Question

基本上，我试图在触发isset（$ _ POST [“submit”]）时强制打开我的一个Jquery移动弹出窗口。

请注意，在这种情况下，在页面加载时弹出窗口加载不起作用，必须在提交表单时激活它。

例如：

<?php
if(isset($_POST['submit']))
{ 
      //other stuff
      //force open popup
}
?>
<form method='post' action='self.php'>
   <input type='submit' name='submit' value='submit' />
</form>
<a href="#popup" data-rel="popup" data-position-to="window" data-role="button" data-inline="true" data-icon="check" data-theme="a" data-transition="pop">popup</a>
<div data-role="popup" id="popupLogin" data-theme="a" class="ui-corner-all">
    <!-- Popup contents -->
</div>

先谢谢！

Answer 1

你能尝试一下吗？

<script type="text/javascript">
 $('#form').on('submit', function () {
    $("#popupLogin").popup("open")
  });
</script>

奥马尔的信誉谢谢你纠正我：）

Answer 2

你能不能回复一些东西让弹出窗口显示出来，或者如果它是正常打开它的a你可以模拟它的点击......

<?php
if(isset($_POST['submit']))
{ 
    //other stuff
?>
<script type="text/javascript">
$("document").ready(function(){
    // Simulate a click on a. I'd recommend giving A an ID or class to get it.
    $("a[href=#popup]").click();
});
</script>

<?php
}
?>

Answer 3

Add id to anchor tag and use its click event when posted. Code below: 

    <form method='post' action='self.php'>
       <input type='submit' name='submit' value='submit' />
    </form>
    <a id="popUp" href="#popup" data-rel="popup" data-position-to="window" data-role="button" data-inline="true" data-icon="check" data-theme="a" 
    <div data-role="popup" id="popupLogin" data-theme="a" class="ui-corner-all">
        <!-- Popup contents -->
    </div>
<?php
    if(isset($_POST['submit']))
    { 
          //other stuff
          //force open popup
          echo "<script>";
          echo "$('#popUp').click();"
          echo "</script>";
    }
    ?>

你如何强制使用PHP打开Jquery Mobile弹出窗口？

3 个答案: