Question

我的MySQL表保存引用。每行都是一个引用，如：

A = {citer，B = cited（即A引用的B）。

我知道如何获得（1）经常引用A的数字和（2）经常引用A的数字：

/* (1) who cited A most often? */
SELECT citer,COUNT(citer) AS citations1 FROM `table` WHERE cited='A' GROUP BY citer ORDER BY citations1 DESC

/* (2) whom did A cite most often? */
SELECT cited,COUNT(cited) AS citations2 FROM `table` WHERE citer='A' GROUP BY cited ORDER BY citations2 DESC

现在我想要得到这两个统计值的总和（citations1 + citations2），这样我就知道谁对A的引用链接最多。

示例：如果B引用A五（5）次，A引用B三（3）次，则A-B链接的总和为八（8）。

使用MySQL公式可以吗？谢谢您的帮助！

Answer 1

您可以这样写：

select person, (sum(citers) + sum(citeds)) as total
from ((select citer as person, count(*) as citers, 0 as citeds
       from citations
       where cited = 'A'
       group by citer
      ) union all
      (select cited, 0, count(*) as citeds
       from citations
       where citer = 'A'
       group by cited
      )
     ) c
group by person
order by total desc;

这个问题有点棘手。如果您尝试使用join，则将排除引用链接最多的人只是“公民”或“被引用”的可能性。

Answer 2

您可以join的结果：

select t1.citer as person, t1.citations1 + t2.citations2 as result
from
(
SELECT citer,COUNT(citer) AS citations1 FROM `table` WHERE cited='A' GROUP BY citer ORDER BY citations1 DESC
) t
join
(
SELECT cited,COUNT(cited) AS citations2 FROM `table` WHERE citer='A' GROUP BY cited ORDER BY citations2 DESC
) t2
on t1.citer = t2.cited

Answer 3

我在子查询中使用了UNION，然后使用了行的总和

SELECT other, SUM(citations)
FROM (
   SELECT citer other,COUNT(*) AS citations 
   FROM citations 
   WHERE cited='A'        
   GROUP BY citer 
   UNION
   SELECT cited, COUNT(*)
   FROM citations 
   WHERE citer='A' 
   GROUP BY cited) AS uni
GROUP BY other

MySQL：两个COUNT的总和（按不同的值）

3 个答案: