Php 如何将两个SQL查询的结果合并为一个

我想合并两个复杂的查询

第一个查询返回以下列：

BOOKID
BOOKNAME
NUM_READ
DID_I_READ_IT

BOOKID
BOOKNAME
NUM_LIKES
DID_I_LIKE_IT

第二个查询返回以下列：

BOOKID
BOOKNAME
NUM_READ
DID_I_READ_IT

BOOKID
BOOKNAME
NUM_LIKES
DID_I_LIKE_IT

我想将这些合并到一个结果中，以得到所有列

BOOKID
BOOKNAME
NUM_ADDED
DID_I_ADD_IT
NUM_LIKEs
DID_I_LIKE_IT

书籍：

| ID | BOOKNAME |
|----|----------|
|  1 |    Book1 |
|  2 |    Book2 |
|  3 |    Book3 |
|  4 |    Book4 |
|  5 |    Book5 |

书籍使用者：

| ID | USERID | BOOKID | STATUS |
|----|--------|--------|--------|
|  1 |    001 |      1 |   Read |
|  2 |    001 |      2 |   Read |
|  3 |    001 |      3 |  Added |
|  4 |    002 |      1 |  Added |
|  5 |    002 |      5 |  Added |
|  6 |    003 |      2 |   Read |
|  7 |    004 |      4 |   Read |

从中，我可以查询书籍列表、阅读人数以及我（用户001）是否阅读过它们

     SELECT a.ID, a.BOOKNAME, a.NUM_READ, 
            CASE WHEN b.BOOKID IS NULL THEN 'No' ELSE 'Yes' END AS DID_I_READ_IT
       FROM (
             select BOOKS.ID, BOOKS.BOOKNAME, 
                    SUM(CASE WHEN USERS_BOOKS.STATUS='Read' THEN 1 ELSE 0 END) AS NUM_READ
               from BOOKS
          LEFT JOIN USERS_BOOKS ON USERS_BOOKS.BOOKID = BOOKS.ID 
           GROUP BY BOOKS.ID
            ) AS a
  LEFT JOIN (
           select DISTINCT USERID, 
                           BOOKID
             FROM USERS_BOOKS
            WHERE STATUS = 'Read'
              AND USERID = '001'
            ) AS b ON a.ID = b.BOOKID

结果:

| ID | BOOKNAME| NUM_READ | DID_I_READ_IT|
|----|---------|----------|--------------|
|  1 |  BOOK1  |    1     |      Yes     |
|  2 |  BOOK2  |    2     |      Yes     |
|  3 |  BOOK3  |    0     |      No      |
|  4 |  BOOK4  |    1     |      No      |
|  5 |  BOOK5  |    0     |      No      |

| ID | BOOKNAME| NUM_LIKES | DID_I_LIKE_IT|
|----|---------|-----------|--------------|
|  1 |  BOOK1  |     2     |      Yes     |
|  2 |  BOOK2  |     1     |      No      |
|  3 |  BOOK3  |     1     |      Yes     |
|  4 |  BOOK4  |     0     |      No      |
|  5 |  BOOK5  |     0     |      No      |

还有一张桌子可以存放类似的书籍：

| ID | USERID | BOOKID | STATUS |
|----|--------|--------|--------|
|  1 |    001 |      1 |   Yes  |   //USER 001 likes BOOK 1
|  2 |    001 |      2 |   No   |   //USER 001 does not like BOOK 2
|  3 |    001 |      3 |   Yes  |
|  4 |    002 |      1 |   Yes  |   //USER 002 likes BOOK 1
|  5 |    002 |      5 |   No   |
|  6 |    003 |      2 |   Yes  |
|  7 |    004 |      4 |   No   |

从中，我可以查询书籍列表、喜欢它们的人数以及我（用户001）是否喜欢它们

     SELECT a.ID, a.BOOKNAME, a.NUM_READ, 
            CASE WHEN b.BOOKID IS NULL THEN 'No' ELSE 'Yes' END AS DID_I_READ_IT
       FROM (
             select BOOKS.ID, BOOKS.BOOKNAME, 
                    SUM(CASE WHEN USERS_BOOKS.STATUS='Read' THEN 1 ELSE 0 END) AS NUM_READ
               from BOOKS
          LEFT JOIN USERS_BOOKS ON USERS_BOOKS.BOOKID = BOOKS.ID 
           GROUP BY BOOKS.ID
            ) AS a
  LEFT JOIN (
           select DISTINCT USERID, 
                           BOOKID
             FROM USERS_BOOKS
            WHERE STATUS = 'Read'
              AND USERID = '001'
            ) AS b ON a.ID = b.BOOKID

结果:

| ID | BOOKNAME| NUM_READ | DID_I_READ_IT|
|----|---------|----------|--------------|
|  1 |  BOOK1  |    1     |      Yes     |
|  2 |  BOOK2  |    2     |      Yes     |
|  3 |  BOOK3  |    0     |      No      |
|  4 |  BOOK4  |    1     |      No      |
|  5 |  BOOK5  |    0     |      No      |

| ID | BOOKNAME| NUM_LIKES | DID_I_LIKE_IT|
|----|---------|-----------|--------------|
|  1 |  BOOK1  |     2     |      Yes     |
|  2 |  BOOK2  |     1     |      No      |
|  3 |  BOOK3  |     1     |      Yes     |
|  4 |  BOOK4  |     0     |      No      |
|  5 |  BOOK5  |     0     |      No      |

我想将这两个查询合并为一个，所以我想要一个图书列表，阅读它们的人数，喜欢它们的人数，我是否阅读它们，我是否喜欢它们

     SELECT a.ID, a.BOOKNAME, a.NUM_READ, 
            CASE WHEN b.BOOKID IS NULL THEN 'No' ELSE 'Yes' END AS DID_I_READ_IT
       FROM (
             select BOOKS.ID, BOOKS.BOOKNAME, 
                    SUM(CASE WHEN USERS_BOOKS.STATUS='Read' THEN 1 ELSE 0 END) AS NUM_READ
               from BOOKS
          LEFT JOIN USERS_BOOKS ON USERS_BOOKS.BOOKID = BOOKS.ID 
           GROUP BY BOOKS.ID
            ) AS a
  LEFT JOIN (
           select DISTINCT USERID, 
                           BOOKID
             FROM USERS_BOOKS
            WHERE STATUS = 'Read'
              AND USERID = '001'
            ) AS b ON a.ID = b.BOOKID

预期结果：

| ID | BOOKNAME| NUM_READ | DID_I_READ_IT| NUM_LIKES | DID_I_LIKE_IT|
|----|---------|----------|--------------|-----------|--------------|
|  1 |  BOOK1  |    1     |      Yes     |     2     |      Yes     |
|  2 |  BOOK2  |    2     |      Yes     |     1     |      No      |
|  3 |  BOOK3  |    0     |      No      |     1     |      Yes     |
|  4 |  BOOK4  |    1     |      No      |     0     |      No      |
|  5 |  BOOK5  |    0     |      No      |     0     |      No      |

---

当然，当某人喜欢一本没有人读或添加到其列表中的书，或者一本书从未被喜欢但被许多人读过时，会出现空值。

这里有一种方法可以使用条件聚合来制定查询。这将使用子查询来遵循原始查询的逻辑：

 SELECT b.ID, b.BOOKNAME, r.NUM_READ, 
        (CASE WHEN r.IReadItFlag > 0 THEN 'Yes' ELSE 'No' END) AS DID_I_READ_IT,
        l.NUM_LIKED,
        (CASE WHEN l.ILikedItFlag > 0 THEN 'Yes' ELSE 'No' END) AS DID_I_LIKE_IT
 FROM books b LEFT OUTER JOIN
      (SELECT ub.bookid, SUM(ub.status = 'Read') AS NUM_READ,
              SUM(ub.status = 'Read' AND ub.userid = '001') as IReadItFlag
       FROM users_books ub
       GROUP BY ub.bookid
      ) r
      on b.id = r.bookid LEFT OUTER JOIN
      (SELECT ul.bookid, SUM(ul.liked = 'Yes') AS NUM_LIKED,
              SUM(ul.liked = 'Yes' AND ul.userid = '001') as ILikedItFlag
       FROM users_likes ul
       GROUP BY ul.bookid
      ) l
      on b.id = l.bookid;

---

是SQL小提琴。

但是有两个人读了第一本书。我是他们中的一员！？！否则，喜欢的数量就没有意义了。你怎么能不喜欢一本你没读过的书呢（除非是杰弗里·阿彻写的书）！？！？用户2阅读第5本书。这是怎么回事？看看我的小提琴。注意：我的模式和数据集与您的略有不同。。。这段代码要简单得多，但如果它和Gordon'sHi Gordon的一样好，我需要尝试一下。第一个查询没有得到我想要的结果，第二个查询由于错误而无法运行。@erdomester。第一个版本有什么问题？以下是两个表：。有时甚至你的第一个代码都不会运行，但是类似的列在哪里呢？谢谢Gordon，干得好！这个代码非常复杂，但再次感谢