Bash preexecute

Posted by Alex_Bender on Super User See other posts from Super User or by Alex_Bender
Published on 2013-10-27T15:34:19Z Indexed on 2013/10/27 15:55 UTC
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I'm trying to write bash command wrapper, which will be patch bash current command on the fly. But i'm faced with the problem. As i'm not a good Shell user, i can't write right expression of variable assignment in string. See bellow:

I'm set trap to preexecute, through this:

 alex@bender:~$ trap "caller >/dev/null ||  xxx \"\${BASH_COMMAND}"\"  DEBUG;

I want change variable BASH_COMMAND, do something like BASH_COMMAND=xxx ${BASH_COMMAND} but i don't know, how i need escaping variables in this string

NOTE: xxx -- my custom function, which must return some value, if in end of command situated word teststr

function xxx(){
# find by grep, if teststr in the end
`echo "$1" | grep "teststr$" >/dev/null`;
# if true ==> do
if [ "$?" == "0" ]; then
    # cut last 6 chars (len('teststr')==6)
    var=`echo "$1" | sed 's/......$//'`;
    echo "$var";
fi    }

How can i do this stuff?:

alex@bender:~$ trap "caller >/dev/null || ${BASH_COMMAND}=`xxx $BASH_COMMAND`"  DEBUG;

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