[SunHELP] Solaris Date commands

[sunhelp at sunhelp.org]

How about:

1.  Reading the timestamp in each file
2.  Write out the timestamp and filename to a temp file
3.  Sort the file using `sort -n` (n is for numeric, but probably doesn't
matter since the date string is uniform in length.
4.  Use the tempfile in a loop, reading a line at a time, either reading the
filename into a variable using read or awk { 'print $2' } to get the
filename.  Do whatever you need to do in the loop.
5.  Remove the tempfile.



-----Original Message-----
From: Gavin Winter [mailto:gwinter at rlo.com.au]
Sent: Sunday, November 04, 2001 5:18 PM
To: sunhelp at sunhelp.org
Subject: [SunHELP] Solaris Date commands


Hello all,
I am new to shell scripting and I am attempting to solve a somewhat complex
problem. On our Solaris 8 box (SPARC) we have a dir with many files. Each
file (flat text) has, in the header record, a timestamp field indicating the
date/time the file was created.  The files have to be processed in timestamp
order, i.e. oldest first.  The unix timestamp against the file is not a
valid indicator of age in this case.

And example of the timestamp field is as follows...
20011025015826
(YYYYMMDDHH24MISS)

I can loop through the files easily enough, reading in the date field but
this make it very difficult to do comparisons between dates to find the
oldest file. I think the easiest method would be to convert the string to a
date format, do a comparison on the date format (perhaps using seconds from
1970 or similar) and proceed forthwith to date the files. However I cannot
find any unix commands to help me do this.

Does anyone have any advice to offer on this?

regards
Gavin Winter

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