Ajax+php无限联动下拉菜单实例
首先是 Ajax.php文件:
这个文件我觉着就是接收数据处理数据的
代码如下 | 复制代码 |
mysql_connect("localhost","root",""); |
上面的这些代码 不用我说都知道是连接数据库的
代码如下 | 复制代码 |
//select 语句 $q=mysql_query($sql); |
上面的1和2的选项是因为 我写了一遍select语句出现报错了 然后我就又写了一遍 结果两个都对了 1 是注释掉了
if(mysql_num_rows($q)!=0){ 判断查找的语句的个数 如果是0的话就代表下面没有分支了 就不会显示了
//记住在$_POST[]加()这是我出现的错误
代码如下 | 复制代码 |
echo " |
输出一个select选择框会添加到后来最终显示的页面的div里面 后面会做介绍
代码如下 | 复制代码 |
while($rs=mysql_fetch_array($q)){ } ?> |
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下面的是Ajax.js文件var xmlhttp;定义一个变量
代码如下 | 复制代码 |
function createxml(){这个部分主要是用来判断浏览器 |
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下面是展示页面liandong.php(原谅我起名字的时候都是很简单的思维)
代码如下 | 复制代码 |
mysql_connect("localhost","root","")or die("链接数据库失败"); ?> |
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随意转载^^但请附上教程地址。

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