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Grepping word based on white space.....

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# 1  
Old 09-20-2011
Error Grepping word based on white space.....
Hi,

I am having a text file with following contents:

[space][space][space]word
[space][space][space][space][space]word

I want to grep the first line i.e. word that is being preceded with three space characters.

So i tried
Code:
sed -n '/ {3}/p' filename
grep " {3}" filename

But is not returning any result. If i don't use {}, then it returns both the patterns.
Last edited by Scott; 09-20-2011 at 05:37 AM.. Reason: .
# 2  
Old 09-20-2011
Code:
 
$ awk ' /^   [A-Za-z0-9]/ ' filename
   ABC
   FGH

# 3  
Old 09-20-2011
Quote:
Originally Posted by itkamaraj
Code:
 
$ awk ' /^   [A-Za-z0-9]/ ' filename
   ABC
   FGH


Yes that will return both the lines, i want only that pattern which is having only three white spaces in front of it.
# 4  
Old 09-20-2011
Code:
grep "^   [^ ]" file

# 5  
Old 09-20-2011
my test file has the below lines.

Code:
 
bash-3.00$ cat test
   ABC
     CDE
   FGH

Code:
 
$ awk ' /^   [A-Za-z0-9]/ ' test     
   ABC
   FGH

it returns only the 3 space lines. Make sure you are giving three space in between the /^ [A
# 6  
Old 09-20-2011
Quote:
Originally Posted by sarbjit
Hi,

I am having a text file with following contents:

[space][space][space]word
[space][space][space][space][space]word

I want to grep the first line i.e. word that is being preceded with three space characters.

So i tried
Code:
sed -n '/ \{3\}/p' filename
grep -E '^ {3}' filename

But is not returning any result. If i don't use {}, then it returns both the patterns.
You could try as above highlighted..
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