Question

我想使用$ _GET []和getdate（）将日期作为查询字符串拉入，形式为：YYYY-mm-dd。如果未输入日期，则获取当前日期。这就是我所做的，但我不知道我是否接近。

if(!isset($_GET['date']))
{
//$date = "2012-09-15"; 
$date = date('Y-m-d');
}
elseif(DateTime::createFromFormat('Y-m-d', $_GET['date']))
{
//$date = DateTime::createFromFormat('Y-m-d',$_GET['date']);
$date = $_GET['date'];
}
else
{
$date = date('Y-m-d');
}
$calendar = DateTime::createFromFormat('Y-m-d', $date);

//This is something else I was attempting

//if(isset($_GET['date']) ? $_GET['date'] : $date = date('Y-m-d')
//$date = explode("-", $_GET['date']);
//$date = date('Y-m-d');
// Create new DateTime object passing the date as a string.
$calendar = DateTime::createFromFormat('Y-m-d', $date);

Answer 1

不确定你的目标是什么，但PHP的日期格式将是这样的

int strtotime ( string $time [, int $now = time() ] )

如何与getdate（）php一起实现$ _GET

1 个答案: