我在连接表时遇到问题，这里是示例表：
表A :( 30行）
╔════╦════════════╦═════════════╗ ║ ID ║ Name ║ Description ║ ╠════╬════════════╬═════════════╣ ║ 1 ║ Type ║ Unicode Art ║ ║ 2 ║ Header ║ Spreadsheet ║ ║ 3 ║ Auto Align ║ Off ║ ╚════╩════════════╩═════════════╝
表B：（100行）
╔════╦════════════╦═════════════╦═════════╗ ║ ID ║ Name ║ Description ║ TableA ║ ╠════╬════════════╬═════════════╬═════════╣ ║ 1 ║ Type ║ Unicode Art ║ 1 ║ ║ 2 ║ Header ║ Spreadsheet ║ 1 ║ ║ 3 ║ Auto Align ║ Off ║ 2 ║ ╚════╩════════════╩═════════════╩═════════╝
表C :( 8000行）
╔════╦════════════╦═════════════╦═════════╗ ║ ID ║ Article ║ Text ║ TableB ║ ╠════╬════════════╬═════════════╬═════════╣ ║ 1 ║ Type ║ Unicode Art ║ 1 ║ ║ 2 ║ Header ║ Spreadsheet ║ 1 ║ ║ 3 ║ Auto Align ║ Off ║ 2 ║ ╚════╩════════════╩═════════════╩═════════╝
表D :( 100 000行和计数）
╔════╦═══════════╦════════════╦═════════════╦═════════╗ ║ ID ║ Date ║ Clicks ║ Impressions ║ TableC ║ ╠════╬═══════════╬════════════╬═════════════╬═════════╣ ║ 1 ║ 20120814 ║ 10 ║ 3 ║ 1 ║ ║ 2 ║ 20120815 ║ 13 ║ 5 ║ 1 ║ ║ 3 ║ 20120816 ║ 15 ║ 10 ║ 2 ║ ╚════╩═══════════╩════════════╩═════════════╩═════════╝
表E :( 200 000行并计数）
╔════╦═══════════╦════════════╦═══════════╦═════════╗ ║ ID ║ Date ║ Views ║ Visitors ║ TableC ║ ╠════╬═══════════╬════════════╬═══════════╬═════════╣ ║ 1 ║ 20120814 ║ 10 ║ 3 ║ 1 ║ ║ 2 ║ 20120815 ║ 13 ║ 5 ║ 1 ║ ║ 3 ║ 20120816 ║ 15 ║ 10 ║ 2 ║ ║ 4 ║ 20120817 ║ 8 ║ 7 ║ 2 ║ ║ 5 ║ 20120818 ║ 9 ║ 4 ║ 2 ║ ╚════╩═══════════╩════════════╩═══════════╩═════════╝
我使用单个sql语句查询此表：
SELECT
A.name,
A.Description,
SUM(D.clicks),
SUM(D.Impressions),
SUM(E.Views),
SUM(E.Visitors)
FROM
A
LEFT JOIN B
ON A.ID=B.TableA
LEFT JOIN C
ON B.ID=C.TableB
LEFT JOIN D
ON C.ID=D.TableC
LEFT JOIN E
ON C.ID=E.TableC
GROUP BY
A.ID
问题是查询返回表D和表E的无效SUM 但是，如果在invidual查询中查询表D和表E，我得到正确的值：
SELECT
A.name,
A.Description,
SUM(D.clicks),
SUM(D.Impressions)
FROM
A
LEFT JOIN B
ON A.ID=B.TableA
LEFT JOIN C
ON B.ID=C.TableB
LEFT JOIN D
ON C.ID=D.TableC
GROUP BY
A.ID
编辑1：
我试过RIGHT JOIN，JOIN，LEFT OUTER JOIN他们都没有工作，
当然，我可能会把它们用在错误的地方
但是在我得到“全部包含”的第一个陈述中，值乘以
比他们真实的数千倍
答案 0 :(得分：9)
你需要弄平D和E表。然后我想A和B只是C的查找，因此不需要在A上进行GROUP BY：http://www.sqlfiddle.com/#!2/fccf1/8
我删除了噪音（A和B），因为我看不到（还有）A和B如何与总结C的信息有关
试试这个：
SELECT
C.Article,
C.Text,
COALESCE(D.ClicksSum,0) AS ClicksSum,
COALESCE(D.ImpressionsSum,0) AS ImpressionsSum,
COALESCE(E.ViewsSum,0) AS ViewsSum,
COALESCE(E.VisitorsSum,0) AS VisitorsSum
FROM
C
LEFT JOIN
(
SELECT TableC, SUM(Clicks) AS ClicksSum, SUM(Impressions) AS ImpressionsSum
FROM D
GROUP BY TableC
) D ON C.ID=D.TableC
LEFT JOIN
(
SELECT TableC, SUM(Views) AS ViewsSum, SUM(Visitors) AS VisitorsSum
FROM E
GROUP BY TableC
) E ON C.ID=E.TableC
输出：
| ARTICLE | TEXT | CLICKSSUM | IMPRESSIONSSUM | VIEWSSUM | VISITORSSUM |
----------------------------------------------------------------------------------
| Type | Unicode Art | 23 | 8 | 23 | 8 |
| Header | Spreadsheet | 15 | 10 | 32 | 21 |
| Auto Align | Off | 0 | 0 | 0 | 0 |
请注意，我没有手动在我的sqlfiddle帖子中输入这些架构，我使用sqlfiddle的文本到DDL
我爱http://sqlfiddle.com，文字到DDL 甚至可以解析你的ASCII艺术数据ツ
在看到更明确的目标（来自您的评论）后，可能是：http://www.sqlfiddle.com/#!2/fccf1/13
SELECT
A.Name, A.Description,
COALESCE(SUM(D.ClicksSum),0) AS ClicksSum,
COALESCE(SUM(D.ImpressionsSum),0) AS ImpressionsSum,
COALESCE(SUM(E.ViewsSum),0) AS ViewsSum,
COALESCE(SUM(E.VisitorsSum),0) AS VisitorsSum
FROM
C
LEFT JOIN
(
SELECT TableC, SUM(Clicks) AS ClicksSum, SUM(Impressions) AS ImpressionsSum
FROM D
GROUP BY TableC
) D ON C.ID=D.TableC
LEFT JOIN
(
SELECT TableC, SUM(Views) AS ViewsSum, SUM(Visitors) AS VisitorsSum
FROM E
GROUP BY TableC
) E ON C.ID=E.TableC
RIGHT JOIN B ON B.ID = C.TableB
RIGHT JOIN A ON A.ID = B.TableA
GROUP BY A.ID
输出：
| NAME | DESCRIPTION | CLICKSSUM | IMPRESSIONSSUM | VIEWSSUM | VISITORSSUM |
----------------------------------------------------------------------------------
| Type | Unicode Art | 38 | 18 | 55 | 29 |
| Header | Spreadsheet | 0 | 0 | 0 | 0 |
| Auto Align | Off | 0 | 0 | 0 | 0 |
上述方法可能仍会产生笛卡尔积，在将其分类为类别（A）之前将子类别（B）展平：http://www.sqlfiddle.com/#!2/fccf1/19
SELECT
A.Name, A.Description,
COALESCE(SUM(B.ClicksSum),0) AS ClicksSum,
COALESCE(SUM(B.ImpressionsSum),0) AS ImpressionsSum,
COALESCE(SUM(B.ViewsSum),0) AS ViewsSum,
COALESCE(SUM(B.VisitorsSum),0) AS VisitorsSum
FROM A
LEFT JOIN
(
SELECT
B.ID, B.TableA,
SUM(C.ClicksSum) AS ClicksSum,
SUM(C.ImpressionsSum) AS ImpressionsSum,
SUM(C.ViewsSum) AS ViewsSum,
SUM(C.VisitorsSum) AS VisitorsSum
FROM B
LEFT JOIN
(
SELECT
C.TableB,
D.ClicksSum,
D.ImpressionsSum,
E.ViewsSum,
E.VisitorsSum
FROM
C
LEFT JOIN
(
SELECT TableC, SUM(Clicks) AS ClicksSum, SUM(Impressions) AS ImpressionsSum
FROM D
GROUP BY TableC
) D ON C.ID=D.TableC
LEFT JOIN
(
SELECT TableC, SUM(Views) AS ViewsSum, SUM(Visitors) AS VisitorsSum
FROM E
GROUP BY TableC
) E ON C.ID=E.TableC
) C ON C.TableB = B.ID
GROUP BY B.ID
) B ON B.TableA = A.ID
GROUP BY A.ID
输出：
| NAME | DESCRIPTION | CLICKSSUM | IMPRESSIONSSUM | VIEWSSUM | VISITORSSUM |
----------------------------------------------------------------------------------
| Type | Unicode Art | 38 | 18 | 55 | 29 |
| Header | Spreadsheet | 0 | 0 | 0 | 0 |
| Auto Align | Off | 0 | 0 | 0 | 0 |
答案 1 :(得分：1)
表D和表E都通过表C链接到其余部分。因此，您的第一个查询给出表D中所有行的笛卡尔积，乘以表E中的所有行，并且SUM函数聚合此笛卡尔积。可能你也必须按表C进行分组，而不仅仅是表A.