使用DESC的SQL查詢無法正常工作

我不明白爲什麼排序不能正確使用「ORDER BY average，votes DESC」，因爲我需要最高的平均值並且票數是從頂部到底部的，但DESC沒有解決什麼問題。我在var_dump http://pastie.org/private/b05smuh0fvw72wwp2w1zq最高entrie的結果是在底部，但我需要從頂部開始並進入底部。使用DESC的SQL查詢無法正常工作

SELECT c.*, r.votes, c.total_comments, 
     ROUND(sumrate/votes) AS average 
FROM catalog c LEFT JOIN 
    (SELECT r.object_id, COUNT(*) as votes, SUM(r.rate) as sumrate 
     FROM ratings r 
     GROUP BY r.object_id 
    ) r 
    ON r.object_id = c.catalog_id LEFT JOIN 
    (SELECT c.catalog_id, COUNT(*) as total_comments 
     FROM comments c 
     GROUP BY c.catalog_id 
    ) c 
    ON c.catalog_id = c.catalog_id 
GROUP BY c.catalog_id 
ORDER BY average, votes DESC;

來源

2016-07-26 SkySonny

請注意，GROUP BY是無效的，並且在更新的MySQL版本中不允許。（如果指定了GROUP BY子句，則SELECT列表中的每個列引用都必須標識分組列或作爲set函數的參數。） – jarlh

你是什麼意思？分組中的簡短表別名是什麼？你能舉個例子嗎？ – SkySonny

因爲您可以爲每列定義訂單，而不僅限於總共order by。默認值是ASC。

您的訂單

ORDER BY average, votes DESC

自動轉成

ORDER BY average ASC, votes DESC

但你正在尋找

ORDER BY average DESC, votes DESC

來源

2016-07-26 10:09:21

你需要指定DESC兩個列。否則，默認情況下它是ASC

SELECT c.*, r.votes, c.total_comments, 
     ROUND(sumrate/votes) AS average 
FROM catalog c LEFT JOIN 
    (SELECT r.object_id, COUNT(*) as votes, SUM(r.rate) as sumrate 
     FROM ratings r 
     GROUP BY r.object_id 
    ) r 
    ON r.object_id = c.catalog_id LEFT JOIN 
    (SELECT c.catalog_id, COUNT(*) as total_comments 
     FROM comments c 
     GROUP BY c.catalog_id 
    ) c 
    ON c.catalog_id = c.catalog_id 
GROUP BY c.catalog_id 
ORDER BY average DESC, votes DESC;

來源

2016-07-26 10:11:32 Madhivanan

使用DESC的SQL查詢無法正常工作

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