Ajax选择框没有填充
[英]Ajax selectbox not populating
问题描述
This is pretty much one of my first ajax scripts so please keep that in mind as I am very much still learning... 
这几乎是我的第一个ajax脚本之一，所以请记住这一点，因为我还在学习......
Im trying to do the following: 我试着做以下事情：
Display results of a certain tournament by allowing user to do the following: 通过允许用户执行以下操作来显示某个锦标赛的结果：
- user selects sport 用户选择运动
- now select tournament based on sport_type 现在根据sport_type选择锦标赛
- now select round based on tournament and sport_type 现在根据锦标赛和sport_type选择回合
My problem 我的问题
The 1st select box (Sport) populated perfectly yet the other select boxes does not populate...any im getting no error messages... 第一个选择框（运动）完全填充，其他选择框不填充...任何我没有得到错误消息...
My Code 我的守则
result.php result.php
$(document).ready(function()
{
$(".sport").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_sport.php",
data: dataString,
cache: false,
success: function(html)
{
$(".tournament").html(html);
}
});
});
$(".tournament").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_round.php",
data: dataString,
cache: false,
success: function(html)
{
$(".round").html(html);
}
});
});
});
</head>
<body>
<center>
<div>
<label>Sport :</label>
<select name="sport" class="sport">
<option selected="selected">--Select Sport--</option>
<?php
var_dump($id);
$sql="SELECT distinct sport_type FROM events";
$result=mysql_query($sql);
while($row=mysql_fetch_array($result))
{
?>
<option value="<?php echo $row['sport_type']; ?>"><?php echo $row['sport_type']; ?></option>
<?php
}
?>
</select>
<label>Tournamet :</label> <select name="tournament" class="tournament">
<option selected="selected">--Select Tournament--</option>
</select>
<label>Round :</label> <select name="round" class="round">
<option selected="selected">--Select Round--</option>
</select>
get_sport.php get_sport.php
include("connect.php");
if($_POST['id'])
{
$id=$_POST['id'];
$sql="SELECT distinct tournament FROM events WHERE sport_type ='$id'";
$result=mysql_query($sql);
?>
<option selected="selected">Select Tournament</option><?php
while($row=mysql_fetch_array($sql)){
?>
<option value="<?php echo $row['tournament'];?>"><?php echo $row['tournament']?></option>
<?php
}
}
get_round.php get_round.php
include('connect.php');
if($_POST['id'])
{
$id=$_POST['id'];
$sql="SELECT DISTINCT round FROM events WHERE tournament='$id'";
$result=mysql_query($sql);
?>
<option selected="selected">Select Round</option><?php
while($row=mysql_fetch_array($result)){
?>
<option value="<?php echo $row['round'] ?>"><?php echo $row['round'] ?></option>
<?php
}
}
?>
Im sure I just forgot to add a statment or something similar Ive been staring at this for the best part of an hour, yet I can not find the error. 我确定我只是忘了添加一个或类似的东西，我已经盯着这一小时的最佳时间，但我找不到错误。 Any help will be greatly appreciated 任何帮助将不胜感激
5 个解决方案
解决方案1
4 2015-06-29 06:33:55
Try dataType : 'html' : 试试dataType : 'html' ：
var dataString = {id:id};
$.ajax({
type : 'POST',
url: "get_sport.php",
dataType : 'html',
data: dataString,
cache: false,
success: function(html)
{
$(".tournament").html(html);
}
});
解决方案2
2 已采纳 2015-06-29 06:47:51
问题是在while循环中的get_sport.php mysql_fetch_array($sql)参数应该是$result ....多么尴尬
解决方案3
2 2015-06-29 07:06:31
Try this by replacing, 通过替换来试试这个，
$(".tournament").change(function()
to 至
$(document).on('change','.tournament',function()
and int get_sport.php replace, 和int get_sport.php替换，
while($row=mysql_fetch_array($sql))
to 至
while($row=mysql_fetch_array($result))
解决方案4
0 2015-06-29 06:36:36
I can't find the problem in your code but the best thing you can do its learn hot to spot this kind of problems. 我无法在你的代码中找到问题，但你可以做的最好的事情是学习热点来发现这类问题。
You can do a few things: 你可以做一些事情：
- Use alerts on Jquery to see if your functions are been called. 在Jquery上使用警报来查看是否已调用您的函数。
- You can debug the PHP or use some Echo or something like that to see if your php pages are been called, or use a tool as fiddler. 您可以调试PHP或使用一些Echo或类似的东西来查看您的php页面是否被调用，或使用工具作为fiddler。
- Look at network in Chrome to see what the php return. 看看Chrome中的网络，看看php返回了什么。
解决方案5
0 2015-06-29 06:59:01
Change the 改变
while($row=mysql_fetch_array($sql)){...
in get_sport.php to 在get_sport.php中
while($row=mysql_fetch_array($result)){...
since you used 因为你用过
$result=mysql_query($sql);
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