PHP IF statement with Date not returning False

Question

I'm trying to learn a little bit of php and can not understand why the 2 second IF is executing here. I have a variable $theDate set to accept a European date 4th May 2010.

$theDate = date('d-m-Y', strtotime("04-05-2010"));
echo "$theDate<br />";

if($theDate > "02-05-2010")
{
 echo "Greater than!<br />";
}

if($theDate > "02-05-2011")
{
 echo "Why am I showing<br />";
}

echo "Endof";

I just want to use IFs for now, no IF-Else etc. But why is the 2nd IF executing when $theDate is NOT greater than 02-05-2011,

thanks in advance,

Joe

Because date() returns a string. You are comparing two string. Not dates. — Phantom, Jul 14 '14 at 11:27
I found it on StackExchange, somewhere. What is the correct method to compare a Date? — user55054, Jul 14 '14 at 11:29
I have submitted simple answer to your question, also please check this for future reference http://www.phptherightway.com/ it may help you with many how-to-do things. — scx, Jul 14 '14 at 11:41
Thanks everyone. Thanks scx - I will try that solution and take a look at your link - cheers, Joe — user55054, Jul 14 '14 at 11:47

score 3 · Accepted Answer · answered Jul 14 '14 at 11:38

As far as I remember its 2014. Why not use DateTime object and do why not do it right way??

$raw = '04-05-2010';
$theDate= \DateTime::createFromFormat('d-m-Y', $raw);
$raw2 = '02-05-2010';
$anotherDate = \DateTime::createFromFormat('d-m-Y', $raw2);

echo 'theDatedate: ' . $theDate->format('m/d/Y') . "<br>";
echo 'anotherDate date: ' . $anotherDate ->format('m/d/Y') . "<br>";

if ($theDate > $anotherDate ) {
    echo "Greater than!<br />";
}

If you are learning php please check this resource, is worth reading.

score 2 · Answer 2 · answered Jul 14 '14 at 11:32

2

To compare dates as string just use the UTC format YYYY-MM-DD. If you want to compare it as integer you have to convert it to seconds or compare starting with the year and then with the month and so on---

answered Jul 14 '14 at 11:32

dhalfageme

1,444
4
21
42

score 1 · Answer 3 · answered Jul 14 '14 at 11:28

It uses string comparison. Thus it first looks at the first character, if that is greater than, true is returned. If less, false is returned. If equal, move to the next character.

You can solve this by formatting the string in reverse: Y-m-d. Or by using a built-in operator:

if($time > strtotime($text))

where $time is specified in seconds.

score 1 · Answer 4 · answered Jul 14 '14 at 11:30

1

You are comparing strings. Convert Date to seconds and then compare, working one for all formats etc.

$theDate = date('d-m-Y', strtotime("04-05-2010"));
echo "$theDate<br />";
if(strtotime($theDate) > strtotime("02-05-2010"))
{
 echo "Greater than!<br />";
}
if(strtotime($theDate) > strtotime("02-05-2011"))
{
 echo "Why am I showing<br />";
}
echo "Endof";

answered Jul 14 '14 at 11:30

vaibhav

396
2
17

Note: This will only work with dates until 2038 - that's already a reachable range today. I suggest using the DateTime class instead: http://stackoverflow.com/questions/5319710/accessing-dates-in-php-beyond-2038 – Piskvor left the building Jul 14 '14 at 11:42

PHP IF statement with Date not returning False

4 Answers4