$thedate = "2014-11-01 00:32";
echo $thedate; = 2014-11-01 00:32 = ok
$pd = date("Y-m-d", strtotime($thedate));
echo $pd; = 2014-11-01 = ok
$dow = date("l",$pd);/$dow = date("w",$pd);
echo $dow; = Wednesday/3 = not OK - 2014-11-01 is Saturday/6
我以为它放错了日期和月份,但1月11日也是周六。怎么了?
您正在将字符串(2014-11-01 in $pd
)作为第二个参数而不是时间戳传递给date
函数
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