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Please help - Command to Subtract two numbers without losing prefix zeros

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Top Forums Shell Programming and Scripting Please help - Command to Subtract two numbers without losing prefix zeros
# 1  
Old 08-02-2017
Please help - Command to Subtract two numbers without losing prefix zeros
Hello,

I have a variable LOGNUM with values 0000095, When i subtract the variable by 1, Its losing its leading zeros. Can you please help me here ?

Code:
LOGNUM=0000095
$OLDLOG=`echo "${LOGNUM}-1"|bc`
$ echo $OLDLOG
94


Am expecting output as
Code:
0000094



Appreciate your help!

Thanks,
Prince

Moderator's Comments:
Mod Comment Please use CODE tags as required by forum rules!
Last edited by RudiC; 08-02-2017 at 05:49 PM.. Reason: Added CODE tags.
# 2  
Old 08-02-2017
Code:
LOGNUM=0000095
printf "%0${#LOGNUM}d\n" $(echo "$LOGNUM-1" | bc)

Last edited by rdrtx1; 08-02-2017 at 05:53 PM..
This User Gave Thanks to rdrtx1 For This Post:
prince1987
# 3  
Old 08-02-2017
Does it have to be bc? Try
Code:
awk -vLNUM=$LOGNUM 'BEGIN {printf "%07d\n", LNUM-1}'
0000094

This User Gave Thanks to RudiC For This Post:
prince1987
# 4  
Old 08-02-2017
When you post questions like this in the future, please tell us what operating system and shell you're using!

If you're using ksh, the easy way to do it is:
Code:
LOGNUM=0000095
OLDLOG=$(printf "%07d" $((LOGNUM - 1)))
echo $OLDLOG

If you're using bash, you could use:
Code:
LOGNUM=0000095
OLDLOG=$(printf "%07d" $(echo "$LOGNUM - 1" | bc))
echo $OLDLOG

You can't use arithmetic expansions for this with bash because bash treats all numbers with a leading 0 as octal (not decimal) values.

If you're using a pure Bourne shell, you could use:
Code:
LOGNUM=0000095
OLDLOG=`echo "$LOGNUM - 1" | bc`
OLDLOG=`printf "%07d" $OLDLOG`
echo $OLDLOG

One of the above is likely to work with any shell that is based on Bourne shell syntax. I make no claims about how to do this with a shell based on csh syntax.
This User Gave Thanks to Don Cragun For This Post:
prince1987
# 5  
Old 08-02-2017
With bash, try
Code:
printf "%0${#LOGNUM}d\n" "$(( 10#$LOGNUM - 1 ))"
0000094

This User Gave Thanks to RudiC For This Post:
prince1987
# 6  
Old 08-02-2017
Hello,

It works if number is below 0000099, If its above 0000100, it is not working as expected.

Code:
$ LOGNUM=0000120
$ printf "%0${#LOGNUM}d\n" "(( $LOGNUM - 1 ))"
0000079

Thanks,

Moderator's Comments:
Mod Comment Please use CODE tags as required by forum rules!
Last edited by RudiC; 08-02-2017 at 06:06 PM.. Reason: Added CODE tags.
# 7  
Old 08-02-2017
See updated post #2.
This User Gave Thanks to rdrtx1 For This Post:
prince1987
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