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#1
03-06-2013
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Chose list of sub directories in a bash script file
Hi, I have a command in my bash script, searchDirectoryName.sh: Code:
DIR_NAME=$(find . -type d)
echo "${DIR_NAME}"Code:
. ./Dir1 ./Dir1/1.0.2.1 ./Dir2 ./Dir2/1.1 ./Dir3 ./Dir3/2.2.1 How can I select only following directory names with second subdirectoies and without first ./ in the DIR_NAME? Code:
Dir1/1.0.2.1 Dir2/1.1 Dir3/2.2.11 Finally substitute "/" to "_" for each name: Code:
Dir1_1.0.2.1 Dir2_1.1 Dir3_2.2.1 Thanks. Kind regards. Last edited by Franklin52; 03-07-2013 at 03:02 AM.. Reason: Please use code tags for data and code samples |
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#2
03-06-2013
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Code:
for DIR_NAME in $(find . -type d)
do
if [[ "$DIR_NAME" =~ \/[0-9]\. ]]
then
DIR_NAME="${DIR_NAME#\.\/}"
DIR_NAME="${DIR_NAME/\//_}"
echo "$DIR_NAME"
fi
done |
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#3
03-06-2013
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Quote:
Cheers. |
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#4
03-07-2013
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hope this helps. Code:
find . -type d |awk '{ gsub(/\.\//,"",$0); gsub(/\//,"_",$0);print }' |
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