Question

这是我将弹出框放入

的地方

echo "<tr><td>".$set['Name']."</td><td>".$set['Position']."</td><td><select name='aaa'><option value='default'>Not Share</option><option value='read'>Read Only</option><option value='edit'>Editable</option></select></td></tr>";

我像$rights=$_POST['aaa'];一样检索它并且我检查它没有发布我的项目，除了每个事情成功的选择框，我发现不仅选择框，任何类型的输入，例如文本框也没有发布。为什么它不起作用？谢谢。

<?
include("../connection/conn.php");
session_start();

if($_SERVER['REQUEST_METHOD'] == "POST"){

$rights=$_POST['aaa'];


}else{?>


<html>
<head>


<script src="../plugin/jquery-1.6.1.min.js"></script>
<script type="text/javascript" src="../plugin/easyui/jquery.easyui.min.js"></script>
<script src="../plugin/jquery.validate.min.js"></script>
<script src="../plugin/jquery.form.js"></script>

<script>
$(document).ready(function(){
$("#addlist").validate();
});

function popup()
{$('#win').window('open');}   //<-- call popup 
</script>

</head>
<body>

<form id="addlist" method="post" action="add.php" >

 <a href="javascript:popup()"> Permission Settings</a>

<div id="win" class="easyui-window" closed="true" title="Login" style="width:400px;height:600px;"> //<==popup box in <div> element

<?


<select name='aaa'><option value='default'>Not Share</option><option value='read'>Read Only</option><option value='edit'>Editable</option></select> //<-- even not in the table , still not works.
<table width="100%" id="viewSub" border="1px;" cellpadding="1" cellspacing="1">
<tr>
<th field="col1" width="40%">Name:</th>
<th field="col2" width="40%">Position:</th>
<th field="col2" width="20%">Permission:</th>
</tr>

<?
foreach ($result as $set) 
{
    echo "<tr><td>".$set['Name']."</td><td>".$set['Position']."</td><td>ss</td></tr>";

}
//".$set['UserID']."
?>
</table>


<input class="submit" type="submit" name="submit" value="Submit"/>


</form>

</body>

</html>

<?
}
?>

Answer 1

将其添加到您的选择框中：

<form action="your-page.php" method="post">
<!-- YOUR POPUP BOX CODE -->

<input type="submit" name="submit" value="Submit" />
</form>

您的<input>和<select> 必须必须在<form>元素内！

<input type="submit">是可选的，因为您使用的是JQuery。

Answer 2

只需在win div

中移动表单标记即可

<script>
    $("#win").window().appendTo($("#addlist"));

</script>
<form id="addlist" method="post" action="add.php" >
<div id="win" class="easyui-window" closed="true" title="Login" style="width:400px;height:600px;">  element

     //<==popup box in


<div>

</form>

如果选择框放在弹出框中，它不会发布？

2 个答案: