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Unable to pass shell script parameter value to awk command in side the same script


 
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Old 01-27-2014
Unable to pass shell script parameter value to awk command in side the same script

Variable I have in my shell script
Code:
diff=$1$2.diff
id=$2
new=new_$diff

echo "My id is $1"
echo "I want to sync for user account $id"

##awk command I am using is as below

cat $diff | awk -F'~' ''$2 == "$id"' {print $0}' > $new

I could see value of $id is not passing to the awk command, i also tried few options as below but no help. Please assist.

Code:
#awk -F'~' -v search=$id '$2 == "search" {print $0} file' nayaka2almadm1.diff > $new

#cat $diff | awk -v "search=$id" -F'~' '$2 == "search" {print $0}' > $new

#cat $diff | awk -F'~' -v search="$id" '$2 ~ "^\"search {print $0}' > $new

#awk -v search="$id" file="$diff" | awk -F'~' '$2 == search {print $0} file' > $new

If I am passing the value directly in place of $id then it is working.

Thanks in advance.

Regards,
Ashutosh
Moderator's Comments:
Mod Comment Use code tags for code please. [code] stuff [/code]

Last edited by Corona688; 01-27-2014 at 01:15 PM..

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