通過ajax創建的帖子變量

Question

我有一個標籤和一個下拉菜單，其中標籤在下拉菜單更改時動態更改。所以我使用ajax來解決這個任務，但我如何將標籤值傳遞給另一個文件的PHP文件？我怎麼能張貼它？

標籤和下拉菜單;

<?php echo '<select name="type" id="category" onchange="changeOwner();"> 

     <option value="Staf DC">Staf DC</option> 
     <option value="Admin">Admin</option> 

    </select></th>'; 

echo "<td align='center'><label id='own'></label></td>"; 

javascript;

<script type="text/javascript" src="jquery.js"></script> 
<script type="text/javascript"> 
function changeOwner() 
{ 
var selname = $("#category option:selected").val(); 
$.ajax({ url: "new_getdata.php", 

    data: {"selname":selname}, 

    type: 'post', 

    success: function(output) { 
     $("#own").html(output); 


    } 

    }); 
} 
window.onload = changeOwner(); 
</script> 

new_getdata.php

if (isset($_POST['selname'])) { 
$selname = $_POST['selname']; 
$query = "SELECT * FROM owner2 where type='$selname'"; 
$res = mysql_query($query); 

    while ($rows = mysql_fetch_assoc($res)) { 
    $name = $rows['owner']; 


    echo $name; 
} 
} 

變量$ name將動態每次更換標籤值的下拉菜單更改。我如何發送標籤值到另一個PHP文件？可以說我想把它發佈到register.php

Answer 1

$.ajax({ url: "new_getdata.php", 
    data: {"selname":selname}, 
    type: 'post', 
    success: function(output) { 
     $("#own").html(output); 

     $.ajax({ 
      data: {label: output}, 
      url: "regisster.php" 
     }); 
    } 
});