要生成不存在的行，可以使用2个集合的CROSS JOIN（笛卡尔积）。这里我们需要将现有车主交叉连接到负特征。这确保了所有车主和所有负代码都存在于结果中。
由于存在一个正特征和一个匹配的负特征，我们需要确保如果当前表中存在一对中的任何一个，我们不会否定现有的正特征，反之亦然。因此，我们将该结果与现有的车主信息连接起来，以识别任何缺失的行，但也要使用not exists检查数据是否已经存在：

INSERT INTO car_features (Owner, Make, Model, Feature)
WITH all_combinations AS (
    SELECT o.Owner, o.Make, o.Model, c.Code as Feature
    FROM car_owners o
    CROSS JOIN car_features_codes c
    WHERE c.Code IN (2,4,6,8)
),
existing_features AS (
    SELECT o.Owner, o.Make, o.Model, f.Feature, CASE WHEN f.Feature % 2 = 0 THEN f.Feature - 1 ELSE f.Feature + 1 END as OppositeFeature
    FROM car_owners o
    JOIN car_features f ON o.Owner = f.Owner AND o.Make = f.Make AND o.Model = f.Model
)
SELECT ac.Owner, ac.Make, ac.Model, ac.Feature
FROM all_combinations ac
WHERE NOT EXISTS (
    SELECT 1 
    FROM existing_features ef 
    WHERE ac.Owner = ef.Owner
    AND ac.Make = ef.Make 
    AND ac.Model = ef.Model
    AND (ac.Feature = ef.Feature OR ac.Feature = ef.OppositeFeature)
    )

字符串
插图后的行：

+-------+---------+---------+---------+
| OWNER |  MAKE   |  MODEL  | FEATURE |
+-------+---------+---------+---------+
|     1 | Ford    | Mustang |       2 |
|     1 | Ford    | Mustang |       3 |
|     1 | Ford    | Mustang |       6 |
|     1 | Ford    | Mustang |       7 |
|     1 | Toyota  | Corolla |       2 |
|     1 | Toyota  | Corolla |       4 |
|     1 | Toyota  | Corolla |       6 |
|     1 | Toyota  | Corolla |       8 |
|     2 | Audi    | A7      |       2 |
|     2 | Audi    | A7      |       4 |
|     2 | Audi    | A7      |       6 |
|     2 | Audi    | A7      |       8 |
|     2 | Porsche | 911     |       2 |
|     2 | Porsche | 911     |       3 |
|     2 | Porsche | 911     |       5 |
|     2 | Porsche | 911     |       7 |
|     2 | Toyota  | Corolla |       2 |
|     2 | Toyota  | Corolla |       4 |
|     2 | Toyota  | Corolla |       6 |
|     2 | Toyota  | Corolla |       8 |
|     3 | Toyota  | Prius   |       1 |
|     3 | Toyota  | Prius   |       4 |
|     3 | Toyota  | Prius   |       5 |
|     3 | Toyota  | Prius   |       8 |
+-------+---------+---------+---------+

型
参见DB2 fiddle

nb此解决方案假设8个代码是成对的，奇数是正数，“+1”是匹配的负数。如果不是这样，那么您可能需要提供每个基本特征的正负“Map表”。