Dan*_*Dan 0 javascript php mysql ajax
我刚从Sandeepan得到了一些很棒的帮助,谢谢!
请任何人都能看到我做错了....
<head>
<script type="text/javascript" src="jquery-1.4.2.js"></script>
<script type="text/javasript">
function addItemToUsersList(itemId)
{
$.ajax({
'url': 'member-bucketadd-execTEST.php',
'type': 'GET',
'dataType': 'json',
'data': {itemid: itemId},
'success': function(data)
{
if(data.status)
{
if(data.added)
{
$("span#success"+itemId).attr("innerHTML","Item added to your personal list");
}
else
{
$("span#success"+itemId).attr("innerHTML","This item is already on your list");
}
}
},
beforeSend: function()
{
$("span#success"+itemId).attr("innerHTML","Adding item to your bucketlist...");
}
'error': function(data)
{
// what happens if the request fails.
$("span#success"+itemId).attr("innerHTML","An error occureed");
}
});
}
</script>
</head>
Run Code Online (Sandbox Code Playgroud)
然后按下按钮激活以下功能:
<a onclick='addItemToUsersList("<?php echo $itemid ; ?>")'> Add<img src='images/plus-green.png' /> </a>
Run Code Online (Sandbox Code Playgroud)
和exec页面:
<?php
if($bucketlist < 1)
{
mysql_query("INSERT INTO membersbuckets (memberbucketid, userid, bucketid, complete)
VALUES ('', '$userid', '$_GET['itemId]', '0')");
return json_encode(array("status" => true, "added" => true));
}
else
{
return json_encode(array("status" => true, "added" => false));
}
//echo "You are being directed to your bucket list, please wait a few moments...<meta http-equiv='Refresh' content='2; URL=mybucketlist.php'/>";
?>
Run Code Online (Sandbox Code Playgroud)
链接显示为链接,但单击它们时没有任何反应!这是我正在研究的测试页面http://olbl.co.uk/showbucketsTEST.php 提前谢谢!
这是我如何弄明白的.打开Firebug,在Firefox中打开页面.打开"控制台".
如果这是一个Javascript错误,你会看到它.Firebug也显示ajax流量,因此您可以打开ajax请求并查看是否出现服务器错误.
| 归档时间: |
|
| 查看次数: |
2449 次 |
| 最近记录: |