it for completeness.

|

Brad Christie · Accepted Answer · 2013-08-22 12:40:23Z

0

You can use the overload of .replace(regex,function):

var txt = 'Some text and XXblaaaaa# and text again XXbl!abla# or XXblalala# or XXbabla#';
var result = .replace(/(XX.*?#)/g,function(m,a){
  // only replace values that do not contain a '!'
  return a.indexOf('!') == -1 ? 'OOTEXT**' : a;
});
// "Some text and OOTEXT** and text again XXbl!abla# or OOTEXT** or OOTEXT**"

I'm not sure if your example was really that simple, so the above is a method with a lot of flexibility. You could also use [^!]*? in place of the .*?, but anything more complex then a single character will need to be more elaborate.

answered Aug 22, 2013 at 12:40

Brad Christie

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3 Comments

Martin Ender Over a year ago

It does have lookaheads.

Brad Christie Over a year ago

@m.buettner: indeed, I think was was thinking of look-behinds. There's one is has and one it doesn't, but never quite remember. ;-)

Martin Ender Over a year ago

Yes, excactly. It doesn't have lookbehinds, but it does have lookaheads.

Martin Ender · Accepted Answer · 2013-08-22 13:01:34Z

0

I think what you want is something like this:

result = input.replace(/XX[^!#]*(?:![^#]*#[^!#]*)*#/g, "OOTEXT**");

How does it work? We match XX, then we consume as many non-!, non-# characters as possible. The next group (?:![^#]#[^!#]*) matches a !, then consumes everything until the next # and continues with non-#, non-! as before. This part is repeated as many times as necessary, until we find a # that was not preceded by !.

answered Aug 22, 2013 at 13:01

Martin Ender

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