在數組中查找值（Postgresql）

Question

以下命令返回“ {3130,3135,2179,3186} ”，鍵入“ 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

如果我試圖通過直接將文本轉換為整數數組來找到數字“3186”，它可以正常工作：

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

但是如果我嘗試對第一個查詢的結果做同樣的操作，我得到一個“ 運算符不存在：integer = integer [] ”錯誤消息：

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
)

我究竟做錯了什么？

我正在使用PostgreSQL 9.5.3，PgAdmin 1.22.1

Answer 1

有兩種形式的ANY() 。 在您的查詢中，它們是混合的。

如果要使用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
    ) 
)

要么

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

如果要使用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
)

Answer 2

您可以顯式轉換為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演示