查詢返回不正確的計數

Question

我有以下查詢。

SELECT a.link_field1 AS journo, count(a.link_id) as articles, AVG(b.vote_value) AS score FROM dan_links a LEFT JOIN dan_votes b ON link_id = vote_link_id WHERE link_field1 <> '' and link_status NOT IN ('discard', 'spam', 'page') GROUP BY link_field1 ORDER BY link_field1, link_id 

此查詢返回列表中第一項的計數爲3。什麼應該返回是

Journo | count | score 
John S | 2 | 6.00 
Joe B | 1 | 4 

不過，對於第一個約翰·S，它返回的3

計數如果我直接查詢

select * from dan_links where link_field1 = 'John S' 

我得到2分返回的記錄，因爲我會期望。我不能爲我的生活找出爲什麼計數是錯誤的，除非由於某種原因它正在計算dan_vote表中的記錄

我如何得到正確的計數，或者是我的查詢完全錯誤？

編輯：表格的內容

dan_links

link_id | link_field1 | link | source | link_status 
1 | John S | http://test.com | test.com | approved 
2 | John S | http://google.com | google | approved 
3 | Joe B | http://facebook.com | facebook | approved 

dan_votes

vote_id | link_id | vote_value 
1 | 1 | 5 
2 | 1 | 8 
3 | 2 | 4 
4 | 3 | 1 

編輯：它看起來像它正指望在投票表中的行出於某種原因

Answer 1

當您在執行條件爲link_id = vote_link的左外連接時_id用於創建的每個匹配的記錄一行，就像

link_id | link_field1 | link | source | link_status|vote_id|vote_value 
1 | John S | http://test.com | test.com | approved|1|5 
1 | John S | http://test.com | test.com | approved|2|8 
2 | John S | http://google.com | google | approved|3|4 
3 | Joe B | http://facebook.com | facebook | approved|4|1 

一些東西現在，當你通過link_field1做組，你得到算作3約翰小號

嵌套查詢可能工作

SELECT journo,count(linkid) as articles,AVG(score) FROM 
(SELECT a.link_field1 AS journo, AVG(b.vote_value) AS score, a.link_id as linkid 
FROM dan_links a 
LEFT JOIN dan_votes b 
ON link_id = vote_link_id 
WHERE link_field1 <> '' 
and link_status NOT IN ('discard', 'spam', 'page') 
GROUP BY link_id 
ORDER BY link_field1, link_id) GROUP BY journo 

上面的查詢會提供不正確的平均值作爲（（N1 + N2）/ 2 + N3）/ 2！=（N1 + N2 + N3）/ 3，所以使用下面查詢

SELECT journo,count(linkid) as articles, SUM(vote_sum)/SUM(count(linkid)) 
FROM 
    (SELECT a.link_field1 AS journo, SUM(b.vote_value) AS vote_sum, a.link_id as linkid, count(a.link_id) as count_on_id 
    FROM dan_links a 
    LEFT JOIN dan_votes b 
    ON link_id = vote_link_id 
    WHERE link_field1 <> '' 
    and link_status NOT IN ('discard', 'spam', 'page') 
    GROUP BY link_id 
ORDER BY link_field1, link_id) GROUP BY journo 

希望這會有所幫助。