CODEWITHSUNDEEP
bash
snripa
snripa

Reputation: 55

bash script with grep and cut : command not found

I'm new to bash, and need some simple script. It runs jar, and has to find "RESPONSE CODE:XXX". I need this response code (just XXX). I've try this:

 URL=$1
echo $URL
callResult=`java -jar RESTCaller.jar $URL`
status=$?
if [ $status -eq 0 ]; then
    result=`$callResult >> grep 'RESPONSE CODE' | cut -d':' -f 2`
else
    echo error
fi

I get ./run.sh: line 7: RESPONSE: command not found

What am I doing wrong?

Upvotes: 2

Views: 4242

Answers (2)

V H
V H

Reputation: 8587

URL=$1
echo $URL
callResult=`java -jar RESTCaller.jar $URL`
status=$?
if [ $status -eq 0 ]; then
    result=$($callResult 2>&1 grep 'RESPONSE CODE' | cut -d':' -f 2)
else
    echo error
fi

You were piping result to some invalid file name >> means write into file adding on..

2>&1 means redirect stderr to stdin - which is all of its output -

Upvotes: 1

Cloud
Cloud

Reputation: 19333

In this line:

result=`$callResult >> grep 'RESPONSE CODE' | cut -d':' -f 2`

You should be piping output to grep, not redirecting. Change it to this:

result=`$callResult | grep 'RESPONSE CODE' | cut -d':' -f 2`

Also, the syntax is a bit off, and you're better off avoiding backticks when possible. This is even better:

result="$(echo ${callResult} | grep 'RESPONSE CODE' | cut -d':' -f 2)"

Upvotes: 2

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