trying to find a way to locate files modified in the last hour in a shell script, unfortunately the command 'find . -mmin -60' is not supported on SunOS 5.10 (works on OpenSolaris 5.11 
)
Does anyone know a method of doing this in shell script on 5.10?
cheers
filetime() # create a filetime x hours ago
{
perl -e '
$mtime = time - (3600 * $ARGV[0]);
# time structure into variables
($sec,$min,$hr,$day,$mon,$yr,$wday,@dntcare) = localtime($mtime);
$yr = ($yr>=70) ? $yr+1900 : $yr+2000;
$mon += 1;
printf ("%d%02d%02d%02d%02d", $yr,$mon,$day,$hr,$min); ' "$1"
}
# get file time in YYYYMMddhhmm format
oldtime=$(filetime 1)
touch -t $oldtime dummy
find /path/to/files -newer dummy -print
Another one:
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' <dirname>
-f is for regular files, you can modify it, if you need something different.
many thanks for these replies guys, is there a way to do this via shell and not perl?
Why? What's the problem with Perl?
non whatsoever, just wondering if it can be accomplished in shell 
---------- Post updated at 03:21 PM ---------- Previous update was at 03:15 PM ----------
#!/usr/bin/ksh
# perl -MFile::Find -le'
# find {
# wanted => sub {
# -f and 1 >= -M and print $File::Find::name;
# }
# }, shift
# ' <dirname>
LOGS=/export/home/richard/share
filetime() # create a filetime x hours ago
{
perl -e '
$mtime = time - (3600 * $ARGV[0]);
# time structure into variables
($sec,$min,$hr,$day,$mon,$yr,$wday,@dntcare) = localtime($mtime);
$yr = ($yr>=70) ? $yr+1900 : $yr+2000;
$mon += 1;
printf ("%d%02d%02d%02d%02d", $yr,$mon,$day,$hr,$min); ' "$1"
}
# get file time in YYYYMMddhhmm format
oldtime=$(filetime 1)
touch -t $oldtime dummy
find $LOGS -newer dummy -print
-bash: ./perl.ksh: /usr/bin/ksh^M: bad interpreter: No such file or directory
Yes,
with zsh:
print -l **/*(Dmh-1)
Thanks 
I guess it's not that easy with ksh then
I suppose it is not, I would use Perl 
I don't understand how to integrate your perl code into a shell script rad, can you give an example? i.e. switching to the directory /richard/share then listing all files in there modified in the last hour?
If that's the only operation, you could simply execute this code on your command line:
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /richard/share
If you need to include that code in a script, just paste it inside:
#!/bin/sh
... other code here ...
filenames="$(
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /richard/share
)"
... continue here ...
is this line here:
-f and 1
which does the actual test to see if it's less than 1 hour? I take it the 1 relates to 1 hour?
when i run it against my directory it shows files which have been modified over an hour ago?
function SearchLogsLastHour
{
cd $URS_LOGS
filenames="$(
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 >= -M and print $File::Find::name;
}
}, shift
' /export/home/richard/share/ipcc/urslogs
)"
echo $filenames | ls -l
}
When i run the above it just shows all the file in the directory - I opened and saved one of them, ran it again (expected it to just list the one file I saved a few seconds ago) but it still lists all the files the directory
Sorry, 1 is one day, so you need 1/24:
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /richard/share
I corrected my posts above.
Working cheers for that rad; you the man 
---------- Post updated at 08:11 PM ---------- Previous update was at 07:25 PM ----------
so say the perl program finds 3 files matching the critera - is there an easy way to modify this so that once the first file is found, we do a grep on that file then output what we need to the temp file, then proceed to the second file, do a grep on that, output to the temp file etc etc?
Of course, just describe what exactly you want to grep and put into the temp file(s).
cd $LOGS
filenames="$(
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /export/home/richard/logs
)"
grep "cheese" $filenames
That seems to work, but I need to introduce an IF into it so that if the perl returns any files of the extension .gz it does a gzgrep instead of a normal grep
You can do it all in Perl, of course, but if you feel more comfortable with shell scripting, you could try something like this (untested):
cd "$LOGS"
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /export/home/richard/logs |
while IFS= read -r; do
case $REPLY in
*.gz ) gzgrep cheese "$REPLY";;
* ) grep cheese "$REPLY" ;;
esac
done
filenames="$(
perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' /export/home/richard/share/logs |
while IFS= read -r; do
case $REPLY in
*.gz ) gzgrep cheese "$REPLY";;
*.log ) grep cheese "$REPLY" ;;
esac
done
)"
The grep only returns one line 
should be about a hundred from one .log file and one .gz file
What happens when you run the code directly, without assigning it to a variable?
yes you can do like
find path -name -print | grep "get the time stamp from date command for the hour only" > file.txt
put it in the cron job which will run every hour and you will get the result