mySQL查询结果以JSON格式返回

$sql="select JSON_OBJECT('dish_name',c.dish_name,'pics',,'product_price',a.product_price,'product_num',a.product_num) as dataJson from (`order_detail` a inner join `order_food` b on a.order_no=b.order_no) inner join `dishes` c on =a.product_id where a.order_no='202206130955455413269876';";

$result=mysqli_query($conn,$sql); //执行SQL语句
if($result && mysqli_num_rows($result) > 0){ //返回执行结果,如果执行成功,则返回true,反之,返回false;
	$i=1;
	$data='[';
	while($rs=mysqli_fetch_assoc($result)){
		if($i==1){
			$data.=$rs['dataJson'];
		}
		else{
			$data.=','.$rs['dataJson'];
		}
		$i++;
	}
	$data.=']';
}

mysqli_free_result($result);
mysqli_close($conn);

echo '{"code":0,"msg":"SUCCESS","data":'.$data.'}';

样例测试结果如图：