Substituting variable with current time


 
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# 1  
Old 11-19-2005
Substituting variable with current time

Hi all

I have a script as follows :-
#!/usr/bin/ksh
IDT=`date +"%OH%M%S"`
while true
do
echo ${IDT}
sleep 1
done

I need the time to show me the current runtime value for the time, however this returns the time as at the start of the script.

Any ideas.
Thanks

JH
# 2  
Old 11-19-2005
#!/usr/bin/ksh
while true
do
date +"%OH%M%S"
sleep 1
done
# 3  
Old 11-19-2005
Dear Unix Daemon

Thanks for the response, however I want to use a variable instead of everytime having to use the date command.

Thanks
# 4  
Old 11-20-2005
How about this?

#!/usr/bin/ksh
while true
do
IDT=`date +"%OH%M%S"`
echo ${IDT}
sleep 1
done

-Mike
# 5  
Old 11-20-2005
There is a technique that might apply. You obtain the current time, compute the seconds after midnight, and set the SECONDS variable. It will now stay synchronized to the clock. Now you can take SECONDS any time you want and compute the time. It looks like a lot of code, but is fast since everything is done internal to the shell. But I have no idea what the OP is really looking for so I don't know if this helps.
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