shell scripting: grepping multiple patterns, logically ANDed

[Tim Daneliuk]

On 06/27/2012 09:25 AM, Aleksandr Miroslav wrote:
> hello,
>
> I'm not sure if this is the right forum for this question, but here
> goes.
>
> I have the following in a shell script:
>
>
>      #!/bin/sh
>
>      if [ "$#" -eq "0" ]; then
>              find /foo
>      fi
>      if [ "$#" -eq "1" ]; then
>              find /foo | grep -i $1
>      fi
>      if [ "$#" -eq "2" ]; then
>              find /foo | grep -i $1 | grep -i $2
>      fi
>      if [ "$#" -eq "3" ]; then
>              find /foo | grep -i $1 | grep -i $2 | grep -i $3
>      fi
>
> Is there an easier/shorter way to do this? If there are 15 arguments
> supplied on the command line, I don't necessarily want to build 15 if
> statements.
>
> Thanks in advance for your answers.

The following solution relies on the fact that you can include multiple
patterns for grep to match with the '-e' argument:


   #!/bin/sh

   PATTERNS=`echo " $*" | sed s/\ /\ -e\ /g`

   find /foo | grep $PATTERNS

Notice that when constructing the $PATTERNS string out of the command line
args, you have to quote them with a prepended space character.  That's because
the subsequent 'sed' substitution needs to find a space *before* each argument
which it then replaces with "-e ".





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Tim Daneliuk