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PORemoving final bash script argument
 

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    <p>I'm trying to write a script that searches a directory for files and greps for a pattern. Something similar to the below <strong>except</strong> the find expression is much more complicated (excludes particular directories and files).</p> <pre><code>#!/bin/bash if [ -d "${!#}" ] then path=${!#} else path="." fi find $path -print0 | xargs -0 grep "$@" </code></pre> <p>Obviously, the above doesn't work because <code>"$@"</code> still contains the path. I've tried variants of building up an argument list by iterating over all the arguments to exclude path such as</p> <pre><code>args=${@%$path} find $path -print0 | xargs -0 grep "$path" </code></pre> <p>or</p> <pre><code>whitespace="[[:space:]]" args="" for i in "${@%$path}" do # handle the NULL case if [ ! "$i" ] then continue # quote any arguments containing white-space elif [[ $i =~ $whitespace ]] then args="$args \"$i\"" else args="$args $i" fi done find $path -print0 | xargs -0 grep --color "$args" </code></pre> <p>but these fail with quoted input. For example,</p> <pre><code># ./find.sh -i "some quoted string" grep: quoted: No such file or directory grep: string: No such file or directory </code></pre> <p>Note that if <code>$@</code> doesn't contain the path, the first script does do what I want.</p> <hr> <p><strong>EDIT</strong> : Thanks for the great solutions! I went with a combination of the answers:</p> <pre><code>#!/bin/bash path="." end=$# if [ -d "${!#}" ] then path="${!#}" end=$((end - 1)) fi find "$path" -print0 | xargs -0 grep "${@:1:$end}" </code></pre>
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