php 如何检查变量类型是否为 DateTime
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how can I check if a variable type is DateTime
提问by rzqr
How can I check if a variable is a datetime type on php, please? I've tried this, but it seems doesn't work.
请问如何检查变量是否是php上的日期时间类型？我试过这个，但它似乎不起作用。
public function setRegistrationDate ($registrationDate) {
if (!is_date($registrationDate, "dd/mm/yyyy hh:mm:ss")) {
trigger_error('registrationDate != DateTime', E_USER_WARNING);
return;
}
$this->_registrationDate = $registrationDate;
}
采纳答案by Daniele Vrut
function validateDate($date, $format = 'Y-m-d H:i:s')
{
$d = DateTime::createFromFormat($format, $date);
return $d && $d->format($format) == $date;
}
Reference:
参考：
回答by Hokusai
I think this way is more simple:
我认为这种方式更简单：
if (is_a($myVar, 'DateTime')) ...
This expression is true if $myVaris a PHP DateTimeobject.
如果$myVar是 PHPDateTime对象，则此表达式为真。
Since PHP 5, this can be further simplified to:
从 PHP 5 开始，这可以进一步简化为：
if ($myVar instanceof DateTime) ...
回答by Nishu Tayal
The simplest answer is : to check with strtotime()function
最简单的答案是：用strtotime()函数检查
$date = strtotime($datevariable);
If it's valid date, it will return timestamp, otherwise returns FALSE.
如果是有效日期，则返回timestamp，否则返回FALSE。
回答by Matteo Tassinari
I would use the following method:
我会使用以下方法：
public function setRegistrationDate ($registrationDate) {
if(!($registrationDate instanceof DateTime)) {
$registrationDate = date_create_from_format("dd/mm/yyyy hh:mm:ss", $registrationDate)
}
if(!($registrationDate instanceof DateTime)) {
trigger_error('registrationDate is not a valid date', E_USER_WARNING);
return false;
}
$this->_registrationDate = $registrationDate;
return true
}
