使用 PHP 的 DateTime 类验证有效日期
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Verify valid date using PHP's DateTime class
提问by user1032531
Below is how I previously verified dates. I also had my own functions to convert date formats, however, now am using PHP's DateTime class so no longer need them. How should I best verify a valid date using DataTime? Please also let me know whether you think I should be using DataTime in the first place. Thanks
以下是我之前验证日期的方式。我也有自己的函数来转换日期格式，但是，现在我使用 PHP 的 DateTime 类，所以不再需要它们。我应该如何最好地使用 DataTime 验证有效日期？还请让我知道您是否认为我应该首先使用 DataTime。谢谢
PS. I am using Object oriented style, and not Procedural style.
附注。我使用的是面向对象的风格，而不是程序风格。
static public function verifyDate($date)
{
//Given m/d/Y and returns date if valid, else NULL.
$d=explode('/',$date);
return ((isset($d[0])&&isset($d[1])&&isset($d[2]))?(checkdate($d[0],$d[1],$d[2])?$date:NULL):NULL);
}
回答by bitWorking
You can try this one:
你可以试试这个：
static public function verifyDate($date)
{
return (DateTime::createFromFormat('m/d/Y', $date) !== false);
}
This outputs true/false. You could return DateTimeobject directly:
这输出真/假。您可以DateTime直接返回对象：
static public function verifyDate($date)
{
return DateTime::createFromFormat('m/d/Y', $date);
}
Then you get back a DateTimeobject or false on failure.
然后你会DateTime在失败时返回一个对象或错误。
UPDATE:
更新：
Thanks to Elvis Ciotti who showed that createFromFormat accepts invalid dates like 45/45/2014. More information on that: https://stackoverflow.com/a/10120725/1948627
感谢 Elvis Ciotti，他表明 createFromFormat 接受无效日期，例如 45/45/2014。更多信息：https: //stackoverflow.com/a/10120725/1948627
I've extended the method with a strict check option:
我使用严格的检查选项扩展了该方法：
static public function verifyDate($date, $strict = true)
{
$dateTime = DateTime::createFromFormat('m/d/Y', $date);
if ($strict) {
$errors = DateTime::getLastErrors();
if (!empty($errors['warning_count'])) {
return false;
}
}
return $dateTime !== false;
}
回答by Faiyaz Alam
With DateTime you can make the shortest date&time validator for all formats.
使用 DateTime，您可以为所有格式制作最短的日期和时间验证器。
function validateDate($date, $format = 'Y-m-d H:i:s')
{
$d = DateTime::createFromFormat($format, $date);
return $d && $d->format($format) == $date;
}
var_dump(validateDate('2012-02-28 12:12:12')); # true
var_dump(validateDate('2012-02-30 12:12:12')); # false
回答by shadyyx
You could check this resource: http://php.net/manual/en/datetime.getlasterrors.php
你可以检查这个资源：http: //php.net/manual/en/datetime.getlasterrors.php
The PHP codes states:
PHP 代码指出：
try {
$date = new DateTime('asdfasdf');
} catch (Exception $e) {
print_r(DateTime::getLastErrors());
// or
echo $e->getMessage();
}
回答by boctulus
Try this:
尝试这个：
function is_valid_date($date,$format='dmY')
{
$f = DateTime::createFromFormat($format, $date);
$valid = DateTime::getLastErrors();
return ($valid['warning_count']==0 and $valid['error_count']==0);
}
回答by bluepinto
$date[] = '20/11/2569';
$date[] = 'lksdjflskdj';
$date[] = '11/21/1973 10:20:30';
$date[] = '21/11/1973 10:20:30';
$date[] = " ' or uid like '%admin%";
foreach($date as $dt)echo date('Y-m-d H:i:s', strtotime($dt))."\n";
Output
输出
1970-01-01 05:30:00
1970-01-01 05:30:00
1970-01-01 05:30:00
1973-11-21 10:20:30
1970-01-01 05:30:00
1970-01-01 05:30:00
回答by julius patta
I needed to allow user input in (n) different, known, formats...including microtime. Here is an example with 3.
我需要允许用户以 (n) 种不同的、已知的格式输入......包括微时间。这是一个例子，3。
function validateDate($date)
{
$formats = ['Y-m-d','Y-m-d H:i:s','Y-m-d H:i:s.u'];
foreach($formats as $format) {
$d = DateTime::createFromFormat($format, $date);
if ($d && $d->format($format) == $date) return true;
}
return false;
}
