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Help undesrtanding what ${0##*/} means


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 PostPosted: Mon Jun 01, 2009 8:06 am   

Joined: Mon Jun 01, 2009 7:49 am
Posts: 1
Hi forum

Greetings to all !
I am a newbie in shell programming and need help understanding some basic expressions in a script.
I have seen the expression ${0##*/} in many scripts but do not understand the principle behind it.
I tried echo ${0##*/} in shell and it returns "bash". When i execute the scripts where it is being used
like this:
---
echo "Usage: ${0##*/}"
....
Then it prints the name of the script.
Please help me understanding how it works and why it works.

Best Regards
Jaswi


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 PostPosted: Sat Jun 13, 2009 9:05 pm   
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Joined: Sat Jun 13, 2009 8:53 pm
Posts: 73
Location: Texas!
What it basically does is returns the name of the program or script you are running. If you just run it in a bash shell, you are running bash, so that's what it will return. The ${ ... } is used to surround a variable, and will return the value of that variable. $0 is a positional parameter (see here: http://bashscripts.org/forum/viewtopic.php?f=15&t=491) and the 0th parameter is the name of the script that was used to execute it, including the path if one was used; that's what the 0 is for. The ## is to match the largest possible substring that matches the following pattern within the preceding variable's value. In this case, the pattern is */ which will match anything that ends with a '/' because * is a wildcard. So, in short, this removes the path preceding the script name that you are running (if it exists).


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