Question

我的查询有问题，应该执行以下操作：

获取特定记录并计算特定时间段的某些值
在另一段时间内计算这些关键字的相同值

这应该发生在1个查询中。我能够写它但是SUM（）返回的错误值远高于正常值。我认为这是因为LEFT JOIN。

SELECT SQL_CALC_FOUND_ROWS table1.id, table1.KeywordId, table1.AccountName, table1.CampaignName, table1.AdGroupName, table1.Keyword, table1.MatchType, SUM(table1.Spend)/SUM(table1.Clicks) AS AverageCpc, SUM(table1.Impressions) AS Impressions, (SUM(table1.Clicks)*table1.revenue_price)/SUM(table1.Impressions) AS Ctr, SUM(table1.Impressions*table1.AveragePosition)/SUM(table1.Impressions) AS AveragePosition, SUM(table1.Clicks) AS Clicks, SUM(table1.Spend) AS Spend, SUM(table1.free_joins) AS FreeJoins, SUM(table1.paid_joins) AS PaidJoins, SUM(table1.paid_joins)*table1.revenue_price AS Revenue, (SUM(table1.paid_joins)*table1.revenue_price)-SUM(table1.Spend) AS Profit, (SUM(table1.paid_joins)*table1.revenue_price)/SUM(table1.Clicks) AS RevPerClick, table1.CurrentMaxCpc, SUM(table2.Impressions) AS Impressions_chg, SUM(table2.Clicks) AS Clicks_chg, SUM(table2.Impressions*table2.AveragePosition)/SUM(table2.Impressions) AS AveragePosition_chg, (SUM(table2.Clicks)*table2.revenue_price)/SUM(table2.Impressions) AS Ctr_chg, SUM(table2.Spend)/SUM(table2.Clicks) AS AverageCpc_chg, table2.CurrentMaxCpc as CurrentMaxCpc_chg, SUM(table2.free_joins) AS FreeJoins_chg, SUM(table2.paid_joins) AS PaidJoins_chg
            FROM keywords_stats_google_naughtymeetings as table1
            LEFT JOIN keywords_stats_google_naughtymeetings as table2
            ON table1.keywordId = table2.keywordId
            WHERE table1.timeperiod >= '2012-05-21 00:00:00' and table1.timeperiod <= '2012-05-27 00:00:00'
            AND table2.timeperiod >= '2012-05-14' and table2.timeperiod <= '2012-05-20'

            GROUP BY table1.KeywordId, table1.MatchType, table1.revenue_price, table2.KeywordId, table2.MatchType, table2.revenue_price
            ORDER BY  FreeJoins
                    asc
            LIMIT 0, 10

有人可以告诉我如何获得正确的SUM结果吗？

Answer 1

我认为你需要INNER JOIN。尝试将LEFT JOIN替换为INNER JOIN。

P.S。我没有得到你想要的，但我认为这个想法应该更简单。

(SELECT id, fields_for_first_period_of_time
  FROM keywords_stats_google_naughtymeetings) t1
  JOIN
(SELECT id, fields_for_second_period_of_time
  FROM keywords_stats_google_naughtymeetings) t2
ON t1.id = t2.id

这只是这个想法的草图。我的意思是通过两个单独的查询获得结果。然后加入他们。这将更容易调试。我希望这可以帮到你。

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