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Re: Dato format converting med PHP

 

> Prøv
>
> $Year = substr($Received, 0, 2);
> $Month = substr($Received, 2, 2);
> $Day = substr($Received, 4, 2);
> $Received = date("Y-m-d", mktime (0,0,0,$Day,$Month,$Year));
>
> Mvh Kim Hemsø Rasmussen
>
>

Det giver forkert resultat.

<?
$ConvDate = "010719";
$Year = substr($ConvDate, 0, 2);
$Month = substr($ConvDate, 2, 2);
$Day = substr($ConvDate, 4, 2);
$Converted = date("Y-m-d", mktime (0,0,0,$Day,$Month,$Year));
echo $Converted;
?>

giver 2002-07-07 ??


Poul



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