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Home/ Questions/Q 8664161
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T17:10:35+00:00 2026-06-12T17:10:35+00:00

I am writing a bash script where i will need to check a directory

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I am writing a bash script where i will need to check a directory for existing files and look at the last 4 digits of the first segment of the file name to set the counter when adding new files to the directory.

Naming Scructure:

yymmddHNAZXLCOM0001.835

I need to put the portion in the example 0001 into a CTR variable so the next file it puts into the directory will be

yymmddHNAZXLCOM0002.835

and so on.

what would be the easiest and shortest way to do this?

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  1. Editorial Team
    Editorial Team
    2026-06-12T17:10:36+00:00Added an answer on June 12, 2026 at 5:10 pm

    You can do this with sed:

    filename="yymmddHNAZXLCOM0001.835"
    first_part=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\1/')
    counter=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\2/')
    suffix=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\3/')
    echo "$first_part$(printf "%04u" $(($counter + 1))).$suffix"
    => "yymmddHNAZXLCOM0002.835"
    

    All three sed calls use the same regular expression. The only thing that changes is the group selected to return. There’s probably a way to do all of that in one call, but my sed-fu is rusty.

    Alternate version, using a Bash array:

    filename="yymmddHNAZXLCOM0001.835"
    ary=($(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\1 \2 \3/'))
    echo "${ary[0]}$(printf "%04u" $((${ary[1]} + 1))).${ary[2]}"
    => "yymmddHNAZXLCOM0002.835"
    

    Note: This version assumes that the filename does not have spaces in it.

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