how to define arg in bash

Posted on 2008-10-20
Last Modified: 2010-04-21
Please tell me what is wrong with this script. I want to define what is passed to +mtime and -mtime command line. So it would run like: 10 20

Then it would list anything older than 10 days (+mtime) and shorter than 60 days.

find . -type f \( -mtime +$args 1 -o -mtime +$ARGS[1] \) -printf "%Tw %p\n" | \
grep -v '^0' | cut -c3- | xargs ls -l
Question by:linuxpig
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Expert Comment

ID: 22764331
find . -type f -mtime +$1-mtime -$2
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Expert Comment

ID: 22764400
shell refer to arguments as positional arguments and they are accessed by names like $1 $2 $3 ...

Author Comment

ID: 22764684
Looking at the find command itself, with the two mtimes, will it keep anything within 14 days and a sunday and remove everything else is the mtimes are 14 and 60 as below?

find . -type f \( -mtime +14 -o -mtime +60 \) -printf "%Tw %p\n" | grep -v '^0' | cut -c3- | xargs rm -rf
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Expert Comment

ID: 22764712


find . -type f \( -mtime +$1 -o -mtime +$2 \) -printf "%Tw %p\n"|grep -v "^0" | cut -c3- | exec ls -l

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Author Comment

ID: 22766389
Thanks tintin,

But that will do what i want it to do right? Using the two mtimes will  keep anything within 14 days and a sunday and remove everything else is the mtimes are 14 and 60 on command line as below? 14 60
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LVL 48

Expert Comment

ID: 22770125

That's correct.

Author Comment

ID: 22772354
If i want to define the arguments below to be entered command line, will what i have below work?
I want to be able to type the command as: 10 60 Sun

Where 10 is the first mtime, 60 is the second, Sun is the day to define in the grep -v '^$3'


find /local/backup/uni* -type f \( -mtime +$1 -o -mtime +$2 \) -printf "%Tw %p\n" | grep -v '^$3' | cut -c3- | xargs ls -la
LVL 48

Accepted Solution

Tintin earned 50 total points
ID: 22772916
I noticed that the mtime logic is incorrect.  It needs to be

-mtime +$1 -a -mtime -$2

To add the 3rd parameter, you can do
find /local/backup/uni* -type f \( -mtime +$1 -a -mtime -$2 \) -printf "%Ta %p\n" | grep -v "^$3" | cut -c5-  | xargs ls -l

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Author Closing Comment

ID: 31508141
Actually -a wont work, -o will, ive tested it already and i also tried the grep you listed, it doesnt do anything but return a 0
LVL 48

Expert Comment

ID: 22773121
I beg to differ with your assessment.

If you call your script with 10 60 Sun

then the logic you have is files that have been modified greater than 10 days ago OR modified greater than 60 days ago will be listed.  The 2nd -mtime is made redundant in this case.

Also, I can not see anyway possible, you would be getting the grep to return 0.  Can you please show an example where that behaviour occurs.


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