Question

我有2个表，1个名为googleimage，1个名为googleimagefound，googleimagefound有数千行，每行有一个image_id列，对应于来自googleimage表（id）的各自图像，googleimage表中的每个图像都可能有数千个从googleimagefound表中对应的行。

我要做的是从googleimage表中选择所有图像，然后按照googleimagefound表中找到的行数对其进行排序。

实施例，

googleimage表有3行，ID为1,2,3

googleimagefound表有100行，其中50行有image_id 1，其中20个有image_id，其中30个有image_id 3

我想说

SELECT * FROM googleimage ORDER BY most rows found in googleimagefound table where id = googleimagefound.image_id

任何帮助都会受到欢迎

Answer 1

尝试这种方式：

SELECT g.*,T1.Found
FROM googleimage g JOIN
(SELECT image_id,COUNT(image_id) as Found
 FROM googleimagefound
 GROUP BY image_id) T1 ON g.image_id=T1.image_id
ORDER BY T1.Found DESC

结果将如下：

id     name    others     Found
--------------------------------
1      NameA   Value1     120
5      NameD   Value4     95
3      NameB   Value2     82

Answer 2

你的意思是这样的：

SELECT count(distinct f.image_id) as mostrowsfound
, i.* FROM googleimage as i
 LEFT JOIN googleimagefound as f ON i.id = f.image_id
 GROUP BY i.id
 ORDER BY mostrowsfound DESC

Answer 3

您可以加入来自googleimagefound的聚合查询，该查询计算每个ID的行数：

SELECT    g.*
FROM      googleimage g
LEFT JOIN (SELECT   image_id, COUNT(*) AS cnt
           FROM     googleimagefound
           GROUP BY image_id) gf ON gf.image_id = g.id
ORDER BY  COALESCE(cnt, 0) DESC

Answer 4

SELECT i.*
  FROM googleimage i
  LEFT
  JOIN googleimagefound f
    ON f.image_id = i.image_id
 GROUP
    BY i.image_id
 ORDER BY COUNT(f.image_id) DESC;

ORDER BY在另一个表中找到的大多数行

4 个答案: