Sqlite 不同的和不不同的值

我有两张桌子、预订和文章：

预订

------------------------------
Id     |   Name   |     City |
------------------------------
 1     |  Mike    | Stockholm
 2     |  Daniel  | Gothenburg
 2     |  Daniel  | Gothenburg
 3     |  Andre   | Gothenburg (Majorna)

-------------------------------------------------------------
ArticleId    |    Name       |    Amount |    ReservationId | 
-------------------------------------------------------------
10           |   Coconuts    |    1      |    1         
10           |   Coconuts    |    4      |    2     
11           |   Apples      |    2      |    2
12           |   Oranges     |    2      |    3

文章

------------------------------
Id     |   Name   |     City |
------------------------------
 1     |  Mike    | Stockholm
 2     |  Daniel  | Gothenburg
 2     |  Daniel  | Gothenburg
 3     |  Andre   | Gothenburg (Majorna)

-------------------------------------------------------------
ArticleId    |    Name       |    Amount |    ReservationId | 
-------------------------------------------------------------
10           |   Coconuts    |    1      |    1         
10           |   Coconuts    |    4      |    2     
11           |   Apples      |    2      |    2
12           |   Oranges     |    2      |    3

我要选择文章名称和文章总数。每个文章的金额。文章ID和预订。城市

我的代码：

SELECT distinct r.ID,a.Name as ArticleName,
       sum(a.Amount) as ArticlesAmount,
       substr(r.City,1,3) as ToCityName 
FROM Reservations r 
INNER JOIN Articles a 
      on r.Id = a.ReservationId 
WHERE  a.Name <> '' 
GROUP BY ToCityName,a.ArticleId,a.Name 
ORDER BY ToCityName ASC

但我想：

Id | ArticleName | ArticlesAmount | ToCityName

2  |  Coconuts   |   4            |   Got 
2  |  Apples     |   2            |   Got 
3  |  Oranges    |   2            |   Got 
1  |  Coconuts   |   1            |   Sto 

我们将不胜感激，请解释：）

---

我再次添加文章以再次选择请求的行。。。这是我的疑问

SELECT DISTINCT
    r.ID,
    a.`Name` AS ArticleName,
    Articles.Amount,
    substr(r.City, 1, 3) AS ToCityName
FROM
    Reservations r
INNER JOIN Articles a ON r.Id = a.ReservationId
INNER JOIN Articles ON a.ReservationId = Articles.ReservationId
AND a.ArticleId = Articles.ArticleId
WHERE
    a. NAME <> ''
GROUP BY
    ToCityName,
    a.ArticleId,
    a. NAME
ORDER BY
    ToCityName ASC

选择DISTINCT
r、 身份证，
a、 “Name”作为ArticleName，
条款。金额，
substr（r.City，1，3）作为ToCityName
从…起
保留区r
r.Id=a.ReservationId上的内部联接项目a
a.ReservationId=Articles.ReservationId上的内部联接项目
a.ArticleId=Articles.ArticleId
哪里
A.名称“”
分组
ToCityName，
a、 ArticleId，
A.名称
订购人
ToCityName ASC

看一看

代码：

选择不同的r.ID，a.Name作为ArticleName，
以条款金额表示的金额，
substr（r.City，1,3）作为ToCityName
来自r
内部连接项目a
on r.Id=a.ReservationId
其中a.名称“
按ToCityName、a.ArticleId、a.Name分组
由ToCityName ASC订购

您希望确保按每个组显示的不同次数对金额求和