此查询将返回您请求的数据，但 s4 丢失。

SELECT a.ars, a.bill, b.codt, a.c4, b.s4
  FROM A a
  LEFT JOIN B b ON a.ars = b.ars AND a.bill = b.bill AND a.codt = b.codt and a.codt <> ''
  order by ars, bill, c4

这将返回：
| 阿尔斯|帐单|科特|碳四|s4|
| --------------|--------------|--------------|--------------|--------------|
| 九百|603121|零|一百二十三|零|
| 九百|603121|零|二百三十四|零|
| 九百|603121| PSP-123|三百四十五|ASD|
| 九百|603121|零|四五六|零|
| 九百|603121|零|五六七|零|
然后，当s4为空时，将使用inner join从 table B 获取 s4：

select DISTINCT s.ars, s.bill, s.codt, s.c4, case when s.s4 is not null then s.s4 else b.s4 end as s4 
from B b
inner join (
  SELECT a.ars, a.bill, b.codt, a.c4, b.s4
  FROM A a
  LEFT JOIN B b ON a.ars = b.ars AND a.bill = b.bill AND a.codt = b.codt and a.codt <> ''
) as s on s.ars = b.ars and s.bill = b.bill
where b.codt = ''
order by ars, bill, c4

结果：
| 阿尔斯|帐单|科特|碳四|s4|
| --------------|--------------|--------------|--------------|--------------|
| 九百|603121|零|一百二十三|公司简介|
| 九百|603121|零|二百三十四|公司简介|
| 九百|603121| PSP-123|三百四十五|ASD|
| 九百|603121|零|四五六|公司简介|
| 九百|603121|零|五六七|公司简介|
Demo here

如果codt codt是一个emtypoyt string = works，则只需要在没有找到匹配的情况下处理NULL

CREATE TABLE A (
  "raw_number" INTEGER,
  "ars" INTEGER,
  "bill" INTEGER,
  "codt" VARCHAR(7),
  "c4" INTEGER
);

INSERT INTO A
  ("raw_number", "ars", "bill", "codt", "c4")
VALUES
  ('1', '900', '603121','', '123'),
  ('2', '900', '603121', 'YPR-003', '234'),
  ('3', '900', '603121', 'Psp-123', '345'),
  ('4', '900', '603121', '', '456'),
  ('5', '900', '603121', '', '567');

CREATE TABLE

INSERT 0 5

CREATE TABLE B (
  "raw_number" INTEGER,
  "ars" INTEGER,
  "bill" INTEGER,
  "codt" VARCHAR(7),
  "s4" VARCHAR(3)
);

INSERT INTO B
  ("raw_number", "ars", "bill", "codt", "s4")
VALUES
  ('1', '900', '603121', 'Psp-123', 'asd'),
  ('2', '900', '603121', '', 'zxc'),
  ('3', '900', '455000', '', 'F');

CREATE TABLE

INSERT 0 3

SELECT a.ars, a.bill, a.codt, a.c4, COALESCE(b.s4,(SELECT s4 FROM B b1 WHERE a.ars = b1.ars AND a.bill = b1.bill and codt = ''))
FROM A a
LEFT JOIN B b ON
a.ars = b.ars AND a.bill = b.bill AND a.codt = b.codt

SELECT 5

fiddle t

正如我所理解的，你想将存在于A中但不存在于B中的codt值与来自B的空字符串codt匹配，你可以在加入之前检查codt在B中的存在性，如下所示：

SELECT a.ars, a.bill, b.codt, a.c4, b.s4
FROM 
  (
    select ars, bill, 
      case 
        when 
         exists(select 1 from B d where d.ars=t.ars and d.bill=t.bill and d.codt= t.codt) then codt 
        else '' 
      end as codt, c4
   from A t
  ) a
LEFT JOIN B b ON
a.ars = b.ars AND a.bill = b.bill AND a.codt = b.codt
ORDER BY c4

See demo