var substitution in a reg expr ?


 
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# 1  
Old 08-10-2005
var substitution in a reg expr ?

In a shell script, how I can achieve substitution of shell script var to a regular
expression, as shown below.


var=`head -1 file1`
awk '$0!~/$var/ {print $0}' file1 > file2


In the case above $var value literally considered for non-exists criteria.
# 2  
Old 08-10-2005
You have to double-quote $var for it to be expanded.

var=`head -1 file1`
awk '$0!~/"$var"/ {print $0}' file1 > file2


vino
# 3  
Old 08-10-2005
Quote:
Originally Posted by videsh77
In a shell script, how I can achieve substitution of shell script var to a regular
expression, as shown below.


var=`head -1 file1`
awk '$0!~/$var/ {print $0}' file1 > file2


In the case above $var value literally considered for non-exists criteria.
Code:
awk "!/$var/" file1 > file2

# 4  
Old 08-10-2005
Thanks futurelet, soln you provided works.
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