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help with expr command in script

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# 1  
Old 02-26-2010
help with expr command in script
Hi,

I am trying to code a unix function to calculate date difference between two date variables. I am stuck at a point where I am trying to convert hours into minutes. Below is the code I am doing.

Code:
 
function get_elapsed_time
{
export PROPS_FILE=temp.properties
export CURRENT_TIME=`date '+%T'`
export PREVIOUS_TIME=`grep ABC $PROPS_FILE |cut -c15-22`
echo $PREVIOUS_TIME
echo START_HR=`echo "$PREVIOUS_TIME" | awk '{print substr($0,1,2)}'`
echo $START_HR
echo END_HR=`echo "$CURRENT_TIME" | awk '{print substr($0,1,2)}'`
echo $END_HR
export START_MINS=`echo \`expr $START_HR \* 60\``
echo $START_MINS
}

Now the PREVIOUS_TIME will have value as say "26-Feb-10 08:16:23".
So the START_HR will have value as "08" and simlarly END_HR will have value as say "15".

The START_MINS should be like 08*60 which is 480. But when trying to run this temp.sh file i get the below error,

START_HR=08
END_HR=15
expr: syntax error

But if I run the same commands individually on the prompt, I get proper result. Its only while running the ksh script that the error is thrown.
#!/usr/bin/ksh

THe same command below runs perfectly fine on individual prompt
$export START_MINS=`echo \`expr $START_HR \* 60\``
$echo $START_MINS
480

Please can you advise.
Thanks
Last edited by Nutan; 02-26-2010 at 01:52 PM.. Reason: missed one line
# 2  
Old 02-26-2010
We don't need the echo.

Code:
export START_MINS=`expr $START_HR \* 60`

# 3  
Old 02-26-2010
May I suggest using bc to do math calculation instead of expr? I am not suggesting that expr may be your problem. In fact i dont the solution to your problem, but I think to math stuff, I usuall use bc, for e.g

Code:
echo "1+2"|bc

# 4  
Old 02-26-2010
Quote:
Originally Posted by methyl
We don't need the echo.

Code:
export START_MINS=`expr $START_HR \* 60`


You don't need expr (or bc or awk), but you do need to remove the leading zero or the number will be taken as octal, and 08 is not a valid octal number:
Code:
export START_MINS=$(( ${START_HR#0} * 60 ))

Last edited by cfajohnson; 02-27-2010 at 12:20 AM..
# 5  
Old 02-27-2010
You indicated that you are using the Korn shell. If it is the latest version of the Korn shell rather than ksh88 or pdksh, the following script may help you
Code:
#!/usr/bin/ksh93
#
# diffdate "Tue, Feb 19, 2008 08:00:02 PM" "Wed, Feb 20, 2008 02:19:09 AM"
#

SDATE=$(printf '%(%s)T' "$1")
FDATE=$(printf '%(%s)T' "$2")

[[ $# -ne 2 ]] && {
   print "Usage: diffdate start-date finish-date"
   exit 1
}

DIFF=$(($FDATE-$SDATE))

SECS=$(($DIFF % 60))
MINS=$(($DIFF % (60 * 60) / 60))
HOURS=$(($DIFF / (60 * 60)))

printf "%02d:%02d:%02d\n" $HOURS $MINS $SECS

exit 0

# 6  
Old 02-28-2010
No need to use command substitute and/or expr or bc. Use builtin, it's enough.
Ksh, bash, zsh, ... but not ex. in dash. (posix not include this, so using ex. dash you need command substitution - thanks for cfajohnson).
Code:
(( SECS= DIFF % 60                  ))
(( MINS= DIFF % (60 * 60) / 60  ))
(( HOURS= DIFF / (60 * 60)        ))

Posix version need command substitution ex. in dash, works also in ksh, bash, ...
Code:
SECS=$((  DIFF % 60                  ))
MINS=$((  DIFF % (60 * 60) / 60  ))
HOURS=$(( DIFF / (60 * 60)        ))

And if you not use ksh (prev. printf using), then use ex. gnu date. Take epoc time and calculate.
Code:
...
SDATE=$( date -d "$1" '+%s' )
FDATE=$( date -d "$2" '+%s' )
(( DIFF = FDATE - SDATE ))
...

Last edited by kshji; 02-28-2010 at 12:11 PM..
# 7  
Old 02-28-2010
Quote:
Originally Posted by kshji
No need to use command substitute and/or expr or bc. Use builtin, it's enough.
Ksh, bash, zsh, dash, ...

This code will not work in dash; it is not POSIX syntax.
Quote:
Code:
(( SECS= DIFF % 60                  ))
(( MINS= DIFF % (60 * 60) / 60  ))
(( HOURS= DIFF / (60 * 60)        ))


The POSIX syntax, which will work in all the shells you mentioned, is:
Code:
SECS=$((  $DIFF % 60   ))
MINS=$((  $DIFF % (60 * 60) / 60  ))
HOURS=$(( $DIFF / (60 * 60) ))
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