SQL自定义函数不按预期返回数据，方法是匹配我的条件。

Question

我正在尝试创建这个函数，但是空值失败。

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，有些条件不起作用，特别是当我传递空调用时，没有给出预期结果的情况是：

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

在我一无所获的时候，我期待着返回(“复习”，-2)。

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)

当我什么都没得到的时候，我期待着返回(，-2)。

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

期望在我得到(“没有匹配”，-4)时返回('IN_PROGRESS'，11)。

简单地说，以下是我试图实现的SQL函数

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

并根据条件返回以下内容

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

我只解释空的情况。您的代码按预期工作。如果你不确定。只需复制粘贴所有工作在sql小提琴。

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)

会回来

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

因为:此执行满足谓词：

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

因为它满足谓词

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