Regexp: lazy, but not lazy enough

Question

I hope there are some regexp experts that can help. I have been searching for hours but can not find an answer.

This is the input string:

parameters:x,y,123,z;parameters:a,b,456,c;

The puzzle is to retreive the last parameters part (a,b,456,c) and I know it starts with "parameters:" and ends with ",c";

So I tried the following regexp:

parameters:(.+?,c);

This matches not the expected last part but the starting from the first parameters. This is the match group:

x,y,123,z;parameters:a,b,456,c

So the ? for doing a lazy match is not lazy enough as it matches more then I want.

Any suggestions?

Why not splitting by semicolon ;, then match?

nhahtdh
– nhahtdh

2013-03-01 16:57:54 +00:00
Commented Mar 1, 2013 at 16:57 — nhahtdh
– nhahtdh, Commented Mar 1, 2013 at 16:57

stema · Accepted Answer · 2013-03-01 16:50:16Z

4

The regex is exactly doing what you have defined: It matches "parameters" and then lazy till the first "c" it finds.

Try this:

parameters:([^;]+,c);

[^;] is a negated character class, that matches every character but ";". It seems that ; can not occur within such a group.

See it here on Regexr

answered Mar 1, 2013 at 16:50

stema

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Comments

Tim Pietzcker · Accepted Answer · 2013-03-01 16:49:42Z

3

The simplest way would be to forbid : and ; from matching:

parameters:([^:;]+,c);

Or you can be more explicit and forbid parameters from matching twice:

parameters:((?:(?!parameters).)+,c);

The reason why your regex wasn't lazy enough is that the regex match starts at the earliest possible position, and .+? matches as much as necessary (which, from the first parameters: in the string, is exactly the string you found it to match).

answered Mar 1, 2013 at 16:49

Tim Pietzcker

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1 Comment

mkoper Over a year ago

Never thought of using a forbid construction. Works perfectly.

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