Amazon Redshift 中 "x is true" 的布尔值评估不一致

Question

使用 Amazon Redshift。我有一个具有 3 个不同值的布尔字段 - true、false 和 null。

当我尝试计算 (x is true) 时，我收到一条 Not implemented 错误消息。不过，尝试评估 (x = true) 效果很好。

dataeng=# with bool_vals as (select distinct val::boolean as x from my_table) select * from bool_vals;
 x
---
 f

 t
(3 rows)

dataeng=# with bool_vals as (select distinct val::boolean as x from my_table) select *, (x is null) from bool_vals;
 x | ?column?
---+----------
 f | f
 t | f
   | t
(3 rows)

dataeng=# with bool_vals as (select distinct val::boolean as x from my_table) select *, (x is true) from bool_vals;
ERROR:  Not implemented
DETAIL:
  -----------------------------------------------
  error:  Not implemented
  code:      1001
  context:   'false' - project naming - 330
  query:     114596
  location:  cg_main.cpp:1265
  process:   padbmaster [pid=7556]
  -----------------------------------------------

dataeng=# with bool_vals as (select distinct val::boolean as x from my_table) select *, (x = true) from bool_vals;
 x | ?column?
---+----------
 t | t
 f | f
   |
(3 rows)

奇怪的是，我无法使用简单的示例案例重现它：

dataeng=# with bools as (select true as x union select false union select null) select *, (x is true) from bools;
 x | ?column?
---+----------
 f | f
 t | t
   | f
(3 rows)

我同意使用 = 作为计算布尔逻辑的运算符，但这种行为不一致似乎很奇怪，我想弄清楚这里发生了什么。

Answer 1

在 Redshift 中，查询的 SELECT 部分不支持 IS {TRUE|FALSE|UNKNOWN} 语法。仅支持作为 WHERE 谓词。

我们在文档中添加了一条注释和一个新示例来阐明这一点。 https://docs.aws.amazon.com/redshift/latest/dg/r_Boolean_type.html

Amazon Redshift 中 "x is true" 的布尔值评估不一致

Inconsistent boolean evaluation with "x is true" in Amazon Redshift

sql

amazon-web-services

amazon-redshift

boolean-expression