Postgresql SQL查询不显示任何结果

我创建了一个视图名-author\u article\u log，如下所示：

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我的工作是获取一天中请求错误的百分比， 我的问题是：

select i.date, 
       ((count (case when i.status = '404 NOT FOUND'
                    then 1
                    else 0
                end)
        * 1.0
        / count (case when i.date = j.date
                      then 1
                      else 0
                 end)
        ) *100
       ) as percentage
from  author_article_log as i, author_article_log as j
group by i.date, j.date
order by percentage desc;

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运行此查询既不会返回任何结果也不会返回任何错误，是否有人可以告诉我发生这种情况的原因？

您可以尝试使用

WITH

子句筛选您的请求

WITH errors AS (
  SELECT date, COUNT(*) AS nb_errs
  FROM author_article_log
  WHERE status = '404 NOT FOUND'
  GROUP BY date
), total AS (
  SELECT date, COUNT(*) AS nb_queries
  FROM author_article_log
  GROUP BY date
)
SELECT e.date, 100.0 * e.nb_errs / t.nb_queries AS percentage
FROM errors e
JOIN total t ON e.date = t.date
ORDER BY percentage DESC

文档：

• 你不需要自我加入
• 您需要

  sum（）

  ，而不是

  count（）

  注意：

  count（0）

  等于

  1

• DATE

  是一个关键字；最好不要将其用作列名

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我不确定是否可以在

COUNT

调用中设置

大小写

，但我想说的是

SUM

函数在您的查询中更合适。顺便问一下：

从author\u article\u log as I，author\u article\u log as j

您是否意识到这两个表引用没有以任何方式连接起来，产生一个Carthesian产品？@joop:我在我的查询中添加了一个join语句，看起来像：从author\u article\u log作为j加入author\u article\u log作为j on I.date=j.date，即使它什么也不返回。

WITH errors AS (
  SELECT date, COUNT(*) AS nb_errs
  FROM author_article_log
  WHERE status = '404 NOT FOUND'
  GROUP BY date
), total AS (
  SELECT date, COUNT(*) AS nb_queries
  FROM author_article_log
  GROUP BY date
)
SELECT e.date, 100.0 * e.nb_errs / t.nb_queries AS percentage
FROM errors e
JOIN total t ON e.date = t.date
ORDER BY percentage DESC

SELECT zdate,
       100.0 * SUM (CASE WHEN status = '404 NOT FOUND' THEN 1 ELSE 0 END)
        / COUNT(*) AS percentage
FROM  author_article_log
GROUP BY zdate
ORDER BY percentage DESC
        ;