[PHP知识库]ThinkPHP接收MUI的ajax的请求

参考：https://blog.csdn.net/qq_30378229/article/details/79433214

1、提交ajax请求

function validate_form(thisform)
		{
			var username = $("#username").val();
			if(username == null || username == ""){
				 mui.alert('请输入姓名');
				 return false;
			}
			
			var mobile = $("#mobile").val();
			if(mobile == null || mobile == ""){
				 mui.alert('请填写手机号码');
				 return false;
			}else if(mobile.length!=11){
				 mui.alert('请认真填写手机号码');
				 return false;
			}
	      
			mui.ajax('/home/agent/addsubmit',{
				data:{
					username:username,
					mobile:mobile
				},
				dataType:'json',
				type:'post',
				timeout:10000,
				headers:{'Content-Type':'application/json'},	              
				success:function(data){
					if(data.code == 0) {
						mui.alert("提交成功");
					}else{
						mui.alert("提交失败");
					}					
				},
				error:function(xhr,type,errorThrown){
					//异常处理；
					console.log(type);
				}
			});
		   
			return false;
		}

2、thinkphp 6 获取ajax请求参数值

//加入代理 提交
	public function addsubmit()
	{
	  //获取post方式提交的表单数据
      //$username = $_POST['username'];	//以通常的$_POST方式获取不到表单数据	
		$input = file_get_contents('php://input'); 
		$object_input = json_decode($input); 
		$username = $object_input->username; 
		$mobile = $object_input->mobile; 
		
		$obj= new \stdClass();
		$obj->code = 0;
		
		$obj->message = 'OK';
		
		echo json_encode($obj);
	}