Question

我有2个数据库表：

表1：

+---------+-------+-------------+
| Page    | Title | Description |
+---------+-------+-------------+
| Apple   | ..... | ........... |
| Orange  | ..... | ........... |
| Pear    | ..... | ........... |
| Grapes  | ..... | ........... |
+---------+-------+-------------+

表2：

+----------+-------------+
|   Link   |    Page     |
+----------+-------------+
| Website1 |    Apple    |
| Website2 |    Orange   |
| Website3 |    Apple    |
| Website4 |    Orange   |
| Website5 |    Apple    |
| Website6 |    Pear     |
| Website7 |    Apple    |
| Website8 |    Grapes   |
| Website9 |    Grapes   |
+----------+-------------+

我想知道/返回表2中引用表1中的页数以及它们被引用的次数。（我不想知道表2中引用了表1中每个页面的次数。）

所以在这个例子中： 1页被引用1次（梨）， 2页被引用2次（Grapes和Orange）＆amp; 1页被引用4次。

我会用什么样的SQL语句来获取它？

Answer 1

以下查询应该做..

SELECT COUNT(1) NoOfPages,CNT ReferencedTimes
FROM
(
    SELECT T2.PAGE,COUNT(1) CNT
    FROM TABLE1 T1 INNER JOIN TABLE2 T2 ON T1.PAGE = T2.PAGE
    GROUP BY T2.PAGE
)T
GROUP BY CNT

Answer 2

我认为以下陈述适合：

SELECT count(*) FROM Table2 WHERE (Table2.Page IN (SELECT Page FROM Table1));

Answer 3

 SELECT (GROUP_CONCAT(DISTINCT page)) AS Page,page_count 
      FROM
          (SELECT table1.Page as page,COUNT(*) as page_count
           FROM table1 INNER JOIN table2 ON table1.Page=table2.Page
          GROUP BY table1.Page)
       as T GROUP BY page_count

希望这有帮助

Answer 4

使用此查询

1) User_Test::test_Query
Unable to find JSON fragment ["data":[{"username":"1137598_update"}]] within [{"current_page":1,"data":[{"updated_at":"2015-10-29 18:57:13","username":"1137598_update"}]}].
Failed asserting that false is true.

Answer 5

如果您要搜索的是X页面被引用了N次，则以下查询将实现：

SELECT COUNT(t1.page), t2.count
FROM table1 t1
INNER JOIN (SELECT page,COUNT(*) AS count FROM table2 GROUP BY page) t2 ON t1.page=t2.page
GROUP BY t2.count

Answer 6

尝试此查询，它将进行左连接并告诉您在table2中引用项目的次数，如果count为零而不是其他表中没有引用

SELECT table1.Page, count(table2.Page) as count
FROM table1
LEFT JOIN table2 ON table2.Page = table1.Page
GROUP BY table1.Page

MYSQL语句：查找其他表中存在多少条记录的次数

6 个答案: