Question

我正在尝试从数据库中提取所有客户信息。 我需要为用户获取所有地址。

我有一个用户表和地址表。地址表为每个用户保存0个地址。我需要加入现有查询的每个用户的所有地址，我不介意每个地址重复的用户信息。

到目前为止，我有这个查询，它返回所有客户数据，与区域表一起获取货币名称，然后与订单联系以获得该用户的已发货订单总数。

SELECT ( CASE
       WHEN u.password IS NULL THEN 'GUEST'
       ELSE 'CUSTOMER'
     END )                                 AS STATUS,
   u.date_created                          AS DateCreated,
   u.NAME                                  AS UserName,
   u.password                              AS Password,
   u.email                                 AS Email,
   r.token                                 AS Currency,
   Cast(u.balance / 100 AS DECIMAL(10, 2)) AS Balance,
   Count(o.user_id)                        AS TotalShippedOrders
FROM   [db].[user] u
   INNER JOIN [db].[region] r
           ON r.currency_id = u.balance_currency
   LEFT JOIN [db].[order] o
          ON o.user_id = u.id
             AND o.status = 'shipped'
 GROUP  BY u.id,
      u.date_created,
      u.NAME,
      u.password,
      u.email,
      r.token,
      u.balance
ORDER  BY TotalShippedOrders DESC;

我试图LEFT JOIN地址表，但它通过TotalShippedOrders计数。

SELECT ( CASE
       WHEN u.password IS NULL THEN 'GUEST'
       ELSE 'CUSTOMER'
     END )                                 AS STATUS,
   u.date_created                          AS DateCreated,
   u.NAME                                  AS UserName,
   u.password                              AS Password,
   u.email                                 AS Email,
   r.token                                 AS Currency,
   Cast(u.balance / 100 AS DECIMAL(10, 2)) AS Balance,
   Count(o.user_id)                        AS TotalShippedOrders, 
   a.*
FROM   [db].[user] u
   INNER JOIN [db].[region] r
           ON r.currency_id = u.balance_currency
   LEFT JOIN [db].[order] o
          ON o.user_id = u.id
             AND o.status = 'shipped'
   LEFT JOIN [db].[address] a
          ON a.user_id = u.id
 GROUP  BY u.id,
      u.date_created,
      u.NAME,
      u.password,
      u.email,
      r.token,
      u.balance
ORDER  BY TotalShippedOrders DESC;

有人能指出我如何制定我的查询的这一部分吗？谢谢！

编辑：我在计数中添加了DISTINCT，这似乎解决了问题，但我还没想到它还在返回所有用户/地址组合。这个简单的查询返回的行数多于上面的行：

select * from [db].[user] u
LEFT JOIN [db].[address] a ON a.user_id = u.id 
order by u.id

我希望每行都能看到用户信息和一个地址。如果用户有多个地址，则会有许多行具有相同的用户信息但地址不同。有些用户没有地址，这就是我进行LEFT JOIN的原因。

有人能指出我查询逻辑出错的地方吗？谢谢！

Answer 1

您的问题表明每个用户有多个地址。也许你只想要一个地址，比如最近的地址。如果是这样，这样的事情应该有效：

SELECT . . .
FROM [db].[user] u INNER JOIN
     [db].[region] r
     ON r.currency_id = u.balance_currency LEFT JOIN
     [db].[order] o
     ON o.user_id = u.id AND o.status = 'shipped' LEFT JOIN
     (SELECT a.*,
             ROW_NUMBER() OVER (PARTITION BY a.user_id ORDER BY a.id DESC) as seqnum
      FROM [db].[address] a
     ) a
     ON a.user_id = u.id AND a.seqnum = 1
. . .

它使用地址表中id最大的地址。

编辑：

如果您需要所有地址，则需要在聚合后进行连接：

with t as (
      <your query here with "u.id as user_id" in the select
     )
select t.*, a.*
from t left join
     addresses a
     on t.user_id = a.user_id;

需要为用户获取所有地址。已经有查询，但无法正确连接地址

1 个答案: