如果我不知道它有多少，如何在 postgres 中创建列？

Question

我的问题是：我有一个返回此表的查询：

Result of my query

解释表格：它显示了一些问题及其数字答案（数字 0 可以表示否，数字 1 可以表示是，数字 2 可以表示一点等等）我想要的是这张表变成这样：

What I want

所以它告诉我每个答案选项中有多少个答案。但是，响应选项的数量会有所不同。在此示例中，我有 0 到 4 的选项，但在某些情况下选项从 0 到 10。那么，如何为每个选项创建列？

我的查询是这样的：

SELECT 
    data,
    id,
    question,
    answer,
    COUNT(*) AS info2,
    COUNT(CASE WHEN answer IS NOT NULL THEN 1 END) AS info3,
    COUNT(CASE WHEN answer IS NULL THEN 1 END) AS info4
FROM table1
JOIN table2
    ON table1.id = table2.id
WHERE table1.variable_y BETWEEN '2020-01-01 00:00:00' AND '2020-05-31 23:59:59'
    AND variable_x = 'ABC123'
    AND variable_z = 'ABC'
GROUP BY 1,2,3,4

Answer 1

您可以执行动态查询。最好的选择是使用编程语言，但也可以使用 sql 语言。

我做了一个简化的例子：

示例数据：

CREATE TABLE t (
answer int
);

INSERT INTO t VALUES (0);
INSERT INTO t VALUES (1);
INSERT INTO t VALUES (2);
INSERT INTO t VALUES (3);
INSERT INTO t VALUES (4);
INSERT INTO t VALUES (3);
INSERT INTO t VALUES (2);
INSERT INTO t VALUES (2);

动态查询：

(SELECT 'SELECT ')

UNION ALL

(SELECT 
    'SUM(CASE WHEN answer = ' || answer || ' THEN 1 ELSE 0 END) AS res' || answer || ','
FROM t
GROUP BY answer
ORDER BY answer)

UNION ALL

(SELECT 'COUNT(*) AS info2,
    SUM(CASE WHEN answer IS NOT NULL THEN 1 ELSE 0 END) AS info3,
    SUM(CASE WHEN answer IS NULL THEN 1 ELSE 0 END) AS info4 
    FROM t;')

输出是你想要的查询：

SELECT 
  SUM(CASE WHEN answer = 0 THEN 1 ELSE 0 END) AS res0,
  SUM(CASE WHEN answer = 1 THEN 1 ELSE 0 END) AS res1,
  SUM(CASE WHEN answer = 2 THEN 1 ELSE 0 END) AS res2,
  SUM(CASE WHEN answer = 3 THEN 1 ELSE 0 END) AS res3,
  SUM(CASE WHEN answer = 4 THEN 1 ELSE 0 END) AS res4,
  COUNT(*) AS info2,
  SUM(CASE WHEN answer IS NOT NULL THEN 1 ELSE 0 END) AS info3,
  SUM(CASE WHEN answer IS NULL THEN 1 ELSE 0 END) AS info4 
FROM t;

查询结果（执行您的愿望查询的结果）：

res0	res1	res2	res3	res4	info2	info3	info4
1	1	3	2	1	8	8	0