SQL 选择时根据大小写填充列中的空值

Question

所以基本上我想要实现的是，如果四列为空，一列包含 1 作为值，一列包含 0，那么我想在一列中输入满足条件的默认值 (0.06077 ).

含义：如果 COL_A、COL_B、COL_C、COL_D 为 NULL 并且 COL_E = 0 AND COL_F = 0 那么用默认值 0.06077 填充 COL_A 中满足这些条件的行。

我尝试了以下方法：

, CASE WHEN COL_A IS NULL AND COL_B IS NULL AND COL_C IS NULL AND COL_D IS NULL 
AND COL_E = 0 AND COL_F = 1 THEN NVL(COL_A, 0.06077) END

以下确实填写了符合我设定条件的词条。有人知道潜在的解决方案吗？

Answer 1

IF COL_A, COL_B, COL_C, COL_D IS NULL and COL_E = 0 AND COL_F = 0 then fill the rows in COL_A which meet these conditions with a default value of 0.06077.

您可能需要这个 case 表达式：

case when
  coalesce(COL_A, COL_B, COL_C, COL_D) is null and COL_E = 0 AND COL_F = 1 
  then  0.06077
  else COL_A end as COL_A

例子

with dt as (
select null COL_A, null COL_B, null COL_C, null COL_D, 0 COL_E, 1 COL_F from dual union all
select 1 COL_A, null COL_B, null COL_C, null COL_D, 0 COL_E, 1 COL_F from dual union all
select null COL_A, null COL_B, null COL_C, null COL_D, 0 COL_E, 0 COL_F from dual
)
select 
case when
  coalesce(COL_A, COL_B, COL_C, COL_D) is null and COL_E = 0 AND COL_F = 1 
  then  0.06077
  else COL_A end as COL_A, 
COL_B, COL_C, COL_D, COL_E, COL_F
from dt;

     COL_A COL_B COL_C COL_D      COL_E      COL_F
---------- ----- ----- ----- ---------- ----------
,06077                                0          1
         1                            0          1
                                      0          0

SQL 选择时根据大小写填充列中的空值

SQL fill empty values in a column based on case when selection

sql

oracle

plsql

case