Vim regex captures part of group pattern at the end of the string

Question

I'm fairly new to Vim and regex. I have a file that looks something like this :

parent = 0001,,,lje1 
parent 0001 = category = 0002,,,ljef
parent 0001 = category = 0003,,,ljex4
parent = 0004,,,lxen 
parent 0004 = category = 0005,,,lxvr

I am looking to get the last set of numbers, commas and the code at the end of each line . If I use the following regex:

/[0-9]*,*l[a-z0-9]*$/

It matches the correct part on each line... numbers, comma and code. But if I try to do a replacement as such:

%s/.*\([0-9]*,*l[a-z0-9]*$\)/\1

I am just left with something that looks like this:

lje1 
ljef
ljex4
lxen 
lxvr

What I would like to be left with is this:

0001,,,lje1 
0002,,,ljef
0003,,,ljex4
0004,,,lxen 
0005,,,lxvr

If anyone has any idea of what I'm doing wrong, please help. I have searched and checked other posts but I can't seem to find a solution.

Cheers

Any reason in particular, Im going to go read up on this now.. — vessel
– vessel, Commented Jul 23, 2015 at 8:43
your regex will also match letter l by itself. * mean 0 to many where + means 1 to many. depending on your need, you can choose one. — nafas
– nafas, Commented Jul 23, 2015 at 8:58

Avinash Raj · Accepted Answer · 2015-07-23 08:40:48Z

4

Just replace the all the chars upto the last space character with an empty string.

/.* /

DEMO

answered Jul 23, 2015 at 8:40

Avinash Raj

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1 Comment

vessel Over a year ago

Good shout, I solved it with the answer below but thanks, didnt even think of this!

Wiktor Stribiżew · Accepted Answer · 2015-07-23 08:39:50Z

2

You need lazy matching with .\{-} instead of .*:

%s/.\{-}\([0-9]*,*l[a-z0-9]*$\)/\1

answered Jul 23, 2015 at 8:39

Wiktor Stribiżew

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2 Comments

vessel Over a year ago

Damn.. so simple.. Thanks so much! will accept this answer as soon as it lets me!

vessel Over a year ago

Thank you, much appreciated! I have only just been allowed to upvote answers as of this very moment, so there you go, you get my first haha!

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