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Removing string between two particular strings in a line

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# 1  
Old 01-06-2012
Removing string between two particular strings in a line
Hi,

I have a file with following format:

1|What is you name (full name)?|Character
2|How far [in kilometers] is your school [from home]?|Numeric

Now I need to remove everything inside brackets () or []. There can be more than one pair of brackets. The output file should look like:
Code:
1|What is you name?|Character
2|How far is your school?|Numeric

Can anybody please help?

I tried with
Code:
sed 's/\[*\]//g'

but not working.
Last edited by Franklin52; 01-06-2012 at 05:48 AM.. Reason: Please use code tags for code and data samples, thank you
# 2  
Old 01-06-2012
Code:
perl -pe 's/ [(\[].+?[)\]]//g' inputfile

This User Gave Thanks to balajesuri For This Post:
ppatra
# 3  
Old 01-06-2012
You were close. Sed's pattern matching is greedy though, so the regex
Code:
s/\[.*\]//g

will match everything from the first opening bracket to the last closing one. What you need is to replace the '.' (which means 'any character'), with 'not closing bracket', that is [^]]:
Code:
sed 's/\[[^]]*\]//g'

should do the trick
This User Gave Thanks to mirni For This Post:
ppatra
# 4  
Old 01-06-2012
Thanks a lot! you guys are great!
The perl one is working exactly what I needed.
But for sed I am becoming fool when trying to to do it for (). Following are not working:
Code:
sed 's/(^*)//g'
sed 's/\((^))*\)//g'

Do we need to handle () in a different way, I guess we don't need to negate () like we need to do for [].

Moderator's Comments:
Mod Comment How to use code tags
Last edited by Franklin52; 01-06-2012 at 06:18 AM.. Reason: Please use code tags for code and data samples, thank you
# 5  
Old 01-06-2012
The difference is that '[' and ']' are metacharacters in regexp.
[^a] means all characters but not 'a'. So literal '[' or ']' needs to be escaped as \[, resp. \].

With parens its actually simpler:
Code:
sed 's/([^)]*)//g'

Since you don't have to escape them. But it's the same logic -- you do use the [^)] to match all characters but not ')'. If you allowed ')' to be matched, it would greedily eat the whole thing between first ( and last ). Not what you want here.
# 6  
Old 01-06-2012
Awesome! Thank you!
# 7  
Old 01-06-2012
Quote:
Originally Posted by mirni
You were close. Sed's pattern matching is greedy though, so the regex
Code:
s/\[.*\]//g

will match everything from the first opening bracket to the last closing one. What you need is to replace the '.' (which means 'any character'), with 'not closing bracket', that is [^]]:
Code:
sed 's/\[[^]]*\]//g'

should do the trick
In fact the last closing bracket does not need to be escaped, it will already be taken as litteral :
Code:
...  sed 's/\[[^]]*]//g'

---------- Post updated at 04:05 PM ---------- Previous update was at 03:56 PM ----------

The right expression to do it one shot is :

Code:
... sed 's/[[(][^])]*[])]//g'

Code:
$ echo 'this (is) a [test]!!!'
this (is) a [test]!!!
$ echo 'this (is) a [test]!!!' | sed 's/[[(][^])]*[])]//g'
this  a !!!

Note that this one shot notation will also match the [...) and (...] blocks :
Code:
$ echo 'this (is) a [darn) funny (and] crazy [test]!!!'
this (is) a [darn) funny (and] crazy [test]!!!
$ echo 'this (is) a [darn) funny (and] crazy [test]!!!' | sed 's/[[(][^])]*[])]//g'
this  a  funny  crazy !!!

Last edited by ctsgnb; 01-06-2012 at 11:11 AM..
This User Gave Thanks to ctsgnb For This Post:
ppatra
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