不解析函数datediff（未知，不带时区的时间戳，不带时区的时间戳）

Question

SELECT "reviewedAt", "createdAt", DATEDIFF('hour', "createdAt"::timestamp, "reviewedAt"::timestamp) as hours_approved from "yadda$prod"."Application" 

错误[42883]错误：不存在函数datediff（未知，没有时区的时间戳，没有时区的时间戳）提示：没有函数与给定的名称和参数类型匹配。您可能需要添加显式类型转换。位置：36

Answer 1

尝试此：

SELECT 
"reviewedAt",
"createdAt",
DATE_PART('day', "reviewedAt"::timestamp - "createdAt"::timestamp) * 24 + DATE_PART('hour', "reviewedAt"::timestamp - "createdAt"::timestamp) AS hours_approved 
FROM "yadda$prod"."Application"

Answer 2

另一个解决方案：

SELECT 
    "reviewedAt",
    "createdAt",
    (EXTRACT(EPOCH FROM "reviewedAt"::timestamp - "createdAt"::timestamp)/3600)::int2 AS hours_approved
FROM "yadda$prod"."Application";