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我有一个标签和一个下拉菜单,当下拉菜单更改时,标签会动态更改。所以我使用 ajax 来解决这个任务,但是如何将标签值传递给另一个文件 php 文件?我怎样才能发布它?

标签和下拉菜单;

<?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;
}
}

每次下拉菜单更改时,变量 $name 都会动态替换标签值。如何将标签值发送到另一个 php 文件?假设我想将其发布到 register.php

4

1 回答 1

2
$.ajax({ url: "new_getdata.php",
   data: {"selname":selname},
   type: 'post',
   success: function(output) {
       $("#own").html(output);

       $.ajax({
          data: {label: output},
          url: "regisster.php"
       });
   }
});
于 2012-05-07T08:14:43.900 回答