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help pulling data string using Awk

Posted on 2016-09-12
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Last Modified: 2016-09-13
I'm working on a script to locate newly changed files and send out an email, I need help pulling just the file name from the resulting find command.

my find command is-
find /mnt/ABC_POOL/ABC_BUILD/ -type f -name .zip -prune -o -mmin -5 -type f -print
the result from the above would be something like this

/mnt/ABC_POOL/ABC_BUILD/XSTORE/ELC/ABC_xyz_7.1.4.17_1.12.3-installx.zip
I would like to use just the zip file name as my final result by using awk or sed to remove the path and leave just the file name, keeping in mind that the full path is not always the same.
I'm hoping there is a way to remove everything from start of string leaving the last part after the last / that would be the file name.

Any idea what would be the best way, I attempted awk -F"\t" but did not work.
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Question by:atorex
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Expert Comment

by:woolmilkporc
ID: 41794877
You could try "basename", a utility exactly meant for that purpose:

find /mnt/ABC_POOL/ABC_BUILD/ -type f -name .zip -prune -o -mmin -5 -type f -print | while read file
  do
    echo $(basename $file)
  done

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If the result of the find command is always just one filename you could put it like this:

FILE=$(basename $(find /mnt/ABC_POOL/ABC_BUILD/ -type f -name .zip -prune -o -mmin -5 -type f -print))

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woolmilkporc earned 500 total points
ID: 41794886
Or simpler with xargs:

find /mnt/ABC_POOL/ABC_BUILD/ -type f -name .zip -prune -o -mmin -5 -type f -print |xargs -i{} basename {}

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Author Comment

by:atorex
ID: 41794891
Awesome, that works perfectly.

much appreciated.

Regards,
Atorex
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Expert Comment

by:woolmilkporc
ID: 41794896
You can of course use "awk", because "awk" can do just everything:

find /mnt/ABC_POOL/ABC_BUILD/ -type f -name .zip -prune -o -mmin -5 -type f -print | awk '{F=split($0,A,"/"); print A[F]}'

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Author Comment

by:atorex
ID: 41796594
I have completed my script, thanks to you woolmilkporc, I have one issue I just noticed, the output is listing all files in sequence with a space separating each file, is there a way to have the output of the command I'm using to list it all like ls -l would in a column?
here is my command as a variable, I use the output of the variable to email the list of files.

FILE=$(basename $(find /mnt/ABC_POOL/ABC_BUILD/ -type f -newerct '5 minute ago' -print))


Regards,
Atorex
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Expert Comment

by:woolmilkporc
ID: 41796664
How do you use the variable?

If it's something like

echo $FILE | mailx ... ...

you can try

echo $FILE | tr " " "\n" | mailx... ...

If you're using a different approach please let me know .
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Author Comment

by:atorex
ID: 41796676
yes I'm using echo $FILE in to a temp file to be used for the body of the email like.

echo $FILE > /mnt/ABC_POOL/ABC_BUILD/newfile

so the below should work?

echo $FILE | tr " " "\n" > /mnt/ABC_POOL/ABC_BUILD/newfile
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Expert Comment

by:woolmilkporc
ID: 41796684
Yes, but if you want the output to go to a file you can simply do this:

find /mnt/ABC_POOL/ABC_BUILD/ -type f -newerct '5 minute ago' -print | xargs -n1 basename > /mnt/ABC_POOL/ABC_BUILD/newfile

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Author Comment

by:atorex
ID: 41796701
Ok, i'm using the variable to validate if a file is found in an if statement, the validation I'm using is
if [ -s $FILE ] and has been working so far.
Thanks for your help, trying to better my limited bash  scripting  capabilities.


Regards,
Atorex
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