My code was like this I'm passing 4 arguments to a script
ex.sh "wavpath" "featpath"
"ex.sh" code is
#!/bin/bash
wavPath=$1
featPath=$2
rm -f $scpFile
echo $wavPath
echo $featPath
for dir in `ls -R $wavPath|grep ":"|cut -d':' -f1`
do
mkdir -p ${dir/$wavPath/$featPath}
done
The error message:
bad substitution
and it is at ${dir/$wavPath/$featPath}
and its showing both the paths
can anyone help
尝试$ {dir} / $ {wavPath} / $ {featPath}
maybe you meant $dir/$wavPath/$featPath
try changing
mkdir -p ${dir/$wavPath/$featPath}
to
echo $dir/$wavPath/$featPath
and see if the output is what you expected for the input of mkdir
.
Also, you're not setting a value for the variable $scpFile
before you use it.
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