why strtotime isn't working on php

Question

I'm trying to compare between two date but unfortunately this isn't working by converting this into UNIX format with strtotime. I'm trying to compare a date to another date.

However this format is working:

if(strtotime("22-04-17") < strtotime("25-05-17")){
    echo 'Date One is smaller than date two';
}

But Many times it's failing. I've seen a lot of examples on the web but I can't figure out anything good!

if(strtotime("22-04-17") < strtotime("04-05-17")){ //passing still the 
    // bigger on but not working
    echo 'Date One is smaller than date two';
}

Possible duplicate of [Comparing two dates](https://stackoverflow.com/questions/3847736/comparing-two-dates) — GrumpyCrouton, Aug 07 '17 at 14:49
@GrumpyCrouton That is actually very complex to understand. I'm just simply trying to compare between two date. So please give me a easier example. — Code Cooker, Aug 07 '17 at 14:52
_Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed._ -- From [strtotime](http://php.net/manual/en/function.strtotime.php) — Milan Chheda, Aug 07 '17 at 14:52
@AbraCadaver The first one example is working. But the second one isn't working. but still the second is also greater. — Code Cooker, Aug 07 '17 at 14:59
2nd is not greater because on this comparison, you are basically testing if year 22 is less than year 04. — Amit Joshi, Aug 07 '17 at 15:05

Matt Gibson · Answer 1 · 2017-08-07T15:07:11.977

From the manual (make special note of the part I've put in bold):

"Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed. If, however, the year is given in a two digit format and the separator is a dash (-, the date string is parsed as y-m-d."

So here's what you're doing, with the dates PHP is interpreting your strings as in comments:

// Is 17 April 2022 earlier than 17 May 2025? Yes.
if(strtotime("22-04-17") < strtotime("25-05-17")){
    echo 'Date One is smaller than date two';
}

// Is 17 April 2022 earlier than 17 May 2004? No.
if(strtotime("22-04-17") < strtotime("04-05-17")){ //passing still the 
    // bigger on but not working
    echo 'Date One is smaller than date two';
}

I hope this makes the problem you're having clear.

As it also says in the manual, use DateTime::createFromFormat/date_create_from_format if you want to avoid ambiguity:

 $date = date_create_from_format('d-m-y', '04-05-17'); // 4 May 2017

Yes, that is elegant to understand the problem. Then how I can solve this? — Code Cooker, Aug 07 '17 at 15:06

score 2 · Answer 2 · answered Aug 07 '17 at 14:58

2

your comparsion is not working because strtotime("22-04-17") actually results to timestamp for this date: 17th April 2022;

Do the following and you will see what I mean. the following code will output '2022-May-17`

$date = "22-05-17";
echo date ("Y-M-d ",strtotime($date))."<br>";

answered Aug 07 '17 at 14:58

Amit Joshi

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I'm not here for negotiation. I want a solution, Please help me if you can – Code Cooker Aug 07 '17 at 15:00
2

I am not trying to negotiate with you. What I am telling you is why your code is not working. since you have chosen a `-` as separator and your year is in 2 digits, PHP assumes the format as `y-m-d`. The solution here is to use the 4 digit year. – Amit Joshi Aug 07 '17 at 15:07

ßiansor Å. Ålmerol · Answer 3 · 2017-08-07T15:06:15.567

2

Try this

$date1 = date('d-m-y',strtotime("22-04-17"));
$date2 = date('d-m-y',strtotime("04-05-17"));;

if((int)strtotime($date1) < (int)strtotime($date2)){ //passing still the  
    echo 'Date One is smaller than date two';
}

Your year format 17 causing the problem in strtotime function

edited Aug 07 '17 at 15:06

answered Aug 07 '17 at 15:02

ßiansor Å. Ålmerol

2,849
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This one is working !! I don't know if there is a bug. Did you use this on production? – Code Cooker Aug 07 '17 at 15:05
@CodeCooker instead of `date('d-m-y',strtotime("22-04-17"));` and `date('d-m-y',strtotime("04-05-17"));` You can just put `strtotime("22-04-17 GMT");` and `strtotime("04-05-17 GMT");` – YaBCK Aug 07 '17 at 15:09
This is kinda misleading because if you do `$date1 = date('d-m-Y',strtotime("22-04-17")); $date2 = date('d-m-Y',strtotime("04-05-17")); if((int)strtotime($date1) < (int)strtotime($date2)){ //passing still the echo 'Date One is smaller than date two'; }` this fails. – Amit Joshi Aug 07 '17 at 15:15
@AmitJoshi Yes It is failing.. ! when I'm using this format `m-d-Y` and giving it like `08-15-17` it is generating 1970 year !! which is incredible. – Code Cooker Aug 07 '17 at 18:17
told you :). So to fix this the best thing you can do is use a 4 digit year. where do you get these dates from anyways? May be you can write a function to turn these into proper days. – Amit Joshi Aug 07 '17 at 19:09

YaBCK · Answer 4 · 2017-08-07T15:09:57.730

0

I know this isn't what you're looking for but have you tried doing this to showing if date greater / smaller?

// Dates
$Date1 = strtotime("22-04-17 GMT");
$Date2 = strtotime("04-05-17 GMT");

// Diff between dates
$Diff = (int)floor(($Date1 - $Date2) / (60*60*24));

if ($Diff < 0) {
   echo "The Diff is negative";
}

Then the other way is like this answer: LINK

$date1 = strtotime("22-04-17 GMT");
$date2 = strtotime("04-05-17 GMT");

if((int)strtotime($date1) < (int)strtotime($date2)){ //passing still the  
    echo 'Date One is smaller than date two';
}

edited Aug 07 '17 at 15:09

answered Aug 07 '17 at 14:52

YaBCK

2,949
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This one also doesn't look like elegant to me. – Code Cooker Aug 07 '17 at 14:55
@CodeCooker This is what i use to work out different between dates and also if one greater or smaller. You could also use `DateTime` object to check what you need to be honest – YaBCK Aug 07 '17 at 14:56
I'm trying to use a date-picker. that could help me to fetch data's from server between two dates in this format `22-04-17` How can I do this, please? – Code Cooker Aug 07 '17 at 15:03
Check the latest question, someone answered in an elegant way. – Code Cooker Aug 07 '17 at 15:07
1

@CodeCooker That is very correct, there is different way to do it – YaBCK Aug 07 '17 at 15:11

score 0 · Answer 5 · answered Aug 07 '17 at 19:54

I wrote a function that takes your datestring and properly return timestmp. Please note that I followed PHP's convention of treating 2 digit years, i.e. 00-69 are mapped to 2000-2069 and 70-99 to 1970-1999

/* functon takes dateString of format dd-dd-yy and return proper timestamp */
function getTimestamp($dateString)
{
    $res = preg_match('/^([0-9]{2})\-([0-9]{2})-([0-9]{2})/', $dateString, $matches);
    if(!$res)
        return 0;
    //00-69 are mapped to 2000-2069 and 70-99 to 1970-1999
    if($matches[3]>=70 &&  $matches[3]<=99 )
        $year = "19".$matches[3];
    else
        $year = "20".$matches[3];
    $formatted_dat_string = $year."-".$matches[2]."-".$matches[1];
    return strtotime($formatted_dat_string);
}
getTimestamp("22-04-99");

So you can now use this function instead of strtotime for comparison.

score 0 · Answer 6 · answered Aug 07 '17 at 20:02

Finally, I got the solution. The strtotime() isn't any good to handle this case. You should go for dateTime object instead.

//First you need to pass the original format then you will need to pass new 
//Format to get this working properly. Hope this will help you guy's
$myDateTimestart = DateTime::createFromFormat('d-m-Y', $dateString);
$startDate = $myDateTimestart->format('m-d-Y');
//with Simply that you can format your date properly

I just Messed my times with this thing it was really bad

why strtotime isn't working on php

6 Answers6