Date: Wed, 23 Mar 2005 20:42:57 +0200 From: Giorgos Keramidas <keramida@ceid.upatras.gr> To: Brian John <brianjohn@fusemail.com> Cc: freebsd-questions@freebsd.org Subject: Re: Simple bash script to grep files for bad keywords Message-ID: <20050323184257.GA78765@orion.daedalusnetworks.priv> In-Reply-To: <4742.209.87.176.4.1111602542.fusewebmail-19592@webmail.fusemail.com> References: <4742.209.87.176.4.1111602542.fusewebmail-19592@webmail.fusemail.com>
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On 2005-03-23 12:29, Brian John <brianjohn@fusemail.com> wrote:
> Hello,
> I am trying to write a simple bash script that will grep all files
> in a directory (except ones that start with "00") for certain bad
> keywords. Here is what I have so far:
> #!/bin/bash
>
> # This is a simple script to check all sql scripts for bad keywords
>
> BAD_KEYWORDS='spool echo timing commit rollback'
>
> for i in $BAD_KEYWORDS;
> do
> echo "*********************************";
> echo "GREPing for bad keyword '$i'"
> echo "*********************************";
> grep $i ./*;
> done
>
> However, I'm not sure how to make it not grep the files that start
> with "00". Can anyone help me with this?
Use xargs, since it will buy you the extra feature of being able to
search through arbitrarily large numbers of files:
for _word in ${BAD_KEYWORDS} ;do
find . | grep -v '^/00' |\
xargs grep "${_word}" /dev/null
done
Tips:
- The quotes in "${_word}" are probably optional, but it's better to
be safe than sorry :-)
- The /dev/null is there so that grep will get at least 2 file
arguments, even if there is just one file in the current directory,
effectively forcing grep(1) to print the filename of this one file
if it happens to match the pattern.
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