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non-greedy sed

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# 1  
Old 04-03-2010
non-greedy sed
I found a sed example that works for me, however, now i would like to understand what it is doing:

Code:
$ echo "<b>foo </b >bar" | sed  's/<[^>]*>//g'
foo bar

the part that i don't understand is the [^>]*

my guess is that ^> means the first instance of > stop matching but why * and how is it able to match everything (including spaces) without .*

thanks!
Last edited by Franklin52; 04-04-2010 at 07:42 AM.. Reason: Please use code tags!
# 2  
Old 04-03-2010
[^>]*

Match 0 or more instances of not ">". See item 3 @ Regular Expressions for more info.

Regards,
Alister
# 3  
Old 04-03-2010
so then do i need the second > (after the *)?

thanks so much!


[edit]

Nevermind -- i just realized what i was asking... basically the [^>] makes the matching stop at the > but doesn't get rid of the >

[/edit]
Last edited by adambot; 04-03-2010 at 11:59 AM.. Reason: dumb question
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