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Chose list of sub directories in a bash script file

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# 1  
Old 03-06-2013
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 04:02 AM.. Reason: Please use code tags for data and code samples
# 2  
Old 03-06-2013
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

# 3  
Old 03-06-2013
Quote:
Originally Posted by bipinajith
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

Thanks bipinajith, terrific.

Cheers.
# 4  
Old 03-07-2013
hope this helps.

Code:
find . -type d |awk '{ gsub(/\.\//,"",$0); gsub(/\//,"_",$0);print }'

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