新手对jQuery不是很懂，求帮个忙
我写了一个的temper.html页面去调用http://hys2013.sinaapp.com/SQL.php页面的内容,我照着网上的例子来怎么不行啊（页面不弹窗）
temper.html:
<html><head> <script type="text/javascript" src="/jquery/jquery.js"></script><script type="text/javascript">$.post("http://hys2013.sinaapp.com/SQL.php",function(data){ alert("Data Loaded: " + data); });</script></head><body></body></html>http://hys2013.sinaapp.com/SQL.php:
<? $mysql = new SaeMysql(); $conn = mysql_connect(SAE_MYSQL_HOST_M.':'.SAE_MYSQL_PORT,SAE_MYSQL_USER,SAE_MYSQL_PASS); //if($conn){ $action="连接成功"; echo "document.write('".$action."');n"; } $sql="SELECT * FROM `temp` WHERE id = ( SELECT MAX( CAST( id AS UNSIGNED ) ) FROM temp )";//获取最大ID的那一行 $row=$mysql->getLine($sql); foreach ($row as $key=>$value) { $insert_id=$value; break;//目前只需要id所以无需遍历所有值 }echo $insert_id;?> 求大神给看看应该怎么改一下
回复讨论(解决方案)
jQuery.post第二个参数是附加参数，第三个参数才是回调吧
$.post("http://hys2013.sinaapp.com/SQL.php" , '', function(data){ alert("Data Loaded: " + data); });
jQuery.post第二个参数是附加参数，第三个参数才是回调吧
$.post("http://hys2013.sinaapp.com/SQL.php" , '', function(data){ alert("Data Loaded: " + data); });
刚才看了一下的确是的，但是改了后怎么还不行啊
语法
jQuery.post(url,data,success(data, textStatus, jqXHR),dataType)
参数 描述
url 必需。规定把请求发送到哪个 URL。
data 可选。映射或字符串值。规定连同请求发送到服务器的数据。
success(data, textStatus, jqXHR) 可选。请求成功时执行的回调函数。
dataType
可选。规定预期的服务器响应的数据类型。
默认执行智能判断（xml、json、script 或 html）。
第 8 行以后不要！
改作 echo $row['id'];
语法
jQuery.post(url,data,success(data, textStatus, jqXHR),dataType)
参数 描述
url 必需。规定把请求发送到哪个 URL。
data 可选。映射或字符串值。规定连同请求发送到服务器的数据。
success(data, textStatus, jqXHR) 可选。请求成功时执行的回调函数。
dataType
可选。规定预期的服务器响应的数据类型。
默认执行智能判断（xml、json、script 或 html）。
第 8 行以后不要！
改作 echo $row['id'];
还是不行啊, .php页面能正常返回值,问题应该就是在.html页面吧，是不是我上面写的这个$.post函数格式不对（根据一楼的改了）
你原来写的，和#1教你写的都一样。都是正确的写法之一
你可以先不要在 SQL.php 做任何事情，而只是
echo ‘ok’;
这时应弹出一个 ok 对话框
你原来写的，和#1教你写的都一样。都是正确的写法之一
你可以先不要在 SQL.php 做任何事情，而只是
echo ‘ok’;
这时应弹出一个 ok 对话框
还是不行啊斑竹
我的PC端不需要安装相关组件吧，我是在dreamweaver下写的
请你用浏览器执行
请你用浏览器执行
我搜狗跟IE都试了
请你用浏览器执行
服务器使用的是新浪SAE
请你用浏览器执行
请你用浏览器执行
请你用浏览器执行
额，找到问题了，实在不好意思，我没下载JQUERY包
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