php简单实现查询数据库返回json数据(返回json数据函数实例)

输出结果：

{"code":1,"message":"\u6570\u636e\u8fd4\u56de\u6210\u529f","data":[{"id":"613","title":"\u81ea\u8003","admin_name":"admin","flid":"1","pid":null,"uid":"0","tid":null,"time":"2017-05-15 17:15:27"},{"id":"614","title":"\u4eba\u529b","admin_name":"admin","flid":"1","pid":null,"uid":"0","tid":null,"time":"2017-05-15 17:24:11"},{"id":"615","title":"\u6559\u5e08\u8d44\u683c\u8bc1","admin_name":"6293092","flid":"1","pid":null,"uid":"0","tid":null,"time":"2017-05-18 10:52:45"}]}

代码：

$code,  
              'message'=>$message,  
              'data'=>$data   
            );  
            //输出json  
            echo json_encode($result);  
            exit;  
}  

	
	
    $sqla = "SELECT * from data_type";  
    $result = mysqli_query($conn, $sqla);
    $dataarr = array();  
    //while($row = mysql_fetch_array($result)){  
	while($row = mysqli_fetch_assoc($result)) {
        $dataarr[]=$row;  
    }  
 echo json(1,'数据返回成功',$dataarr);  
 
//    if($id==1){  
//       echo json(1,'数据返回成功',$dataarr);  
//    }else if($id==2){  
//        echo json(0,'失败');  
//    } 

?>