Php 无法正确实现联接

我想弄明白

（图书编号、分行编号、总份数、可用数量）

以下是模式：

以下是我所做的：

SELECT
  TITLE,
  A.BOOK_ID,
  A.BRANCH_NAME,
  A.BRANCH_ID,
  NO_OF_COPIES,
  COUNT(BOOK_LOANS.BOOK_ID) AS NUM_OUT,
  NO_OF_COPIES - COUNT(BOOK_LOANS.BOOK_ID) AS NUM_AVAIL,
  AUTHOR_NAME                                      
FROM
  (SELECT
    TITLE,
    BOOK.BOOK_ID,
    BOOK_COPIES.BRANCH_ID,
    BRANCH_NAME,
    NO_OF_COPIES,
    BOOK_AUTHORS.AUTHOR_NAME                                          
  FROM
    BOOK,
    BOOK_COPIES,
    LIBRARY_BRANCH,
    BOOK_AUTHORS                                          
  WHERE
    BOOK_COPIES.BOOK_ID=BOOK.BOOK_ID 
    AND BOOK_AUTHORS.BOOK_ID = BOOK.BOOK_ID 
    AND (
      BOOK.TITLE LIKE :title 
      OR BOOK.BOOK_ID = :book_id 
      OR BOOK_AUTHORS.AUTHOR_NAME LIKE :author_name
    )                                          
    AND BOOK_COPIES.BRANCH_ID=LIBRARY_BRANCH.BRANCH_ID) AS A 
LEFT OUTER JOIN
  BOOK_LOANS 
    ON A.BOOK_ID=BOOK_LOANS.BOOK_ID 
    AND A.BRANCH_ID=BOOK_LOANS.BRANCH_ID 
GROUP BY
  A.BOOK_ID,
  A.BRANCH_ID

但我遇到的问题是，因为BOOK_AUTHORS表中的一本书有多个作者。所以我能够理解，只在book_id上加入book和book_作者，而我应该能够完全使用book_id和author_名称

因此，对于帐簿贷款中的单个条目，它返回计数2次

e、 g.如果图书id为1234的图书签出一次，且其总拷贝数为2，则应将剩余拷贝数返回为1。但它返回0

请帮助我了解如何修复此问题

---

注意：我使用的是php prepare语句。

我没有注意到您的查询（对我来说太复杂了）

• COUNT（不同的BOOK\u LOANS.BOOK\u ID）

  而不是

  COUNT（BOOK\u LOANS.BOOK\u ID）

  。如果查询除了计数之外运行良好，DISTINCT将非常有用
• 业绩很重要。我想内联视图太复杂了。我不知道是否有这样的变化

  SELECT
      DISTINCT BOOK.BOOK_ID
  FROM
      BOOK INNER JOIN BOOK_AUTHORS USING(BOOK_ID)                                     
  WHERE
        BOOK.TITLE LIKE :title 
        OR BOOK.BOOK_ID = :book_id 
        OR BOOK_AUTHORS.AUTHOR_NAME LIKE :author_name
  

  SELECT
  TITLE,
  BOOK.BOOK_ID,
  LIBRARY_BRANCH.BRANCH_NAME,
  LIBRARY_BRANCH.BRANCH_ID,
  BOOK_COPIES.NO_OF_COPIES,
  COUNT(distinct BOOK_LOANS.BOOK_ID) AS NUM_OUT,
  BOOK_COPIES.NO_OF_COPIES - COUNT(distinct BOOK_LOANS.BOOK_ID) AS NUM_AVAIL,
  AUTHOR_NAME  
  
  FROM
    (
      SELECT
          DISTINCT BOOK.BOOK_ID
      FROM
          BOOK,
          BOOK_AUTHORS                                          
      WHERE
            BOOK.TITLE LIKE :title 
            OR BOOK.BOOK_ID = :book_id 
            OR BOOK_AUTHORS.AUTHOR_NAME LIKE :author_name
    ) AS A 
  INNER JOIN BOOK USING(BOOK_ID)
  INNER JOIN BOOK_AUTHORS USING (BOOK_ID)
  INNER JOIN BOOK_COPIES USING(BOOK_ID) 
  INNER JOIN LIBRARY_BRANCH USING(BRANCH_ID)
  LEFT OUTER JOIN
    BOOK_LOANS 
      ON BOOK.BOOK_ID=BOOK_LOANS.BOOK_ID 
      AND LIBRARY_BRANCH.BRANCH_ID=BOOK_LOANS.BRANCH_ID 
  GROUP BY
    BOOK.BOOK_ID,
    LIBRARY_BRANCH.BRANCH_ID,
    LIBRARY_BRANCH.BRANCH_NAME,
    BOOK_COPIES.NO_OF_COPIES,
    AUTHOR_NAME  
  

• 最后的查询如下所示

  SELECT
      DISTINCT BOOK.BOOK_ID
  FROM
      BOOK INNER JOIN BOOK_AUTHORS USING(BOOK_ID)                                     
  WHERE
        BOOK.TITLE LIKE :title 
        OR BOOK.BOOK_ID = :book_id 
        OR BOOK_AUTHORS.AUTHOR_NAME LIKE :author_name
  

  SELECT
  TITLE,
  BOOK.BOOK_ID,
  LIBRARY_BRANCH.BRANCH_NAME,
  LIBRARY_BRANCH.BRANCH_ID,
  BOOK_COPIES.NO_OF_COPIES,
  COUNT(distinct BOOK_LOANS.BOOK_ID) AS NUM_OUT,
  BOOK_COPIES.NO_OF_COPIES - COUNT(distinct BOOK_LOANS.BOOK_ID) AS NUM_AVAIL,
  AUTHOR_NAME  
  
  FROM
    (
      SELECT
          DISTINCT BOOK.BOOK_ID
      FROM
          BOOK,
          BOOK_AUTHORS                                          
      WHERE
            BOOK.TITLE LIKE :title 
            OR BOOK.BOOK_ID = :book_id 
            OR BOOK_AUTHORS.AUTHOR_NAME LIKE :author_name
    ) AS A 
  INNER JOIN BOOK USING(BOOK_ID)
  INNER JOIN BOOK_AUTHORS USING (BOOK_ID)
  INNER JOIN BOOK_COPIES USING(BOOK_ID) 
  INNER JOIN LIBRARY_BRANCH USING(BRANCH_ID)
  LEFT OUTER JOIN
    BOOK_LOANS 
      ON BOOK.BOOK_ID=BOOK_LOANS.BOOK_ID 
      AND LIBRARY_BRANCH.BRANCH_ID=BOOK_LOANS.BRANCH_ID 
  GROUP BY
    BOOK.BOOK_ID,
    LIBRARY_BRANCH.BRANCH_ID,
    LIBRARY_BRANCH.BRANCH_NAME,
    BOOK_COPIES.NO_OF_COPIES,
    AUTHOR_NAME  
  

---

就模式而言，我认为应该有一个名为

authors\u master

的表，它有两列

author\u id，author\u name

，这样就可以使用一个表

book\u author

将其映射到book，该表只有两列

book\u id，author\u id

，这可能是一个解决方案，但这正是我要做的。更正了打字错误，但A.BRANCH\u ID=BOOK\u LOANS.BRANCH\u ID不起作用，因为A里面没有

BRANCH\u ID

。@user2961121抱歉。固定的。它是复制粘贴的。@user2961121“返回两行”无法消除。当一本书有3位作者时，选择“始终返回3行”。我想你需要做两个查询，一个是图书和贷款信息，一个是作者信息。哦，好的，我需要找到一种方法来清楚地显示它们。我成功了！但是你能解释一下你做了什么吗？太多了！