/usr/local/bin/expr function not working

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# 1  
/usr/local/bin/expr function not working


I am not able to set "expr" function in ksh script.
Below is the sample code i used, and output is as "Syntax error"
Please help me to come out of it.

DATE1="`date -u +%Y`"
DATE2="`date -u +%m%d`"
print "CurrentDate = $CurrentDate"
cutoffdate=`/usr/local/bin/expr $CurrentDate - 3`
print "cutoffdate = $cutoffdate"
cat /tmp/loadDSdata.tmpl|while read BASENAME
Len=`/usr/local/bin/expr length $BASENAME`
Len=`/usr/local/bin/expr $Len + 2`
Len2=`/usr/local/bin/expr $Len + 7`

print "Len = $Len"
print "Len2 = $Len2"


CurrentDate = 20120315
cutoffdate = 20120312
expr: syntax error
expr: syntax error
expr: syntax error
Len =
Len2 =

Last edited by sdosanjh; 03-15-2012 at 06:07 AM..
# 2  
what is there in /tmp/loadDSdata.tmpl ?

---------- Post updated at 02:53 PM ---------- Previous update was at 02:48 PM ----------

if the BASENAME is numeric then you can try the below


# 3  
cutoffdate=`/usr/local/bin/expr $CurrentDate - 3`

You can't do date arithmetic with expr.
# 4  
What OS your using..? Maybe you can try as Len=`/usr/local/bin/expr "$BASENAME" : '.*'` instead of length expression.
This User Gave Thanks to michaelrozar17 For This Post:
# 5  
$BASENAME is not numeric..its say "India"

@Franklin52, but i am getting value in cutoffdate variable

@itkamaraj; in /tmp/loadDSdata.tmpl i have contents like below

Then it will go to directory search for Kenya_*.csv file and remove.

---------- Post updated at 01:43 AM ---------- Previous update was at 01:42 AM ----------


Thanks dear, that worked.

CurrentDate = 20120315
cutoffdate = 20120312
Len = 7
Len2 = 14
---------- Post updated at 01:44 AM ---------- Previous update was at 01:43 AM ----------

Originally Posted by michaelrozar17
What OS your using..? Maybe you can try as Len=`/usr/local/bin/expr "$BASENAME" : '.*'` instead of length expression.

# 6  
Originally Posted by sdosanjh
@Franklin52, but i am getting value in cutoffdate variable
And what should be the result if you subtract 3 from a date like 20120101?

20120101 - 3

# 7  
Originally Posted by Franklin52
And what should be the result if you subtract 3 from a date like 20120101?

20120101 - 3

Right... so it will fail every year for 1st 3 days...?
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