条件SQL计数 [英] Conditional SQL count

问题描述

创建计算表中数据出现次数的列的最佳方法是什么?该表需要按一列分组吗?
我的数据库是PostgreSQL.

What is the best way to create columns which count the number of occurrences of data in a table? The table needs to be grouped by one column?
My database is PostgreSQL.

我看过:

SELECT
    sum(CASE WHEN question1 = 0 THEN 1 ELSE 0 END) AS ZERO,
    sum(CASE WHEN question1 = 1 THEN 1 ELSE 0 END) AS ONE,
    sum(CASE WHEN question1 = 2 THEN 1 ELSE 0 END) AS TWO,
    category
FROM reviews
    GROUP BY category

其中 question1 的值可以为 0 ， 1 或 2 .

我还看到了使用 count(在问题1 = 0然后是1时的情况)

但是，随着 question1 可能值的增加，编写此代码变得更加麻烦.是否有一种方便的方式来编写此查询，可能会优化性能?

However, this becomes more cumbersome to write as the number of possible values for question1 increases. Is there a convenient way to write this query, possibly optimizing performance?

推荐答案

在Postgres 9.4或更高版本中使用汇总 FILTER 选项:

Use the aggregate FILTER option in Postgres 9.4 or later:

SELECT category
     , count(*) FILTER (WHERE question1 = 0) AS zero
     , count(*) FILTER (WHERE question1 = 1) AS one
     , count(*) FILTER (WHERE question1 = 2) AS two
FROM   reviews
GROUP  BY 1;

FILTER 子句的详细信息:

• 聚合带有其他(不同)过滤器的列

如果要短:

SELECT category
     , count(question1 = 0 OR NULL) AS zero
     , count(question1 = 1 OR NULL) AS one
     , count(question1 = 2 OR NULL) AS two
FROM   reviews
GROUP  BY 1;

可能的变体概述:

• 对于绝对性能，SUM是更快还是COUNT?

crosstab()可获得最佳性能，并且对于较长的选项列表而言较短:

crosstab() yields the best performance and is shorter for longer lists of options:

SELECT * FROM crosstab(
     'SELECT category, question1, count(*) AS ct
      FROM   reviews
      GROUP  BY 1, 2
      ORDER  BY 1, 2'
   , 'VALUES (0), (1), (2)'
   ) AS ct (category text, zero int, one int, two int);

详细说明:

• PostgreSQL交叉表查询

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