Question

select * from order
-------------------
|orderID|productID|
-------------------
|  1    |  234    |
|  2    |  234    |
|  3    |  123    |
-------------------

select * from product_supplier
-------------------------------------------
|ID|supplierID|productID|cost_price|latest|
-------------------------------------------
|1 |  1       | 234     | 1.00     | 0    |
|2 |  1       | 234     | 0.50     | 1    |
|3 |  2       | 123     | 0.75     | 1    |
-------------------------------------------


desired result
------------------------------
|orderID|productID|cost_price|
------------------------------
| 1     | 234     | 1.00     |
| 2     | 234     | 1.00     |
| 3     | 123     | 0.75     |
------------------------------

我正在查看上面的两个表格，以获得给定orderID的{{1}}，productID和最大cost_price。

productID

给出SELECT orderID, productID, cost_price FROM order LEFT JOIN product_supplier ON order.productID=product_supplier.productID AND MAX(cost_price);

如何将ERROR 1111 (HY000): Invalid use of group function ed表限制为每join行一行，加上order中对应的最大cost_price值？

Answer 1

接近此方法的规范方法是使用子查询从product_supplier表中标识产品及其最高价格，然后将此子查询加入order以获取所需的结果集

SELECT t1.orderID,
       t1.productID,
       COALESCE(t2.cost_price, 0.0) AS cost_price  -- missing products will appear
FROM order t1                                      -- with a zero price
LEFT JOIN
(
    SELECT productID, MAX(cost_price) AS cost_price
    FROM product_supplier
    GROUP BY productID
) t2
    ON t1.productID  = t2.productID AND
       t1.cost_price = t2.cost_price

Answer 2

当您需要使用聚合时，您必须使用group by和compare condtion使用拥有

new Promise((resolve, reject) => {
  const a = 1;
  resolve(Promise.all([
    Promise.resolve("a"),
    Promise.resolve("b"),
    a,
  ]))
})
.then(([resultFromPromise_1, resultFromPromise_2, a]) => {
   console.log(a);
})

从另一个表中加入具有MAX值的表

2 个答案: