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# 1  
Old 08-11-2013
Help with stat command

Hi Experts,

I am here with very simple request:


a=`stat -c %y log1.csv`

echo  $a

and this stat command returning value as
2013-08-11 05:42:10.000000000 -0400:

But I want to see in mm/dd/yyyy format?
any help is highly appreciated thank you

---------- Post updated at 05:47 PM ---------- Previous update was at 05:42 PM ----------

I know we do this using date:

$ date "%m-%d-%Y"

can I do this with a date variable?
$a "%m-%d-%Y"

Last edited by parpaa; 08-11-2013 at 09:12 AM.. Reason: modified
# 2  
Old 08-11-2013
You're lucky you are using bash. Read man bash about printf and try:
printf "%(%m/%d/%Y)T\n" $(stat -c %Y file)

# 3  
Old 08-11-2013
invalid format character

Thank you so much for your reply

unfortunately it error out with this message

 printf: `(': invalid format character

Last edited by Scott; 08-11-2013 at 09:31 AM.. Reason: Code tags
# 4  
Old 08-11-2013
RudiC's answer is using syntax specific to Korn Shell 93.

Here is one way of doing what you want using the Bash shell:
stat -c %y log1.csv | date +%m/%d/%Y

# 5  
Old 08-11-2013
Thank you sssssssssssssssssssso much guys


Thank you so much guys!!! its working Smilie
# 6  
Old 08-11-2013
Originally Posted by fpmurphy
stat -c %y log1.csv | date +%m/%d/%Y

Are you sure that it works?
I haven't yet met a date that reads from stdin...
# 7  
Old 08-11-2013
Originally Posted by fpmurphy
RudiC's answer is using syntax specific to Korn Shell 93.
Sorry - No.
GNU bash, version 4.2.45(1)-release (x86_64-pc-linux-gnu)
 printf "%(%m/%d/%Y)T\n" $(stat -c %Y file)

---------- Post updated at 18:23 ---------- Previous update was at 18:19 ----------

Originally Posted by MadeInGermany
Are you sure that it works?
I haven't yet met a date that reads from stdin...
It does not.
stat -c %y file | date +%m/%d/%Y

That's today's date, nothing to do with file. All three file's dates are 08/09/2013.

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