为什么我的 Ajax 发布脚本不起作用？
我几乎没有创建一个脚本（因为我只是 ajax 的初学者），事实上我已经从某个地方复制了它的某些部分（ajax 脚本）。但是，当我单击链接（投票赞成/投票反对）时，没有任何反应，甚至 Mysql 数据库中的值也没有更改，但是单击提交按钮时，我的 MySql 数据库中发生了更改！这是我的代码 --
-:::- HTML PART (test.php) -:::-
<html>
<title>
TEST
</title>
<head>
<script type="text/javascript" src="jquery.1.4.4.js"></script>
<script type="text/javascript">
function vote(type)
{
$.ajax({
'url': 'test.func.php',
'type': 'POST',
'dataType': 'json',
'data': {type: type},
'success': function(data)
{
if(data.status)
{
if(data.voted)
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","You have voted up!");
});
}
else
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","You have voted Down!");
});
}
}
},
beforeSend: function()
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","Voting....");
});
},
'error': function(data)
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","An error occureed");
});
}
});
}
</script>
</head>
<body>
<a href="#" onclick="vote('up')" > Vote Up </a>
<span id="status_up" ></span>
<br>
OR
</br>
<a href="#" onclick="vote('down')" > Vote Down </a>
<span id="status_down" ></span>
</body>
</html>
-:::- PHP PART (test.func.php) -:: :-
<?php
function db_connect($i)
{
if(isset($i))
{
if(mysql_connect('localhost', 'root', 'root'))
{
if(mysql_select_db($i))
{
return;
}
else
{
echo 'ERROR';
}
}
else
{
echo 'ERROR';
}
}
else
{
echo 'ERROR';
}
}
if($_POST)
{
db_connect('tests');
$vote_type=$_POST['type'];
$post_id = '123';
$query = mysql_query("SELECT * FROM test WHERE post_id='$post_id'");
$cur_vote_get = mysql_fetch_array($query);
$vote_up = $cur_vote_get['votes']+1;
$vote_down = $cur_vote_get['votes']-1;
if($vote_type=='up')
{
mysql_query("UPDATE test SET votes='$vote_up' WHERE post_id='$post_id'");
return json_encode(array("status" => true, "voted" => true));
}
elseif($vote_type=='down')
{
mysql_query("UPDATE test SET votes='$vote_down' WHERE post_id='$post_id'");
return json_encode(array("status" => true, "voted" => false));
}
}
?>
JavaScript 错误已解决！
一切都已解决！
由于我对 Ajax 很陌生，所以我无法找到任何解决方案。
I have barely created a script (as I'm just a beginner in ajax), in fact I have copied some part (ajax script) of it from somewhere. But when I click on the the link (vote up / vote down) nothing happens not even the value in Mysql database changes, but on click the submit button I got a change in my MySql Database! Here is my code --
-:::- HTML PART (test.php) -:::-
<html>
<title>
TEST
</title>
<head>
<script type="text/javascript" src="jquery.1.4.4.js"></script>
<script type="text/javascript">
function vote(type)
{
$.ajax({
'url': 'test.func.php',
'type': 'POST',
'dataType': 'json',
'data': {type: type},
'success': function(data)
{
if(data.status)
{
if(data.voted)
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","You have voted up!");
});
}
else
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","You have voted Down!");
});
}
}
},
beforeSend: function()
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","Voting....");
});
},
'error': function(data)
{
$(document).ready(function () {
$("span#status"+type).attr("innerHTML","An error occureed");
});
}
});
}
</script>
</head>
<body>
<a href="#" onclick="vote('up')" > Vote Up </a>
<span id="status_up" ></span>
<br>
OR
</br>
<a href="#" onclick="vote('down')" > Vote Down </a>
<span id="status_down" ></span>
</body>
</html>
-:::- PHP PART (test.func.php) -:::-
<?php
function db_connect($i)
{
if(isset($i))
{
if(mysql_connect('localhost', 'root', 'root'))
{
if(mysql_select_db($i))
{
return;
}
else
{
echo 'ERROR';
}
}
else
{
echo 'ERROR';
}
}
else
{
echo 'ERROR';
}
}
if($_POST)
{
db_connect('tests');
$vote_type=$_POST['type'];
$post_id = '123';
$query = mysql_query("SELECT * FROM test WHERE post_id='$post_id'");
$cur_vote_get = mysql_fetch_array($query);
$vote_up = $cur_vote_get['votes']+1;
$vote_down = $cur_vote_get['votes']-1;
if($vote_type=='up')
{
mysql_query("UPDATE test SET votes='$vote_up' WHERE post_id='$post_id'");
return json_encode(array("status" => true, "voted" => true));
}
elseif($vote_type=='down')
{
mysql_query("UPDATE test SET votes='$vote_down' WHERE post_id='$post_id'");
return json_encode(array("status" => true, "voted" => false));
}
}
?>
JavaScript Error solved!
Everything Solved!
As I'm very new to Ajax so I'm not able to find any solution to this.
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评论（2）
如果这完全是您的代码（我的意思是，如果您只是从真实的 .php 文件中复制/粘贴它），则第一个 .php 文档的第二个脚本块中存在拼写错误。
javasript-> javasCript
尝试一下。
If that's STRICTLY your code, (I mean, if you've just literally copy/pasted it from your real .php file), there's a typo error in the second script block of the first .php document.
javasript -> javasCript
Give it a try.
您不是回显 json，而是回显随机文本（“投票”），这会破坏 ajax json 解析器。
删除
echo 'Voted!';并尝试echo json_encode(array("status" => true, "voted" => false));。您也可以跳过
SELECT查询并执行：UPDATE test SET votes=votes+1 WHERE post_id='$post_id'You're not echoing the json, you're echoing random text ('Vote up'), which will break the ajax json parser.
Remove the
echo 'Voted!';and tryecho json_encode(array("status" => true, "voted" => false));.Also you can skip the
SELECTquery and do :UPDATE test SET votes=votes+1 WHERE post_id='$post_id'