Question

我是MYSQL的新手，并且正在努力形成一个连接两个表并返回唯一产品行的查询。

表 product：

╔══════════════════════════════╦══════════╦════════════╦════════════════════╗
║             ref              ║  brand   ║    mpn     ║        sku         ║
╠══════════════════════════════╬══════════╬════════════╬════════════════════╣
║ 0001___DOGICLON___912-101242 ║ DOGICLON ║ 912-101242 ║ 000000000001082649 ║
║ 0002___DOGICLON___912-101242 ║ DOGICLON ║ 912-101242 ║ 912-101242         ║
║ 0003___Dogiclon___912-101242 ║ Dogiclon ║ 912-101242 ║ 912-101242(R400)   ║
║ 0005___Dogiclon___912-101242 ║ Dogiclon ║ 912-101242 ║ MILT-R400          ║
╚══════════════════════════════╩══════════╩════════════╩════════════════════╝

表 inventory：

╔══════════════════════════════╦═══════╦═════════╦══════════╗
║             ref              ║ scost ║ instock ║ location ║
╠══════════════════════════════╬═══════╬═════════╬══════════╣
║ 0001___DOGICLON___912-101242 ║ 53.68 ║      24 ║ WA       ║
║ 0001___DOGICLON___912-101242 ║ 53.68 ║       0 ║ CA       ║
║ 0002___DOGICLON___912-101242 ║ 61.00 ║     121 ║ WA       ║
║ 0003___Dogiclon___912-101242 ║ 53.53 ║     100 ║ WA       ║
║ 0003___Dogiclon___912-101242 ║ 53.53 ║       0 ║ NY       ║
║ 0003___Dogiclon___912-101242 ║ 53.53 ║      20 ║ MA       ║
║ 0003___Dogiclon___912-101242 ║ 53.53 ║       2 ║ CA       ║
║ 0005___Dogiclon___912-101242 ║ 56.00 ║       5 ║ IN       ║
║ 0005___Dogiclon___912-101242 ║ 56.00 ║       5 ║ MA       ║
║ 0005___Dogiclon___912-101242 ║ 56.00 ║       5 ║ WA       ║
║ 0005___Dogiclon___912-101242 ║ 56.00 ║       5 ║ NY       ║
║ 0005___Dogiclon___912-101242 ║ 56.00 ║       2 ║ CA       ║
╚══════════════════════════════╩═══════╩═════════╩══════════╝

我猜pseduo代码将是：

SHOW all products
WHERE 
    instock (any location) > 0 AND
    (cost > 10 AND cost < 2000)
ORDER BY
    cost asc

备注：

ref是每个供应商的唯一商品
brand和mpn查找需要不区分大小写

预期结果：

╔══════════╦══════════╦════════════╦══════════════╦══════════════╦═══════════════════════════╗
║   ref    ║  brand   ║    mpn     ║     sku      ║    scost     ║          instock          ║
╠══════════╬══════════╬════════════╬══════════════╬══════════════╬═══════════════════════════╣
║ whatever ║ Dogiclon ║ 912-101242 ║ based on ref ║ based on ref ║ based on ref and location ║
╚══════════╩══════════╩════════════╩══════════════╩══════════════╩═══════════════════════════╝

这就是我正在尝试的：

SELECT DISTINCT
    product.ref,
    product.brand,
    inventory.scost,
    inventory.instock
FROM
    product
    JOIN inventory ON inventory.ref = product.ref
WHERE
    inventory.instock > 1 
    AND ( app.inventory.scost >= 10 AND app.inventory.scost <= 2000 ) 
GROUP BY
    product.ref

Answer 1

如果您按产品中的唯一列进行分组，似乎是ref，则distinct将不会按照您的意愿执行。 DISTINCT负责每个返回行的唯一性（返回所有值）。在这种情况下，为了获得每个产品参考结果的一行，您需要删除不同的部分，将product.mfr添加到group by并聚合非分组的列，如下所示：

SELECT
    product.ref,
    product.mfr,
    group_concat(inventory.scost) as scost,
    group_concat(inventory.instock) as instock,
FROM
    product
    JOIN inventory ON inventory.ref = product.ref
WHERE
    inventory.instock > 1 
    AND ( app.inventory.scost >= 10 AND app.inventory.scost <= 2000 ) 
GROUP BY
    product.ref, product.mfr

如果你想确保scost和instock的正确排序，只需在group_concat中包含ordered子句即：

group_concat(column order by column1, [...])

通过公共列

1 个答案: