Bash Script : Unwanted Output

Question

I have this simple bash script: I run ns simulator on each file passed in argument where last argument is some text string to search for.

#!/bin/bash

nsloc="/home/ashish/ns-allinone-2.35/ns-2.35/ns"
temp="temp12345ashish.temp"

j=1

for file in "$@"
do
        if [ $j -lt $# ]
        then
                let j=$j+1

                `$nsloc $file > $temp 2>&1`

                if grep -l ${BASH_ARGV[0]} $temp
                then
                        echo "$file Successful"

                fi

        fi
done

I expected:

   file1.tcl Successful

I am getting:

   temp12345ashish.temp
   file1.tcl Successful

When i run the simulator command myself on the terminal i do not get the file name to which output is directed.

I am not getting from where this first line of output is getting printed. Please explain it.

Thanks in advance.

aks · Accepted Answer · 2013-04-22 21:15:19Z

3

See man grep, and see specifically the explanation of the -l option.

In your script (above), you are using -l, so grep is telling you (as instructed) the filename where the match occurred.

If you don't want to see the filename, don't use -l, or use -q with it also. Eg:

grep -ql ${BASH_ARGV[0]} $temp

answered Apr 22, 2013 at 21:15

aks

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Comments

perreal · Accepted Answer · 2013-04-22 20:08:44Z

0

Just silence the grep:

 if grep -l ${BASH_ARGV[0]} $temp &> /dev/null

answered Apr 22, 2013 at 20:08

perreal

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