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I wrote this small unix shell script to validate the date format. My script should check whether the string is in YYYYMMDDHHMISS format . If it is not in this format , it should send an error message .

For example, I declared an variable a and assigned an value to it.

a=20150620223405
date "+%Y%m%d%H%M%S" -d $a > /dev/null 2>&1
if [ $? -ne 0 ]
then
echo "Invalid format"
else
echo "Valid format"
fi

It always shows "Invalid" .I want to know what is the mistake here and how to proceed.. thanks

share|improve this question
    
If you're using GNU date, here is documentation on the date formats it accepts: gnu.org/software/coreutils/manual/html_node/… –  Mark Plotnick Jun 21 at 3:07

1 Answer 1

up vote 1 down vote

The reason your test fails is because the condition test uses a value of $? that is non-zero. The reason it's non-zero is because date is producing a non-zero exit status. If you temporarily stop discarding date's stderr with the > /dev/null 2>&1 you'll get to see the error message it's producing. That will help you identify the issue.

date: invalid date ‘20150620223405’

What date is saying is that your date format is not acceptable.

You could try this:

a=20150620223405
b=$(echo "$a" | sed 's/^\(....\)\(..\)\(..\)\(..\)\(..\)\(..\)$/\1-\2-\3 \4:\5:\6/')
c=$(date "+%Y%m%d%H%M%S" -d "$b" 2>/dev/null)
if test $? -eq 0 -a "$c" = "$a"
then
    echo ok
else
    echo not ok
fi
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1  
thanks.. but when I give 20152006223405(YYYYDDMMHHMISS) , instead of saying invalid it says "Valid" and displays an date in 2016 ... –  Pari Sairam Mohan Jun 20 at 18:27
    
Actually , the 'date' tries to convert the given date to the format specified which results in a date in 2016.So I think I must use an egrep ... –  Pari Sairam Mohan Jun 20 at 19:13
    
@Pari a comparison of date and the original string would pick that up; I'll modify my code shortly –  roaima Jun 21 at 14:10

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