Bash Script to grep string in file and output to another file

Question

I wrote a bash script to grep for a string and succeeding few lines and then output to another file.

Here is my bash script

#!/bin/bash

echo "$#"

if [ "$#" -ne 4 ]; then

        echo "Usage : ./redirectTrace.sh searchText inputFile number_of_Lines_succeeding_the_searchText outputFile"

else

        searchText=$0
        inputFile=$1
        lines=$2
        outputFile=$3

        echo "$0 , $2, $3, $4"

        grep -i -A "$lines" "$searchText" $inputFile > $outputFile
fi

When I execute this as ./redirectTrace.sh Drag:: log/Test.log 10 Drag1.log , I get

grep: log/Test.log: invalid context length argument

If I don't pass number of lines and execute with a fixed value, I get another error.

Script :

grep -i -A 10 "$searchText" $inputFile > $outputFile

Execution : ./redirectTrace.sh Drag:: log/Test.log Drag1.log`

I get following error :

grep: Drag::: No such file or directory

Also log/Test.log file becomes empty.

Why?

davidsheldon · Accepted Answer · 2016-08-23 09:45:55Z

3

A major part is because $0 is the name of the script not the first argument to the script. The first argument is $1, the fourth argument is $4, etc.

So your assignments should be:

searchText=$1
inputFile=$2
lines=$3
outputFile=$4

edited Aug 23, 2016 at 9:45

davidsheldon

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answered Aug 23, 2016 at 9:41

Some programmer dude

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