$datef = (date("Y-m-d",mktime(0, 0, 0, $_POST[ month ], $_POST[ day ], $_POST[ year ])));
出于兴趣,如果使用上述代码从我的手工艺中挑选记录,那么就不得不改变这些记录,以规定时间不光是月、日、年。 该栏的类型是技经评估组,如果是,
Thanks
$datef = (date("Y-m-d",mktime(0, 0, 0, $_POST[ month ], $_POST[ day ], $_POST[ year ])));
出于兴趣,如果使用上述代码从我的手工艺中挑选记录,那么就不得不改变这些记录,以规定时间不光是月、日、年。 该栏的类型是技经评估组,如果是,
Thanks
你们需要改变:
如果看你操纵日期,请见:
然而,问题并不明确。 如果你试图取得<条码>日标/代码”的价值,并在你的座位上使用,你可能想这样做:
$day=$_POST[ day ];
$month=$_POST[ month ];
$year=$_POST[ year ];
$hour=12;
$minute=00;
$second=00;
$datef=strtotime($year.":".$month.":".$day." ".$hour.":".$minute.":".$second);
如果你想工作,你将不得不修改你的我的问答,否则,你只能用你规定的准确时间(例如:中午12:00,而不是中午12:0001):
SELECT * FROM rec或ds WHERE datefield BETWEEN 2010-11-01 07:00:00 AND 2010-11-01 07:59:59
或
SELECT * FROM rec或ds WHERE datefield LIKE 2010-11-01 07% ;
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