Sql server SQL查询以获取SQL Server中所有类别相册的相册计数

我想显示

AlbumCategories

，每个类别的

AlbumCategories

计数如下所示

预期结果：

Sports(2)
Education(3)
Community(0)
Something(0)

查询：

SELECT 
    AC.ACID, MAX(AC.Name) as Name,
    COUNT(*) as Frequency 
FROM 
    Album A
INNER JOIN 
    AlbumCategories AC ON A.ACID = AC.ACID
GROUP BY 
    AC.ACID  

此查询仅返回具有上述查询的唱片样本输出的唱片类别

Sports(2)
Education(3)

表专辑类别：

ACID    Name
1       Sports
2       Education
3       Community
4       Something

AID     ACID    Name
1       1       Footbal season 2015
2       1       Footbal season 2016
3       2       Album one
4       2       Album Two
5       2       Album Three

PID     AID Image

表相册：

ACID    Name
1       Sports
2       Education
3       Community
4       Something

AID     ACID    Name
1       1       Footbal season 2015
2       1       Footbal season 2016
3       2       Album one
4       2       Album Two
5       2       Album Three

PID     AID Image

表相册照片：

ACID    Name
1       Sports
2       Education
3       Community
4       Something

AID     ACID    Name
1       1       Footbal season 2015
2       1       Footbal season 2016
3       2       Album one
4       2       Album Two
5       2       Album Three

PID     AID Image

---

您需要使用

右连接

<代码>内部联接仅提取联接表之间的匹配记录。从

相册

表格中计算列数不要使用

*

架构设置

CREATE TABLE AlbumCategories
    ([ACID] int, [Name] varchar(9))
;

INSERT INTO AlbumCategories
    ([ACID], [Name])
VALUES
    (1, 'Sports'),
    (2, 'Education'),
    (3, 'Community'),
    (4, 'Something');


CREATE TABLE albums
    ([AID] int, [ACID] int, [Name] varchar(19))
;

INSERT INTO albums
    ([AID], [ACID], [Name])
VALUES
    (1, 1, 'Footbal season 2015'),
    (2, 1, 'Footbal season 2016'),
    (3, 2, 'Album one'),
    (4, 2, 'Album Two'),
    (5, 2, 'Album Three');

SELECT AC.ACID,
       AC.NAME       AS NAME,
       Count(A.ACID) AS Frequency
FROM   albums A
       RIGHT JOIN AlbumCategories AC
               ON A.ACID = AC.ACID
GROUP  BY AC.ACID,
          AC.NAME 

查询

CREATE TABLE AlbumCategories
    ([ACID] int, [Name] varchar(9))
;

INSERT INTO AlbumCategories
    ([ACID], [Name])
VALUES
    (1, 'Sports'),
    (2, 'Education'),
    (3, 'Community'),
    (4, 'Something');


CREATE TABLE albums
    ([AID] int, [ACID] int, [Name] varchar(19))
;

INSERT INTO albums
    ([AID], [ACID], [Name])
VALUES
    (1, 1, 'Footbal season 2015'),
    (2, 1, 'Footbal season 2016'),
    (3, 2, 'Album one'),
    (4, 2, 'Album Two'),
    (5, 2, 'Album Three');

SELECT AC.ACID,
       AC.NAME       AS NAME,
       Count(A.ACID) AS Frequency
FROM   albums A
       RIGHT JOIN AlbumCategories AC
               ON A.ACID = AC.ACID
GROUP  BY AC.ACID,
          AC.NAME 

结果：

╔══════╦═══════════╦═══════════╗
║ ACID ║   NAME    ║ Frequency ║
╠══════╬═══════════╬═══════════╣
║    1 ║ Sports    ║         2 ║
║    2 ║ Education ║         3 ║
║    3 ║ Community ║         0 ║
║    4 ║ Something ║         0 ║
╚══════╩═══════════╩═══════════╝

---

您需要使用

右连接

<代码>内部联接仅提取联接表之间的匹配记录。从

相册

表格中计算列数不要使用

*

架构设置

CREATE TABLE AlbumCategories
    ([ACID] int, [Name] varchar(9))
;

INSERT INTO AlbumCategories
    ([ACID], [Name])
VALUES
    (1, 'Sports'),
    (2, 'Education'),
    (3, 'Community'),
    (4, 'Something');


CREATE TABLE albums
    ([AID] int, [ACID] int, [Name] varchar(19))
;

INSERT INTO albums
    ([AID], [ACID], [Name])
VALUES
    (1, 1, 'Footbal season 2015'),
    (2, 1, 'Footbal season 2016'),
    (3, 2, 'Album one'),
    (4, 2, 'Album Two'),
    (5, 2, 'Album Three');

SELECT AC.ACID,
       AC.NAME       AS NAME,
       Count(A.ACID) AS Frequency
FROM   albums A
       RIGHT JOIN AlbumCategories AC
               ON A.ACID = AC.ACID
GROUP  BY AC.ACID,
          AC.NAME 

查询

CREATE TABLE AlbumCategories
    ([ACID] int, [Name] varchar(9))
;

INSERT INTO AlbumCategories
    ([ACID], [Name])
VALUES
    (1, 'Sports'),
    (2, 'Education'),
    (3, 'Community'),
    (4, 'Something');


CREATE TABLE albums
    ([AID] int, [ACID] int, [Name] varchar(19))
;

INSERT INTO albums
    ([AID], [ACID], [Name])
VALUES
    (1, 1, 'Footbal season 2015'),
    (2, 1, 'Footbal season 2016'),
    (3, 2, 'Album one'),
    (4, 2, 'Album Two'),
    (5, 2, 'Album Three');

SELECT AC.ACID,
       AC.NAME       AS NAME,
       Count(A.ACID) AS Frequency
FROM   albums A
       RIGHT JOIN AlbumCategories AC
               ON A.ACID = AC.ACID
GROUP  BY AC.ACID,
          AC.NAME 

结果：

╔══════╦═══════════╦═══════════╗
║ ACID ║   NAME    ║ Frequency ║
╠══════╬═══════════╬═══════════╣
║    1 ║ Sports    ║         2 ║
║    2 ║ Education ║         3 ║
║    3 ║ Community ║         0 ║
║    4 ║ Something ║         0 ║
╚══════╩═══════════╩═══════════╝

---

使用

LEFT JOIN

代替。使用

LEFT JOIN

代替。我的意思是使用RIGHT JOIN获取所有记录，每个记录的频率为1。

COUNT（*）

可能会给你

1

，但这不是这个答案建议你使用的。仔细阅读答案。@Learning-这可能是因为您仍在使用

*

count aggregate。谢谢你。它可以工作，您可以指出我所犯的一些错误。我的意思是使用RIGHT JOIN get me所有记录，每个记录的频率为1。

COUNT（*）

可能会为您提供

1

，但这不是此答案建议您使用的方法。仔细阅读答案。@Learning-这可能是因为您仍在使用

*

count aggregate。谢谢你。它起作用了，你指出了我犯的几个错误。