SQL 自定义函数未通过匹配我的条件按预期返回数据

Question

我正在尝试创建此函数，但空值失败

CREATE OR REPLACE function func(inspection_date date, strike_time date, taskaction_display varchar, last_known_status varchar, severity int)
RETURNS TABLE(last_known_task_status varchar, taskaction int) AS $$
    SELECT CASE
        WHEN inspection_date IS null THEN (taskaction_display, severity)
        WHEN strike_time >= inspection_date THEN (taskaction_display, severity)
        WHEN ((strike_time BETWEEN inspection_date - INTERVAL '6' MONTH AND inspection_date) AND last_known_status IS NOT null) THEN (last_known_status, 
             CASE
                WHEN last_known_status = 'IN_PROGRESS' THEN 11
                WHEN last_known_status = 'PENDING' THEN 12
                WHEN last_known_status = 'COMPLETE' THEN 13
                WHEN last_known_status = '' THEN -1
                ELSE -2
             END)
        WHEN ((strike_time BETWEEN inspection_date::date - INTERVAL '6' MONTH AND inspection_date) AND last_known_status IS null) THEN ('REVIEWED'::text, -3)
        WHEN (strike_time < inspection_date - INTERVAL '6' MONTH) THEN ('REVIEWED'::text, -2)
        ELSE ('NO MATCH'::text, -4)
    END
$$
language sql stable;

但是有些条件不工作，特别是当我传递 null 时没有给出预期结果的调用是：

SELECT * FROM func('2022-03-22', '2022-01-22',
'strike in six months before inspection + last_known_status = null',
null, 20)

以上期望为 return ('REVIEWED', -2) 而我一无所获

SELECT * FROM func(null, '2021-01-22',
'strike time older than inspection date by more than 6 months but inspection_date is null',
'IS_NULL', 11)

以上期望为 return ( , -2) 而我一无所获

SELECT * FROM func('2022-03-22', '2021-01-22',
'strike in six months before inspection + last_known_status = IN_PROGRESS',
'IN_PROGRESS', 20)

期望以上 return ('IN_PROGRESS', 11) 当我得到 ("NO MATCH", -4)

简而言之，以下是我想要实现的目标接受

的 SQL 函数

func(inspection date, strike date, action varchar, status varchar, severity int)

和return根据条件

if(inspection == null) return (action, severity)
else if(strike >= inspection) return (action, severity)
else if(strike >= inspection-6 months) {
    if(status == null) return ('REVIEWED', -1)
    else if(status == 'IN_PROGRESS') return (status, 11)
    else if(status == 'PENDING') return (status, 12)
    else if(status == 'COMPLETE') return (status, 13)
    else ('NO STATUS MATCH', -2)
}
else if(strike < inspection-6 months) return ('REVIEWED', -3)
else return ('NO MATCH', -4)

Answer 1

我只解释 null 的情况。您的代码按预期工作。如果你不确定。只需复制粘贴 sql fiddle.
中的所有工作 demo

SELECT * FROM func(null, '2021-01-22',
'strike time older than inspection date by more than 6 months but inspection_date is null',
'IS_NULL', 11)

将return

                               last_known_task_status                                  | taskaction
------------------------------------------------------------------------------------------+------------
 strike time older than inspection date by more than 6 months but inspection_date is null |         11

因为：此执行符合谓词：

WHEN inspection_date IS null THEN (taskaction_display, severity)

SELECT * FROM func('2022-03-22', '2022-01-22',
'strike in six months before inspection + last_known_status = null',
null, 20);

会得到

 last_known_task_status | taskaction
------------------------+------------
 REVIEWED               |         -2

因为它符合谓词

WHEN ((strike_time BETWEEN inspection_date::date - INTERVAL '6' MONTH AND inspection_date)
                  AND last_known_status IS null) THEN ('REVIEWED'::text, -2)

SQL 自定义函数未通过匹配我的条件按预期返回数据

SQL Custom Function is not returning data as expected by matching to my conditions

sql

postgresql

date

function