Question

我真的尝试了一切，我从日期选择器中得到了这样的日期：05/10/2016

然后我用/替换了 - 。使用strtotime但我总是得到01-1-1970。

这是我的代码：

 if (isset($_POST['CampaignName']) && isset($_POST['CampaginBudget']) && isset($_POST['start_date']) && isset($_POST['end_date']) && $email  )
 {
   // receiving the post params
    $CampaignName= $_POST['CampaignName'];
    $CampaignBudget = $_POST['CampaginBudget'];
    $start_date= str_replace('/', '-', $_POST['start_date']);
    $end_date = explode('/', $end_date);
    $start_date = date('Y-m-d', strtotime($start_date));
    $end_date = date('Y-m-d', strtotime(implode('-', strtotime($end_date))));
    //$end_date = date('Y-m-d', strtotime($end_date2));
    echo $start_date;  echo $end_date;

编辑：startdate工作正常，但结束日期有问题。即使它们完全相同

html代码：

Campaign Duration :<input type="text" name="start_date" class="form-control" id="datetimepicker" placeholder="mm/dd/yyyy"/  required> <span> TO </span>  <input type="text" name="end_date" class="form-control" id="datetimepicker2" placeholder="mm/dd/yyyy"/   required>

Answer 1

你试过这个吗？

$originalDate = $_POST['start_date'];
echo $newDate = date("Y-m-d", strtotime($originalDate));

Answer 2

你在结束日期的双倍时间上做了stint strtotime，这就是为什么会出现问题：

结束日期只需：

$end_date = explode('/', $end_date);
$end_date = date('Y-m-d', strtotime(implode('-', $end_date))); // remove strtotime from implode function

完整代码：

if (isset($_POST['CampaignName']) && isset($_POST['CampaginBudget']) && isset($_POST['start_date']) && isset($_POST['end_date']) && $email  )
 {
   // receiving the post params
    $CampaignName= $_POST['CampaignName'];
    $CampaignBudget = $_POST['CampaginBudget'];
    $start_date= str_replace('/', '-', $_POST['start_date']);
    $start_date = date('Y-m-d', strtotime($start_date));
    $end_date = $_POST['end_date']; // add this line
    $end_date = explode('/', $end_date);
    $end_date = date('Y-m-d', strtotime(implode('-', $end_date))); // remove strtotime from implode function
    echo $start_date;  echo $end_date;

strtotime总是返回01-1-1970

2 个答案: