在数组中查找值（Postgresql）答案

【问题标题】：Find value in array (Postgresql)在数组中查找值（Postgresql）
【发布时间】：2019-02-11 20:02:14
【问题描述】：

以下返回“{3130,3135,2179,3186}”，输入“INTEGER[]”：

SELECT
array_agg(c) AS c
FROM
(
SELECT 3130::INTEGER AS c
UNION
SELECT 3135::INTEGER AS c
UNION
SELECT 2179::INTEGER AS c
UNION
SELECT 3186::INTEGER AS c
) AS v

如果我尝试通过直接将文本转换为整数数组来查找数字“3186”，它就可以正常工作：

SELECT 3186::INTEGER = ANY
(
'{3130,3135,2179,3186}'::INTEGER[]
)

但如果我尝试对第一个查询的结果执行相同操作，我会收到“operator does not exist: integer = integer[]”错误消息：

SELECT 3186::INTEGER = ANY
(
    SELECT
    array_agg(c) AS c
    FROM
    (
    SELECT 3130::INTEGER AS c
    UNION
    SELECT 3135::INTEGER AS c
    UNION
    SELECT 2179::INTEGER AS c
    UNION
    SELECT 3186::INTEGER AS c
    ) AS v
)

我做错了什么？

我正在使用 PostgreSQL 9.5.3，PgAdmin 1.22.1

【问题讨论】：

标签： sql arrays postgresql select

【解决方案1】：

ANY() 有两种形式。在您的查询中，它们是混合的。

如果你想使用ANY (array expression):

SELECT 3130 = ANY(
    ARRAY(
        SELECT 3130::INTEGER AS c
        UNION
        SELECT 3135::INTEGER AS c
        UNION
        SELECT 2179::INTEGER AS c
        UNION
        SELECT 3186::INTEGER AS c
    ) 
)

或

SELECT 3130 = ANY(array_agg(c))
FROM (
    SELECT 3130::INTEGER AS c
    UNION
    SELECT 3135::INTEGER AS c
    UNION
    SELECT 2179::INTEGER AS c
    UNION
    SELECT 3186::INTEGER AS c
    ) alias

如果你想使用ANY (subquery):

SELECT 3130 = ANY(
    SELECT 3130::INTEGER AS c
    UNION
    SELECT 3135::INTEGER AS c
    UNION
    SELECT 2179::INTEGER AS c
    UNION
    SELECT 3186::INTEGER AS c
)

【讨论】：

【解决方案2】：

您可以显式转换为INTEGER[]：

SELECT 3186::INTEGER = ANY(
(
    SELECT array_agg(c) AS c
    FROM (
       SELECT 3130::INTEGER AS c
       UNION
       SELECT 3135::INTEGER AS c
       UNION
       SELECT 2179::INTEGER AS c
       UNION
       SELECT 3186::INTEGER AS c
    ) AS v
)::INTEGER[]);

DBFiddle Demo

【讨论】：