Regex (regular expressions), replace the second occurence in javascript
This is an example of the string that's being worked with:
xxxxxx[xxxxxx][7][xxxxxx][9][xxxxxx]
I'm having a little trouble matching the second occurrence of a match, I want to return the 2nd square brackets with a number inside. I have some regex finding the first square backets with numbers in a string:
\[+[0-9]+\]
This returns [7], however I want to return [9].
I'm using Javascript's replace function, the following regex matches the second occurrence (the [9]) in regex testeing apps, however it isn't replaced correctly in the Javascript replace function:开发者_Go百科
(?:.*?(\[+[0-9]+\])){2}
My question is how do I use the above regex to replace the [9] in Javasctipt or is there another regex that matches the second occurrence of a number in square brackets.
Cheers!
If xxx
is just any string, and not necessarily a number, then this might be what you want:
(\[[0-9]+\]\[.*?\])\[([0-9]+)\]
This looks for the second number in []
. Replace it with $1[<replacement>]
. Play with it on rubular.
Your regular expression fails to work as intended because groups followed by +
only end up holding the last [xxx]
.
Try
result = subject.replace(/(\[\d\]\[[^\]]+\])\[\d\]/, "$1[replace]");
As a commented regex:
( # capture the following in backref 1:
\[\d\] # first occurrence of [digit]
\[ # [
[^\]]+ # any sequence of characters except ]
\] # ]
) # end of capturing group
\[\d\] # match the second occurence of [digit]
If the number of [xxx]
groups between the first and second [digit]
group is variable, then use
result = subject.replace(/(\[\d\](?:\[[^\]]+\])*?)\[\d\]/, "$1[replace]");
By surrounding the part that matches the [xxx]
groups with (non-capturing) parentheses and the lazy quantifier *?
I'm asking the regex engine to match as few of those groups as possible, but as many as necessary so the next group is a [digit]
group.
console.log( "xxxxxx[xxxxxx][7][xxxxxx][9][xxxxxx]".replace(
/^(.*\[[0-9]+\].*)(\[[0-9]+\])(.*)$/,
'$1[15]$3')); // replace with matches before ($1) and after ($3) your match ($2)
returns:
// xxxxxx[xxxxxx][7][xxxxxx][15][xxxxxx]
It will match where [n] is preceeded by 1 set of brackets with numbers inside.
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