Question

我的PHP页面中有不同的标签。我在每个标签中都有每个表单。现在我想提交表格。但是当我提交表格时，它会转到第一个标签。我只是希望它会保留在提交的相同标签上，但它永远不会。请帮我，代码如下......

   <html>
<head>
    <script type="text/javascript" src="http://code.jquery.com/jquery-1.11.1.js"></script>
    <script type="text/javascript">
        $(function(){
            $("#tab-container").on("click", ".tab-lbl", function(){
                var that = $(this);
                var tabid = that.data("tab");

                $(".tab").each(function(k, v){
                    $(this).hide();
                });

                $(tabid).show();
            });
        });
    </script>
</head>

<body>

<div id="header">
    <div class="logo"><a href="#"><span>TAJWEED</span></a></div>
</div>

<div id="container">
   <div class="sidebar">
       <div id="nav">
       <ul id="tab-container">
        <li><a href="#" class="selected tab-lbl" data-tab="#tab-dashboard">Dashboard</a></li>
        <li><a href="#" class="tab-lbl" data-tab="#tab-menu">Menue</a></li>
        <li><a href="#" class="tab-lbl" data-tab="#tab-slider">Slider</a></li>
        <li><a href="#" class="tab-lbl" data-tab="#tab-gallery">Gallery</a></li>
        <li><a href="#" class="tab-lbl" data-tab="#tab-pictures">Pictures</a></li>

       </ul>

       </div>

   </div>
   <div class="content">

        <div id="tab-menu" class="tab" style="display: none;">

  <?php

$connection = new mysqli('localhost','root','','Tajweed');// Establishing Connection with Server

if(isset($_POST['submitv'])){ // Fetching variables of the form which travels in URL
$name = $_POST['pname'];
$email = $_POST['plink'];


//Insert Query of SQL
$sql=$connection->query("INSERT INTO main_page(pname, plink) values ('$name', '$email')");



}
?>

<form name="myForm" action="admin.php" method="POST" onsubmit=" return validateForm()" >
Page Name: <input  placeholder="page name :" name="pname" type="text"  />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="submit" value="submit"  name="submitv">
</form>






</div>
        <div id="tab-slider" class="tab" style="display: none;">


  <?php

$connection = new mysqli('localhost','root','','Tajweed');// Establishing Connection with Server

if(isset($_POST['submitv'])){ // Fetching variables of the form which travels in URL
$name = $_POST['pname'];
$email = $_POST['plink'];


//Insert Query of SQL
$sql=$connection->query("INSERT INTO main_page(pname, plink) values ('$name', '$email')");



}
?>

<form name="myForm" action="admin.php" method="POST" onsubmit=" return validateForm()" >
Page Name: <input  placeholder="page name :" name="pname" type="text"  />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="submit" value="submit"  name="submitv">
</form>






</div>
        <div id="tab-gallery" class="tab" style="display: none;"><h1>Gallery</h1></div>
        <div id="tab-pictures" class="tab" style="display: none;"><h1>Pictures</h1></div>
</div>

</body>
</html>
</div>

Answer 1

为链接添加ID：

<li><a id="tab1" href="#" class="tab-lbl" data-tab="#tab-menu">Menue</a></li>

向表单添加新的隐藏变量：

<form name="myForm" action="admin.php" method="POST" onsubmit=" return validateForm()" >
Page Name: <input  placeholder="page name :" name="pname" type="text"  />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="hidden" name="link" value="tab1">
<input type="submit" value="submit"  name="submitv">
</form>

添加js代码：

<?php if(isset($_POST['link'])): ?>
$(document).ready(function(){
$("#<?php echo $_POST['link']?>").click();
});
<?php endif; ?>

<html>
<head>
    <script type="text/javascript" src="http://code.jquery.com/jquery-1.11.1.js"></script>
    <script type="text/javascript">
        $(function(){
            $("#tab-container").on("click", ".tab-lbl", function(){
                var that = $(this);
                var tabid = that.data("tab");

                $(".tab").each(function(k, v){
                    $(this).hide();
                });

                $(tabid).show();
            });
        });

        <?php if(isset($_POST['link'])): ?>
        $(document).ready(function(){
        $("#<?php echo $_POST['link']?>").click();
        });
        <?php endif; ?>
    </script>
</head>
<body>
<div id="header">
<div class="logo"><a href="#"><span>TAJWEED</span></a></div>
</div>
<div id="container">
<div class="sidebar">
<div id="nav">
<ul id="tab-container">
<li><a href="#" class="selected tab-lbl" data-tab="#tab-dashboard">Dashboard</a></li>
<li><a href="#" id="tab1" class="tab-lbl" data-tab="#tab-menu">Menue</a></li>
<li><a href="#" id="tab2" class="tab-lbl" data-tab="#tab-slider">Slider</a></li>
</ul>
</div>
</div>
<div class="content">
<div id="tab-menu" class="tab" style="display: none;">
<form name="myForm" action="admin.php" method="POST" onsubmit=" return validateForm()" >Form1:
Page Name: <input  placeholder="page name :" name="pname" type="text"  />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="hidden" name="link" value="tab1">
<input type="submit" value="submit"  name="submitv">
</form>
</div>
<div id="tab-slider" class="tab" style="display: none;">Form2:
<form name="myForm" action="admin.php" method="POST" onsubmit=" return validateForm()" >
Page Name: <input  placeholder="page name :" name="pname" type="text"  />
Page Link: <input type="text" placeholder="Page Link :" name="plink" />
<input type="hidden" name="link" value="tab2">
<input type="submit" value="submit"  name="submitv">
</form>
</body>
</html>

当我提交表单时，它不会返回到PHP中的相同选项卡

1 个答案: