Question

我写这篇文章是为了询问有关以下架构的查询的帮助：

想象一下有两种格式，例如PDF和电子书。我的目标是获得以下结果：

PushiblerName  |  Number_PDF   |  Numerber_BOOK  |
--------------------------------------------------
ExampleName1   |    2          |   0
ExampleName2   |    3          |   1

有人会帮我写这个查询吗？提前谢谢。

Answer 1

select name,(select count(*) from t_book where IdPublisher = t_publisher.Id) as Numerber_BOOK,
(select count(*) from t_book where IdPublisher = t_publisher.Id and IdFormat = 3)  as Number_PDF from t_publisher

用Pdf格式的Id替换3

Answer 2

我想你可以尝试这样的事情

首先设置一些示例数据：

declare @t_book table (Id int, IdPublisher int)
declare @t_publisher table (Id int, Name varchar(50))
declare @t_book_format table (IdBook int, idFormat int)
declare @t_format table (Id int, Description varchar(50))

insert into @t_book values (1, 1), (2, 1), (3, 2), (4, 2), (5, 2), (6, 2)
insert into @t_publisher values (1, 'examplename1'), (2, 'examplename2')
insert into @t_book_format values (1, 1), (2, 1), (3, 2), (4, 1), (5, 1), (6, 1)
insert into @t_format values (1 ,'pdf'), (2, 'book')

现在查询：

select p.Name as PublisherName,
       f.Description,
       count(f.Id) as numbers
from   @t_book b
  inner join @t_publisher p on b.idPublisher = p.Id
  inner join @t_book_format bf on b.Id = bf.idBook
  inner join @t_format f on bf.idFormat = f.Id
group by p.Name, f.Description

但如果没有来自所有表的样本数据，这只是猜测

结果采用不同的格式，但如果在t_book_format和t_format

中添加更多格式，它仍然有效

PublisherName   Description numbers 
-------------   ----------- ------- 
examplename1    pdf         2      
examplename2    book        1   
examplename2    pdf         3

如果你想要列出你的问题，你可以做一个支点但是当有新格式时，这种新格式不会出现在结果中，因此每次添加新格式时都必须修改查询

select * from ( select p.Name,
                       f.Description,
                       count(f.Id) as numbers
                from   @t_book b
                  inner join @t_publisher p on b.idPublisher = p.Id
                  inner join @t_book_format bf on b.Id = bf.idBook
                  inner join @t_format f on bf.idFormat = f.Id
                group by p.Name, f.Description
              ) t
pivot
( sum(numbers)
  for Description in (pdf, book)
) piv

结果现在看起来像这样

Name            pdf book    
----            --- ----    
examplename1    2   null    
examplename2    3   1

我想你可以把它放在一个存储过程中并在动态sql中构建最后一个查询，所以当你添加新格式时它仍会继续工作

Answer 3

试试这个：

select P.Name as PublisherName,
       SUM(case when F.Code = 'PDF' then 1 else 0 end) as Number_PDF,
       SUM(case when F.Code = 'BOOK' then 1 else 0 end) as Number_BOOK
from T_BOOK as B
join T_BOOK_FORMAT as BF on B.Id = BF.BookId
join T_FORMAT as F on BF.IdFormat = F.Id
join T_PUBLISHER as P on B.IdPublisher = P.Id
group by P.Name

我认为，这本书的格式保存在T_FORMAT列Code中（F.Code部分）。

Answer 4

您可以使用此代码

    SELECT publisher.Name as PublisherName,
        (SELECT COUNT(*)
         FROM T_BOOK book
         JOIN T_BOOK_FORMAT bookformat ON book.Id = bookformat.IdBook
         JOIN T_FORMAT format ON bookformat.IdFormat = format.Id
         WHERE book.IdPublisher = publisher.Id and format.Code='PDF'
         ) AS Number_PDF,
        (SELECT COUNT(*)
         FROM T_BOOK book
         JOIN T_BOOK_FORMAT bookformat ON book.Id = bookformat.IdBook
         JOIN T_FORMAT format ON bookformat.IdFormat = format.Id
         WHERE book.IdPublisher = publisher.Id and format.Code='EBOOK'
         ) AS Number_BOOK
    FROM T_PUBLISHER publisher
    GROUP BY publisher.Id

Answer 5

你走了：

CREATE TABLE Author (
    AuthorID INT IDENTITY (1,1) PRIMARY KEY NOT NULL,
    AuthorName VARCHAR (25) NOT NULL
    );
CREATE TABLE Books (
    BookID INT IDENTITY (1,1) PRIMARY KEY NOT NULL,
    BookName VARCHAR (25) NOT NULL,
    BookAothor INT NOT NULL,
    CONSTRAINT PK_AuthorOfBook FOREIGN KEY (BookAothor) REFERENCES Author (AuthorID)
    );
CREATE TABLE Formats (
    FormatID INT IDENTITY (1,1) PRIMARY KEY NOT NULL,
    FormatName VARCHAR (10) NOT NULL
    );
CREATE TABLE BookFormat (
    BookID INT NOT NULL,
    FormatID INT NOT NULL,
    CONSTRAINT PK_BookFormats PRIMARY KEY (BookID, FormatID),
    CONSTRAINT FK_BookRef FOREIGN KEY (BookID) REFERENCES Books (BookID),
    CONSTRAINT FK_FormatRef FOREIGN KEY (FormatID) REFERENCES Formats (FormatID)
    );
INSERT INTO Author VALUES
('Author1'), ('Author2'), ('Author3');
INSERT INTO Books VALUES
('Book1', 1), ('Book2', 1),
('Book3', 2), ('Book4', 2),
('Book5', 2), ('Book6', 2);
INSERT INTO Formats VALUES
('PDF'), ('E-Book');
INSERT INTO BookFormat VALUES
(1, 1), (2,1),
(3,1), (4,1), (5,1), (6,2);

SELECT A.AuthorName,
       ISNULL(Pdf.PdfBooks, 0) AS PdfBooks,
       ISNULL(EBook.EBooks, 0) AS EBooks
FROM
Author A  LEFT JOIN
    (
        SELECT A.AuthorID, COUNT(B.BookID) AS PdfBooks
        FROM Author A JOIN Books B ON A.AuthorID = B.BookAothor
             INNER JOIN BookFormat BF ON B.BookID = BF.BookID
        WHERE BF.FormatID =1
        GROUP BY A.AuthorID
    ) Pdf ON A.AuthorID = Pdf.AuthorID
     FULL JOIN
    (
        SELECT A.AuthorID, COUNT(B.BookID) AS EBooks
        FROM Author A JOIN Books B ON A.AuthorID = B.BookAothor
             INNER JOIN BookFormat BF ON B.BookID = BF.BookID
        WHERE BF.FormatID =2
        GROUP BY A.AuthorID    
    ) EBook 
    ON Pdf.AuthorID = EBook.AuthorID;

结果：

+------------+----------+--------+
| AuthorName | PdfBooks | EBooks |
+------------+----------+--------+
| Author1    |        2 |      0 |
| Author2    |        3 |      1 |
| Author3    |        0 |      0 |
+------------+----------+--------+

这是 demo

如何使用Count SQL Server Query

5 个答案: