Days in 2 Dates

Hello,

I have 2 dates lets say: 1-1-2005 and 5-1-2005. Now there are 5 days in those dates. I want to get the days from the 2 specified days.

How do i do that ? Please help.

Thanks.

Hi,

I was trying this query but it does not seems to be working and i am getting an error instead:

$q3 = "SELECT DATEDIFF ('" . $a1["sdate"] . " 23:59:59', '" . $a1["fdate"] . "')";
$r3 = mysql_query($q3) or die(mysql_error());
echo mysql_result($r3, 0);


The error i am getting is:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '('2005-05-10 23:59:59', '2005-06-18')' at line 1

Please help.

Thanks

Quote:

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Originally Posted by AvisSoft

Hello,

I have 2 dates lets say: 1-1-2005 and 5-1-2005. Now there are 5 days in those dates. I want to get the days from the 2 specified days.

How do i do that ? Please help.

Thanks.

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Manavo11 gave you a link in this thread which shows you exactly how to do this.

Hi,

Thanks for that code..but i got around with that using the following code. Here it is maybe it will help others also who don't wanna get involved in mysql:

PHP Code:

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                $sVar_dateString1 = $a1["sdate"];
                $sVar_dateString2 = $a1["fdate"];

                $aVar_date1 = explode('-', $sVar_dateString1);
                $aVar_date2 = explode('-', $sVar_dateString2);

                $iVar_time1 = mktime(0,0,0,$aVar_date1[1],$aVar_date1[2], $aVar_date1[0]);
                $iVar_time2 = mktime(0,0,0,$aVar_date2[1],$aVar_date2[2], $aVar_date2[0]);

                $iVar_timeDifference = $iVar_time2 - $iVar_time1;
                $iVar_dayDifference = ($iVar_timeDifference / (60 * 60 * 24)) + 1;
                echo $iVar_dayDifference; 


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Thanks