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need bash command help

Hello everyone. I am new to bash shell scripting and I have this problem I need to solve

Basically I have around 50,000 or so files in a directory of the format <department id>-<date>-<time>-<file id>.<ext>

I need to write a command that will output a list of department ID's and their counts as well as a count of files without department ID's.

I am stuck how to make this as a one command that won't break. Any help is greatly appreciated

With that many files using flags with ls becomes a problem.

However you can usually use ls itself.

Counts can be gotten with the wc command. "wc -l" shows number of lines. (Type "man wc" for more details of the command.)

Count of files irrespective of department ID is simple:

Code:

ls |wc -l

Count of department IDs and their files becomes a bit more problematical but you can do it with a loop:

Code:

for DEPID in $(ls -R |awk -F\- '{print $1}' |sort -u)

do DEPCNT=$(ls ${DEPID}-* |wc -l)

  echo Department ID $DEPID has $DEPCNT files

done

Putting it all together in one script:

Code:

#!/bin/bash

echo Total files is" $(ls |wc -l)

echo Break down by Department ID follows:

for DEPID in $(ls -R |awk -F\- '{print $1}' |sort -u)

do DEPCNT=$(ls ${DEPID}-* |wc -l)

  echo Department ID $DEPID has $DEPCNT files

done

Quote:

Originally Posted by MensaWater (Post 4703359)

Code:

ls |wc -l

Count of department IDs and their files becomes a bit more problematical but you can do it with a loop:

Code:

for DEPID in $(ls -R |awk -F\- '{print $1}' |sort -u)

do DEPCNT=$(ls ${DEPID}-* |wc -l)

  echo Department ID $DEPID has $DEPCNT files

done

Putting it all together in one script:

Code:

#!/bin/bash

echo Total files is" $(ls |wc -l)

echo Break down by Department ID follows:

for DEPID in $(ls -R |awk -F\- '{print $1}' |sort -u)

do DEPCNT=$(ls ${DEPID}-* |wc -l)

  echo Department ID $DEPID has $DEPCNT files

done

Worked out beautifully, thank-you very much for your help

Glad I could help. Please go to thread tools and mark this as Solved as it helps others find solutions more quickly when doing web searches in future.