Php MySQL语句连接两个表并求和相同的项目数量

我有两张桌子，客户和交易

Customer:
--------------------------------------------
ID | Name            | Tel
--------------------------------------------
1  | Peter           | 123 4567
2  | John            | 456 1234
3  | Alice           | 789 4561
4  | Amy             | 741 8525

Transaction:
--------------------------------------------
CustID | Books | Pens  | Ruler
--------------------------------------------
  1    |   2   |  0    |   1
  2    |   1   |  0    |   0
  1    |   0   |  3    |   0
  1    |   0   |  0    |   1
  2    |   1   |  1    |   1
  3    |   0   |  2    |   2

我需要以下信息

Results:
-------------------------------------------------------------------
ID | Name          | Tel            | Books | Pens  | Ruler
-------------------------------------------------------------------
1  | Peter         | 123 4567       |   2   |   3   |   2
2  | John          | 456 1234       |   2   |   1   |   1
3  | Alice         | 789 4561       |   0   |   2   |   2
4  | Amy           | 741 8525       |   0   |   0   |   0

基本上，它将汇总同一客户的书籍、钢笔和尺子

我试过：

                        $sql = "select 
                                    `customer`.id,
                                    `custmaster`.name,
                                    `custmaster`.tel,
                                    `transaction`.id,
                                    `transaction`.books,
                                    `transaction`.pens,
                                    `transaction`.ruler,
                                from `customer` 
                                left join `transaction` 
                                on `customer`.id=`transaction`.custid 
                                ORDER BY `customer`.id ASC";

---

但是没有显示（我知道我在什么地方需要sum（）函数。任何人都可以帮忙吗？

使用

sum

和

分组方式

SELECT c.id, c.name, c.tel, SUM(t.books) as books, SUM(t.pens) AS pens, SUM(t.ruler) AS ruler
FROM customer AS c
LEFT JOIN transactions AS t ON c.id = t.custid
GROUP BY c.id
ORDER BY c.id

使用
SUM
和
GROUP BY

SELECT c.id, c.name, c.tel, SUM(t.books) as books, SUM(t.pens) AS pens, SUM(t.ruler) AS ruler
FROM customer AS c
LEFT JOIN transactions AS t ON c.id = t.custid
GROUP BY c.id
ORDER BY c.id

这样试试

select 
    `customer`.id,
    `custmaster`.name,
    `custmaster`.tel,
    `transaction`.id,
    sum(`transaction`.books) as books,
    sum(`transaction`.pens) as pens,
    sum(`transaction`.ruler) as ruler,
from `customer` 
left join `transaction` 
on `customer`.id=`transaction`.custid
Group by `customer`.id,`customer`.Name
ORDER BY `customer`.id ASC";

这样试试

select 
    `customer`.id,
    `custmaster`.name,
    `custmaster`.tel,
    `transaction`.id,
    sum(`transaction`.books) as books,
    sum(`transaction`.pens) as pens,
    sum(`transaction`.ruler) as ruler,
from `customer` 
left join `transaction` 
on `customer`.id=`transaction`.custid
Group by `customer`.id,`customer`.Name
ORDER BY `customer`.id ASC";

很好…为便于参考而设置表名快捷方式。很好…为便于参考而设置表名快捷方式。我需要2个分组依据吗？还是只需要按
customer
进行分组。id就可以了。我对这两者进行了某种测试，两者都给出相同的结果。选择
transaction.id
，没有意义，因为您正在聚合所有事务。@SykesTangSQL要求您必须在
GROUP BY
子句中列出所有未聚合的列。但是MySQL允许您省略它们，因此您只需要定义组的列。我需要2个GROUP BY吗？或者我可以只按
customer
进行分组。id就可以了。我会对这两个列进行排序测试，两者都会给出相同的结果。选择
没有意义>transaction.id
，因为您正在聚合所有事务。@SykesTang Strict SQL说您必须在
GROUP BY
子句中列出所有未聚合的列。但是MySQL允许您将它们省略，所以您只需要定义组的列。