Question

美好的一天。我真的遇到了大麻烦，因为我无法弄清楚我的陈述有什么问题。它如下：

$sumitem = mysqli_query($conn,"SELECT SUM(AMOUNT) AS TOTAL FROM sum_query WHERE MAIN_ID = 1 GROUP BY ITEM_ID");
$sum = mysqli_fetch_assoc($sumitem);

我正在尝试打印$ sum [＆＃39; TOTAL＆＃39;]但它只显示第一列。以下是我的样本表：

MAIN_ID | ITEM_ID | AMOUNT |  DESC  | OTHER_DETAILS 
   1    |     1   |  500   | Item 1 |  a 
   1    |     2   |  5000  | Item 2 |  a 
   1    |     2   |  5000  | Item 2 |  b

我想要的结果应该是5500，但它只返回500。非常感谢你。

Answer 1

如果您希望ans为6500，请删除group by子句

SELECT SUM(AMOUNT) AS TOTAL FROM sum_query WHERE MAIN_ID = 1

Answer 2

SELECT SUM(g.AMOUNT)  AS TOTAL 
    FROM sum_query g
   WHERE g.MAIN_ID = 1 
GROUP BY g.ITEM_ID

此查询正在尝试试用此

Answer 3

你有2条记录第2项。我不知道你为什么要保留＆a;和b＆＃39;在OTHER_DETAILS。

您的要求不明确

如果您只想拍摄与OTHER_DETAILS＆＃39; a＆＃39;匹配的记录，请使用以下查询

SELECT SUM(AMOUNT) AS TOTAL FROM sum_query WHERE MAIN_ID = 1 and OTHER_DETAILS = 'a'

与WHERE和GROUP BY CLAUSE相加

3 个答案: