Would appreciate if someone can explain the ${0##/} line. What does it do?
I am aware that $0 is the script name, $# is number of arguments passed in, $ is all the arguments. With the curly brackets {} added in, what's the eventual effect?
Does ${0##/} actually equals $0$#$? (something like factorising?)
myscript.sh
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LOGFILE=$MYLOG/${0##*/}.logfile
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What does your code return ... I just test on my shell ... and I get
[~]$ cat try.sh
#!/bin/bash
MYLOG="mylog"
LOGFILE=$MYLOG/${0##*/}.logfile
echo $LOGFILE
Output:
[~]$ ./try.sh
mylog/try.sh.logfile
Do you get different output ??
This is a parameter expansion implemented by ksh, bash and possibly other shells and documented in these shell respective manual pages which I recommend you to read.
It means here remove all directory components that might appear in $0. It is equivalent to using `basename $0` but faster.
that is shell's string operations.
yeah, if you see the result it is same as basename $0
${0##*/} is removing everything before "/" from the variable $0.
so its returning just the script name itself, without absolute path or ./ ( if executed from the current dir).
built in shell operations are always faster than external commands eg basename as per my knowledge.
see man pages for shell for more details.