Question

源表示例

Table1
id      id_parent   product_id    
===     =========   ========    
154     104389        1043891       
155     104389        1995137331

然后我有两个其他库存表如下：

(table2)
product_id   location_id       on_hand     transit
1043891          2           12      1
1043891          5           33      0
1043891          6           7       0
1043891          4           3       0
1995137331       2           8       5
1995137331       8           50      9
1995137331       5           95      0
1995137331       11          22      5
1995137331       13          93      1

(table3)
product_id   location_id    on_hand      transit
1043891           1         25       1
1995137331        1         29       0

我有以下查询：

; with 
O as (select Product_Id, SUM(OnHand) AS AllStore from Table2 group by Product_Id),
S as (select Product_Id, SUM(OnHand) as WH from Table3 group by Product_Id)

SELECT
   L1.id_parent as parent
  ,L1.product_id as child
  , O.AllStore
  , S.WH
FROM Table1 L0 JOIN Table1 L1 ON L0.id_parent = L1.id_parent 

LEFT OUTER JOIN S on S.Product_ID = L0.Product_Id
LEFT OUTER JOIN O on O.Product_ID = L0.Product_ID
group by L1.id_parent, L1.product_id, S.WH, O.AllStore

它有点工作，但重复记录。目前的结果：

parent  child       AllStore    WH
1043892 1043891     104         102
1043892 1043891     242         123
1043892 1995137331  104         102
1043892 1995137331  242         123

预期结果：

parent  child       AllStore    WH
1043892 1043891     104         102
1043892 1995137331  242         123

有人可以看看它，看看我做错了什么，或者是否有更好的方法。

感谢。

Answer 1

问题来自于您将Table1加入到

的条件下

L0.id_parent = L1.id_parent

这种连接有效地将表1中的2行104389与2行连接起来。此连接条件将产生4行。

SELECT
   *
FROM 
    @Table1 L0
    INNER JOIN @Table1 L1 ON
        L0.id_parent = L1.id_parent

id          id_parent   product_id  id          id_parent   product_id
----------- ----------- ----------- ----------- ----------- -----------
154         104389      1043891     154         104389      1043891
154         104389      1043891     155         104389      1995137331
155         104389      1995137331  154         104389      1043891
155         104389      1995137331  155         104389      1995137331

根据您提供的示例，此连接完全没必要，因为您需要的数据已经在L0

中

由于您没有运行任何聚合，因此group by语句也是不必要的。将其保留的最终结果与在整个查询中放置DISTINCT相同。

因此：产生所描述的“预期结果” 查询可以修改为以下

; with 
O as (select Product_Id, SUM(on_hand) AS AllStore from @Table2 group by Product_Id),
S as (select Product_Id, SUM(on_hand) as WH from @Table3 group by Product_Id)

SELECT
   L0.id_parent as parent
  ,L0.product_id as child
  ,O.AllStore
  ,S.WH
FROM
    @Table1 L0

    LEFT OUTER JOIN S ON
        S.Product_ID = L0.Product_Id

    LEFT OUTER JOIN O ON
        O.Product_ID = L0.Product_ID

注意：如果L0和L1连接是拼写错误，并且有第4个表正在播放但仍然产生重复，则可以将L0和L1连接提取到子查询中并对其进行DISTINCT。

聚合问题具有连接和父子级别的mssql

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