postgresql - sql:如何在HAVING子句之后从一列 bool 值中选择具有真值的行

HI有3个产品表，每个表有3列，分别是customer name、boolean optout和blacklist。Having子句之后，每个客户名称将有3行（假设他拥有所有3个产品）。
如果任何布尔列包含true，如何输出true。我用下面的cast操作解决了这个问题，但是我认为应该有一个更优雅的解决方案。

SELECT customer_name,
       cast(int4(sum(cast(optout     As int4))) As Boolean) As optout,
       cast(int4(sum(cast(blacklist  As int4))) As Boolean) As blacklist
FROM
(SELECT * FROM product1
UNION SELECT * FROM product2
UNION SELECT * FROM product3) AS temp1
GROUP BY customer_name, optout, blacklist
HAVING optout=true or blacklist=true;

最佳答案

尝试bool_or聚合函数，听起来就像您正在寻找的：

SELECT customer_name,
       bool_or(optout)    As optout,
       bool_or(blacklist) As blacklist
FROM
(SELECT * FROM product1
UNION SELECT * FROM product2
UNION SELECT * FROM product3) AS temp1
GROUP BY customer_name, optout, blacklist
HAVING optout=true or blacklist=true;

关于postgresql - sql:如何在HAVING子句之后从一列 bool 值中选择具有真值的行，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/5700269/