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# 1  
Old 08-24-2010
Regular Expressions question
Hi
I have a multiple line file in the following format:
....
Code:
  1095     1000  0011  0000  1000  0101  0100  1100  1111
  1096     1000  0011  0000  1000  0101  0100  1100  1111
  1111     1000  0011  0010  1000  1001  0100  1100  1101
  1112     1000  0011  0000  1000  0001  0100  1100  1101
  1114     1000  0011  0000  1000  1101  0100  1100  1111

....

I need to grep the line which has "1111" at the begining (marked in bold)
I was trying to use the following regular expression to address this, but it fails:
Code:
cat /bb/data/fme.txt  | awk '/^\s 1111 /'

Thanks a lot for help.
Last edited by vbe; 08-24-2010 at 01:39 PM.. Reason: code tags...
# 2  
Old 08-24-2010
Code:
 grep -E '^1{4}' /bb/data/fme.txt

# 3  
Old 08-24-2010
Thanks, but unfortunatly it fails as well:
$ cat /bb/data/fme.txt | grep -E '^1{4}'
grep: illegal option -- E
Usage: grep -hblcnsviw pattern file . . .
$
# 4  
Old 08-24-2010
Quote:
Originally Posted by aoussenko
Thanks, but unfortunatly it fails as well:
$ cat /bb/data/fme.txt | grep -E '^1{4}'
grep: illegal option -- E
Usage: grep -hblcnsviw pattern file . . .
$
The use of cat is superfluous:
Code:
grep '^1111' /bb/data/fme.txt

# 5  
Old 08-24-2010
The problem is that the line starts with the blanks, so the above grep fails as well
Code:
$grep '^1111' /bb/data/fme.txt
$

# 6  
Old 08-24-2010
try this,
Code:
 perl -nle 'if (/^\s+1111/) { print $_; }' inputfile

# 7  
Old 08-24-2010
Quote:
Originally Posted by aoussenko
The problem is that the line starts with the blanks, so the above grep fails as well
Code:
$grep '^1111' /bb/data/fme.txt
$

Try this:
Code:
grep '^ *1111' /bb/data/fme.txt
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