计算列中具有相同值的行数

Question

我有这两个实体的属性:

book - book_id,book_name,author_id,editor_id,subject_id,isbn

作者 - author_id,fn,ln

我必须在这里查询具有最多行数的作者属性和实体:

| book_id | book_name   | author_id |  editor_id |  isbn     |
--------------------------------------------------------------
|       1 | Book1 Title |  Author1  |  Editor1   | 8000-9000 |
|       2 | Book2 Title |  Author2  |  Editor1   | 8000-9001 |
|       1 | Book1 Title |  Author1  |  Editor1   | 8000-9002 |
|       3 | Book2 Title |  Author2  |  Editor1   | 8000-9003 |

| author_id |  fn    |    ln    |
---------------------------------
|       1   | name1  |  lname1  |
|       2   | name2  |  lname2  |
|       3   | name3  |  lname3  | 

这是我的代码:

       mysqli_query($con,"SELECT author.author_id, author.fn, author.ln, 
        COUNT(DISTINCT book.editor_id) as num
        FROM `editor`, `book`
        GROUP BY `editor_id`
        LIMIT 0,1")or die ('Query Failed');

但是当然我不能得到我想要的东西.输出必须是:

| author_id |  fn    |    ln    |
---------------------------------
|       1   | name1  |  lname1  |

因为"author_id = 1"在书上有2个条目.

Answer 1

试试这一个,但请注意,我转换的author_id来自varchar于int):

SELECT book_count,a.author_id,a.fn, a.ln 
FROM author a
JOIN
(
  SELECT c.author_id,COUNT(*) book_count FROM book c
  GROUP BY c.author_id
) b 
ON a.author_id = b.author_id
ORDER BY book_count DESC LIMIT 1

这是SQLFiddle.