Simple bash script to grep files for bad keywords

[Brian John]

> On 2005-03-23 12:29, Brian John <brianjohn at 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.
>
Cool, I think I get it for the most part.  However, what exactly am I
doing when I am piping to xargs?  I can see that the filenames not
starting with '00' will be piped, but what does the '\' do?  Sorry, I am
really new to scripting and *nix in general.  But I am a programmer so I
learn fast.

Thanks!

/Brian