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Getting sysdate - 2 by an unix command


 
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# 1  
Getting sysdate - 2 by an unix command

How can I get an equivalent in unix for the following Oracle SQL command:

select sysdate - 2 from dual;


Thanks

Josť
# 2  

Here is one way:


#!/usr/bin/sh
_day=`date +%d`
_month=`date +%m`
_year=`date +%y`
newday=`expr $_day - 2`
echo "Two days ago was $_month/$newday/$_year "


I know the GNU date provides an easier way, but this is more portable...

your to_char function is up to you :P
# 3  
Except if you run it on the 1st, then 1-2 = -1. Oops.

A portable solution is to convert the current date to a julian value, subtract 2 and then convert back to a gregorian value. A script to convert to/from julian date is here: http://droflet.net/julian.txt

Last edited by PxT; 08-28-2001 at 03:49 PM..
# 4  
this login does not work with number greater than the days passed in the going month.
plz try this script it will fail .

#!/usr/bin/sh
echo "Plz enter the number of diff required"
read _diff
_day=`date +%d`
_month=`date +%m`
_year=`date +%y`
newday=`expr $_day - $_diff`
echo "Two days ago was $_month/$newday/$_year "
# 5  

Yeah, that is the shortcoming that PxT pointed out in my script also. I like his idea better of converting to julian date, subtracting the number you need, then converting back to regular date. Please check out his script - it's much better, and is less likely to contain the simple errors I have included (as features, of course) in mine. Smilie

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