Question

我有以下查询：

select transactions.timestamp, products.productHash
from transactions
inner join transactionItems on transactionItems.transactionHash = transactions.transactionHash
inner join products on products.productHash = transactionItems.productHash
where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
    and (products.type = 'applier' and products.gender = 'f')
    and products.additionalData->'$."skintone"' in ('-1', -1)
    and exists
    (
        select 1 from materialAppliers
        where materialAppliers.productHash = products.productHash
            and materialAppliers.applierType = 'skin'
    )
group by transactionItems.productHash

给我以下结果：

+---------------------+----------------------------------------+
|      timestamp      |              productHash               |
+---------------------+----------------------------------------+
| 2014-12-30 23:37:32 | cHJvZF81YWZhZmJkY2JiMzQ0OC4yMTg5MTY3OQ |
| 2014-11-30 19:17:47 | cHJvZF81YWZiZDVlOTQ3ZjM5Mi44NTc2Mjc0MQ |
+---------------------+----------------------------------------+

我必须与materialAppliers匹配的productHash加入：

+----------------------------------------+------------------------------------------+-------+
|              productHash               |               applierHash                | asset |
+----------------------------------------+------------------------------------------+-------+
| cHJvZF81YWZhZmJkY2JiMzQ0OC4yMTg5MTY3OQ | bWF0QXBwXzVhZmFmYmRjY2E1ZTExLjE3NDA3NjYx | val1  |
| cHJvZF81YWZhZmJkY2JiMzQ0OC4yMTg5MTY3OQ | bWF0QXBwXzVhZmFmYmRjY2Q1MmE0LjI3NTA4Nzcx | val2  |
| cHJvZF81YWZiZDVlOTQ3ZjM5Mi44NTc2Mjc0MQ | bWF0QXBwXzVhZmJkNWU5NDdmNWQ0LjU1MzQ1NDg5 | val3  |
| cHJvZF81YWZiZDVlOTQ3ZjM5Mi44NTc2Mjc0MQ | bWF0QXBwXzVhZmJkNWU5NmY2MTA0LjQyOTkxNzY5 | val4  |
| cHJvZF81YWZiZDVlOTQ3ZjM5Mi44NTc2Mjc0MQ | bWF0QXBwXzVhZmJkNWU5OTZlNGU3LjcxNTI1MDY1 | val5  |
+----------------------------------------+------------------------------------------+-------+

这样每个materialApplier都有自己的行，由timestamp组成，取自上一个查询;来自applierHash的{{1}}和asset：

materialAppliers

我该如何做到这一点？

Answer 1

只需将materialAppliers加入您的查询。

SELECT *
       FROM
(
select transactions.timestamp, products.productHash
from transactions
inner join transactionItems on transactionItems.transactionHash = tra nsactions.transactionHash
inner join products on products.productHash = transactionItems.productHash
where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
    and (products.type = 'applier' and products.gender = 'f')
    and products.additionalData->'$."skintone"' in ('-1', -1)
    and exists
    (
        select 1 from materialAppliers
        where materialAppliers.productHash = products.productHash
            and materialAppliers.applierType = 'skin'
    )
group by transactionItems.productHash
) x
LEFT JOIN materialAppliers
          ON materialAppliers.productHash = x.productHash;

但是列列中有一列，您没有GROUP BY，并且您没有应用任何聚合函数。虽然MySQL接受较低版本或某些设置并不是一件好事。我建议修理这个。使用例如max()获取最新的时间戳。

SELECT *
       FROM
(
select max(transactions.timestamp), products.productHash
from transactions
inner join transactionItems on transactionItems.transactionHash = tra nsactions.transactionHash
inner join products on products.productHash = transactionItems.productHash
where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
    and (products.type = 'applier' and products.gender = 'f')
    and products.additionalData->'$."skintone"' in ('-1', -1)
    and exists
    (
        select 1 from materialAppliers
        where materialAppliers.productHash = products.productHash
            and materialAppliers.applierType = 'skin'
    )
group by transactionItems.productHash
) x
LEFT JOIN materialAppliers
          ON materialAppliers.productHash = x.productHash;

Answer 2

看起来很简单：使用materialAppliers添加一个内部联接，然后在select ...和group中添加两个选中的值，我错过了什么？

select transactions.timestamp, products.productHash, ma.applierHash, ma.asset
from transactions
inner join transactionItems on transactionItems.transactionHash = transactions.transactionHash
inner join products on products.productHash = transactionItems.productHash
inner join materialAppliers ma on ma.productHash = products.productHash
where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
    and (products.type = 'applier' and products.gender = 'f')
    and products.additionalData->'$."skintone"' in ('-1', -1)
    and exists
    (
        select 1 from materialAppliers
        where materialAppliers.productHash = products.productHash
            and materialAppliers.applierType = 'skin'
    )
group by transactionItems.productHash

您应该能够摆脱存在，只需添加and ma.applierType = 'skin'即可。我说应该是因为我不完全理解applierType和producthash之间的关系，但它看起来应该有效。

如：

select transactions.timestamp, products.productHash, ma.applierHash, ma.asset
from transactions
inner join transactionItems on transactionItems.transactionHash = transactions.transactionHash
inner join products on products.productHash = transactionItems.productHash
inner join materialAppliers ma on ma.productHash = products.productHash
where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
  and products.type = 'applier' 
  and products.gender = 'f'
  and products.additionalData->'$."skintone"' in ('-1', -1)
  and ma.applierType = 'skin'

Answer 3

我会尝试为整个第一个查询设置别名，然后在之后将最终连接添加到materialAppliers表中

Select * from (select transactions.timestamp, products.productHash
    from transactions
    inner join transactionItems on transactionItems.transactionHash = transactions.transactionHash
    inner join products on products.productHash = transactionItems.productHash
    where userKey = '8eb6a8a9-4f1a-4402-89ab-4d1fbffd1284'
        and (products.type = 'applier' and products.gender = 'f')
        and products.additionalData->'$."skintone"' in ('-1', -1)
        and exists
        (
            select 1 from materialAppliers
            where materialAppliers.productHash = products.productHash
                and materialAppliers.applierType = 'skin'
        )
    group by transactionItems.productHash) as x

inner join materialAppliers as ma on x.producthash = ma.productHash

SQL连接允许给定列的多行

3 个答案: