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#1
July 22nd, 2006, 03:45 PM
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Unix script question, passing parameters to a script
I call the code below as follows:
testscript.sh part1 when it runs: It prints: parameter is part1 But in perl line it doesnt replace $1 with part1, any ideas? #!/usr/bin/ksh perl -le 'print join("\n", grep {((stat($_))[7] ne "0")}<$1*>)' print "parameter is $1"
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#2
July 22nd, 2006, 03:53 PM
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that's because it's in single quotes; you probably want to leave it out of the single quotes:
Code:
#!/usr/bin/ksh
perl -le 'print join("\n", grep {((stat($_))[7] ne "0")}<'$1'*>)'
print "parameter is $1"
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