Time comparison

Shell Programming and Scripting
1

hi friends,
I am new to shell scripting and i am using KSH shell .I would like to automate our daily routine manual tasks .first i ll explain the situation .I will list out the contents of directory named "log" using "ls " and verify whether all the listed files time differences is not more than 30 minutes.if not, i will send the mail. Now , i need to automate this . i think i can take the time[for eg: 00:00 ,00:30,01:00,01:30....23:00] alone by using "awk" command or "date" function how can i compare the time values because i am having approximately[not exactly] 48 files[24/0.5] per day i need to check the time difference between above 48 files[for every 30 minutes once one file will be created ] .based on that i will do some other tasks...

           My time stamp is "Jan 28 18:48"
  
           Eg:  ls -l log
                 file 1 00:30                    file 2 - file 1 is 30 minutes
                 file 2 01:00                     
                 file 3 01:30                    file 3 - file 2 is 30 minutes

                 can anyone help me ? Thanks in advance .....
2

Here is one approach which works across midnight time change-

#!/bin/ksh

filetime()  # file time in epoch seconds
{
    perl  -e '          
          print (stat $ARGV[0])[9];
         ' $1
}

istooold()  # return 0 if too old, 1 if okay
{
	now=$(date +%s)       # epoch time in seconds
	ftime=$(filetime "$1")   # file age in seconds
	now=$( now - 1800 )   # 30 minutes ago 
	if [[ $ftime -ge $now ]] ; then
		print 1
	else
	    print 0
	fi	
}

for filename in  $(ls log)
do
	ok=$( istooold $filename)
	if [[ ok -eq 1 ]] ; then
	   echo "$filename is okay"
	else
	   echo "$filename is too old"
	fi   
done
3

thanks guru .. i will try and let you know what is happening

4

That will break if any filenames contain spaces. But it won't work anyway, because the filenames will not contain the directory.

for filename in log/*