timestamp difference of two files in unix shell

I have two files. I need to calculate the timestamp difference between the two files.

I need the timestamp difference between test1 and test2: -rw-r--r-- 1 root root 1 Aug 16 16:26 test1 -rw-r--r-- 1 root root 2 Dec 13 2010 test2

I need the timestamp difference between test3 and test4.

-rw-r--r--   1 root     root           3 Aug 16 16:26 test3
-rw-r--r--   1 root     root           4 Aug 16 17:34 test4

Please let me know how we can achieve it in Solaris . I am using Solaris and my machine info is :

- Sol 5.10 Generic_127128-11 i86pc i386 i86pc. 

If you need to know anything else please let me know.

I have got the below answer date conversion to seconds in Solaris.

truss 开发者_JS百科/usr/bin/date 2>&1 | grep ^time | awk -F"= " '{print $2}'

but this is for the current date .. how we can do it for file (for example test3, as above)?

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The stat(1) command can show the time stamp in seconds-since-epoch:

stat --printf '%Y\n' foo

Thus something like the following might work:

(stat --printf '%Y' test2; printf ' - '; stat --printf '%Y\n' test4) | bc -lq

Tweak as required.

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Try something like this to get the date in seconds

date +%s -d "`ls -l test1 | awk -F " " '{ print $6,$7 }'`"

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$ ls -l .bashrc*
-rw-r--r-- 1 max max 360 Dec 14  2010 .bashrc
-rw-r--r-- 1 max max 359 Dec  2  2010 .bashrc~

$ echo $((`stat --format=%Y .bashrc` - `stat --format=%Y .bashrc~`))
1018884

The $(()) syntax is bash arithmetic expression expansion. What happens above is that file timestamps in seconds since Unix epoch get subtracted and the difference is printed.