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This command in my .sh script is supposed to look in /home/backup/VBtest for .sql.gz files and remove any older than 5 days:

find /home/backup/VBtest/*.sql.gz -mtime +5 -exec rm {} \;

But it doesn't. It gives me no warnings, no errors, just a silent fail. I am running it on CentOS 6.6.

EDIT - after comment suggestions, I also tried this find '/home/backup/VBtest' -name '*.sql.gz' -mtime +5 rm {} \; and got find: paths must precede expression: rm. Also what is the argument for file creation time (as opposed to file modified time)?

leylandski
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    What output do you get if you don't use -exec? Do you really mean to pass a literal list of file paths to find? – Michael Homer Sep 04 '15 at 09:12
  • What do you mean older? -mtime only check for modification time. – cuonglm Sep 04 '15 at 09:14
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    That's not how you use find. It should be more like find /home/backup/VBtest' -name '*.sql.gz' ... (test before you put in the exec part). – Petr Skocik Sep 04 '15 at 09:15
  • I don't use unix/linux very often, but when I do I get the command syntax wrong. I will try all your suggestions now. – leylandski Sep 04 '15 at 09:17
  • Are those .sql.gz files in /home/backup/VBtest/or in a subdirectory? I suspect that the way you used the command, find won't try to search in subdirectories - it just looks at those files directly in /home/backup/VBtest/ –  Sep 04 '15 at 09:26
  • The files it's supposed to remove are all in that VBtest folder. – leylandski Sep 04 '15 at 09:30
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    You need the -exec between +5 and rm: find /home/backup/VBtest -name "*.sql.gz" -mtime +5 -exec rm {} \; – Lambert Sep 04 '15 at 09:32

2 Answers2

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It should be more like:

find /home/backup/VBtest/  -name '*.sql.gz' -mtime +5 # -exec rm {} \;

(remove the # from the exec part if this gives the correct results) This will scan the whole directory tree.

/home/backup/VBtest/*.sql.gz by itself would get expanded by the shell (almost equivalent to the above find command with -maxdepth 1) as you can learn by doing 

echo /home/backup/VBtest/*.sql.gz

You can go down a pure-shell route if you want to. All posix shells can compare timestamps ( [ + either -nt for "newer than" or -ot for "older than") so you just need a reference timestamp and then filter the expanded glob by that:

touch /tmp/5dAgo --date '5 days ago'
trap 'rm -f /tmp/5dAgo' exit
for file in /home/backup/VBtest/*.sql.gz; do
   #remove the echo if this give the expected results
   [ "$file" -ot /tmp/5dAgo ] && echo rm "$file" 
done
Petr Skocik
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From find's man page:

    Numeric arguments can be specified as

   +n     for greater than n,
   -n     for less than n,
    n     for exactly n.

  -mtime n
          File's data was last modified n*24 hours ago.  See the comments for 
          -atime to understand how rounding  affects  the  interpretation  of
          file  modification times.

   -atime n
          File was last accessed n*24 hours  ago.   When  find  figures  out  
          how  many 24-hour  periods  ago  the  file  was  last  accessed, any 
          fractional part is ignored, so to match -atime +1, a file has to have 
          been accessed at least two days ago.

So, -mtime +5 will find those files last modified more than 5*24h ago and -mtime -5 will find those files last modified less than 5*24h ago. It clearly tells that if you have modified the file/s your command will not remove those files. Also try modifying the command as:

find /home/backup/VBtest/ -maxdepth 1 -name "*.sql.gz" -mtime +5 -exec rm {} \;
TPS
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