Bash - String comparison does not work

Question

I am trying to compare two variables in a .sh file but it never works ;-(
How can I compare them?

curDate=2014-03-09
nextDate=2014-04-17

if [ “$nextDate” = “$curDate” ]; then
   echo $curDate = $nextDate
else
    echo $curDate != $nextDate
fi

The code works just as expected - for the case you provide it prints 2014-03-09 != 2014-04-17 and when I change the curDate to the same value as nextDate it prints: 2014-04-17 = 2014-04-1. Also on ideone: ideone.com/FSFMof and ideone.com/Qu59WJ — Ivaylo Strandjev
– Ivaylo Strandjev, Commented Sep 16, 2015 at 10:10
Are you actually using Bash, or some other shell? Do you have #!/bin/bash at the top of your script? — John Zwinck
– John Zwinck, Commented Sep 16, 2015 at 10:16
Are the fancy quotes “ and ” actually in your script? They should be normal double quotes ". — Tom Fenech
– Tom Fenech, Commented Sep 16, 2015 at 10:40

Ivan · Accepted Answer · 2015-09-16 11:07:08Z

1

Everything that is supposed to be a string should better be quoted:

echo "$curDate != $nextDate"

instead of

$curDate != $nextDate

Consider quoting date assignments too.

Try running this:

var=asd das
for thing in $var; do
echo $thing
done

and then with var="asd das"

Asides from that - use " instead of “ in if [ “$nextDate” = “$curDate” ]; then

answered Sep 16, 2015 at 11:07

Ivan

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Comments

Botacco · Accepted Answer · 2015-09-16 11:19:38Z

0

I found the solution. In Mac OSX this solution works:

#!/bin/bash
  a=2014-03-09
  b=2014-03-09
if [[ $a == $b ]] ; then
  echo Found.
else
  echo not found
fi

answered Sep 16, 2015 at 11:19

Botacco

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Your original code would work as well, if you used "..." instead of “...” to quote the parameter expansions.