Question

怎么用Thinkphp+jquery实现ajax的提交，接收处理和返回

Answer 1

<span class="jijin_btn" id="upgradebrokers">我要升级</span></li>

<script>
//我要升级
$(function(){
$("#upgradebrokers").click(function(){
var ajaxurl="{:U('Upgradebroker/upgrade')}";//ajax提交路径 提交给哪个php文件里的哪个方法 这里是提交给upgrade方法去处理
$.ajax({
type:"post",
url:ajaxurl,
//async:true,
data:{},
dataType:"json",
success:function(data){
if (data.status == 1) {

//alert(data.msg);
$(".main-bg").show();
$(".jinji_fnBox2").show();

}else if(data.status == 2){
// alert(data.msg);
$(".main-bg").show();
$(".jinji_fnBox1").show();

}else if(data.status == 0){
alert(data.msg);
//window.location.reload();
}else{
alert(data.msg);
//window.location.reload();
}
}
});
});
});
</script>

php里面upgrade方法
public function upgrade(){

//数据处理
.....
...
...

//判断结果

if(!$res){
$lastid=$Upgrade->add($updata);

if($lastid)
$this->ajaxReturn(array('status'=>1,'msg'=>'申请成功'));
else
$this->ajaxReturn(array('status'=>-1,'msg'=>'申请失败'));
}else{

$this->ajaxReturn(array('status'=>0,'msg'=>'您已申请过了，请等待审核...'));

}

}

Answer 2

首先，需要引入Jquery包
其次，需要写如下格式:(举个例子)
<script type="text/javascript">

<script>
$("#btn").click(function(){
$.ajax({
'url':"demo3.php",
'data':{'username':'科学家','age':18},
'success':function(msg){
$('#span').html(msg);
},
'dataType':'html',
'type':'post',
});
});
</script>
后台是这样写的
<?php

echo "我要做个科学家";

echo "<pre>";
var_dump($_POST);
echo "</pre>";