SQL查询值未传递给PHP对象

时间:2014-08-13 12:37:25

标签: php mysql

我是从SQL查询创建对象的。

$result = mysqli_query($dbconnect,"SELECT SUM(value) AS total FROM `sales` WHERE `tdate` = CURDATE();"); 

我认为它应该将[' value]或[' total']传递给对象,但它似乎只会返回

{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null} 

这让我觉得查询本身返回NULL,因为没有找到结果,但是如果我在DB中运行查询,它会正确返回。有什么想法吗?

1 个答案:

答案 0 :(得分:2)

您忘记了fetch结果,请参阅下面的示例代码

$result = mysqli_query($dbconnect,"SELECT SUM(value) AS total FROM `sales` WHERE `tdate` = CURDATE();"); 
$data = mysqli_fetch_assoc($result);
var_dump($data);