关于PHP读取时间号码问题
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2024-02-02 21:14:58
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public function pai3(&$actionNo, &$actionTime, $time=null){ $this->setTimeNo($actionTime, $time); //echo $actionTime,' ',date('Y-m-d H:i:s', $time); if($actionTime >= date('Y-m-d H:i:s', $time)){ $actionNo = date('Yz', $time)-6; }else{ $actionNo = date('Yz', $time)-6; $actionTime=date('Y-m-d 18:30', $time); } }
上面代码读取的年和号,如201456我想改成2014056增加一位,请问怎么改?
回复讨论(解决方案)
$year = date('Y', time());$day = str_pad(date('z',time()),3,'0',STR_PAD_LEFT);echo $year.$day;
$year = date('Y', time());$day = str_pad(date('z',time()),3,'0',STR_PAD_LEFT);echo $year.$day;
能加到我代码上发给我下吗
public function pai3(&$actionNo, &$actionTime, $time=null){ $this->setTimeNo($actionTime, $time); //echo $actionTime,' ',date('Y-m-d H:i:s', $time); $year = date('Y', $time); $day = str_pad(date('z',$time),3,'0',STR_PAD_LEFT); $times = $year.$day; if($actionTime >= date('Y-m-d H:i:s', $time)){ $actionNo = $times-6; }else{ $actionNo = $times-6; $actionTime=date('Y-m-d 18:30', $time); } }
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