Shell script parameter substitution
Can someone tell me what the difference is between these two conventions for getting the name of the currently running script?
#!/bin/sh
echo $0
echo $开发者_开发百科{0##*/}
Does the second version strip the path from the script name?
Thanks!
Your guess is correct. From the bash
docs on my machine:
${parameter#word} ${parameter##word} The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches the beginning of the value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the ``#'' case) or the longest matching pattern (the ``##'' case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.
Yes, second version uses bash extension to cut longest prefix of $0 which match to */ regexp, so if $0 is "/bin/bash", then ${0##*/} will be "bash". You can use `basename $0` to achieve the same result.
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