条件 SQL 计数

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?表格需要按一列分组吗?
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.

我见过:

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 的值可以是 0、1 或 2.

where question1 can have a value of either 0, 1 or 2.

我也看到过使用 count(CASE WHEN question1 = 0 THEN 1)

但是，随着 question1 的可能值数量的增加，编写起来会变得更加麻烦.有没有一种方便的方法来编写这个查询，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，使用聚合 FILTER 选项.通常是最干净和最快的:

In Postgres 9.4 or later, use the aggregate FILTER option. Typically cleanest and fastest:

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:

• 具有附加(不同)过滤器的聚合列

如果你想要它短:

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;

更多语法变体:

• 对于绝对性能，SUM 是更快还是 COUNT?

crosstab() 产生最佳性能并且对于长选项列表更短:

crosstab() yields the best performance and is shorter for long 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

这篇关于条件 SQL 计数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！