>> I have the file_date like this 20100930.
>> How will i convert it to absolute time ??
Just like you always when you chewed of more than you can handle: A couple of bytes at a time
Joseph give a fine bit of parsing and reconstructing using F$ELEMENT. I like and often use that. But then you have to glue in the dashes.
So for this case I prefer F$EXTRACT and ditch the intermediate variable... those are for wimps!
$ vmsdate = f$extract(6,2,cdt) + -
f$extract( 4 * f$extract(4,2,cdt), 5, "NULL-JAN-FEB-MAR-APR-MAY-JUN-JUL-AUG-SEP-OCT-NOV-DEC-") + -
f$extract(0,4,cdt)
$
$ write sys$output cdt, " = ", vmsdate, " = ", f$cvtime(vmsdate,,"DATE")
>> I need to find out all the files from a certain logical, which are older than 1 day from the current date and rename them.
Ah! Here is what we really want to to (Jan!)
Is that a simple rename like directory, extension or name only, or complex replacing pieces of string in the name?
If it is complex rename, then you may want to consider a perl 'one-liner'
Just for fun....
First you test with 'print qq'
$ perl -le "for () { if ( -M > 1 ) { $old = $_; s/V2B/done/; print qq($old, $_ ) }}"
for () ... GLOB returning matching file names in $_
if ( -M > 1 ) ... Modification data older than 1.0 days?
$old = $_; ... Remember the old name
s/V2B/done/; ... Substitute a piece of string in $_
print qq($old, $_ ) ... print what rename details.
Now make it real by just replacing "print qq" with "rename"
$ perl -le "for () { if ( -M > 1 ) { $old = $_; s/V2B/done/; rename ($old, $_ ) }}"
This assumes that the file modification date matched the file name, which may be a stretch
And it assumes that you really wanted more than 24 hours = 1.0 days old, not just 'yesterday'.
If you want yesterday use DIR/BEF=YES.
Perl can of course also suck on that piece of time string in the filename and turn it into a real time.
Just for your entertainment....
One of many ways to pick apart the string:
$ perl -le "($y,$m,$d) = unpack(q(a4a2a2),q(20100930)); print qq(y=$y m=$m d=$d)"
y=2010 m=09 d=30
And now turn that into a time (seconds since 1-jan-1970, 86400 per day) and back:
$ perl -le "use Time::Local; ($y,$m,$d) = unpack q(a4a2a2),q(20100930); print scalar localtime timelocal(0,0,0,$d,$m-1,$y)"
Thu Sep 30 00:00:00 2010
Hope this helps some,
Hein
