使用 SELF JOIN 从 LEFT JOIN 中删除重复行

Question

在 SQLite 中，我想从此结果中删除重复项（gist，sqlime.org demo）：

有些功能适用于所有产品，有些功能仅适用于某些产品。结果有不正确的行，如第二个

2024   B12XBE  A1C B1  B1 hardened steal   UT gehärteter Stahl

。两个 feature_text 都应以

UT

开头。但它包含重复的行。如何避免重复行？我想了解底层逻辑（笛卡尔积或类似）。

样品：

DROP TABLE IF EXISTS product_features;
DROP TABLE IF EXISTS feature_text;
CREATE TABLE product_features (
    year INTEGER, -- (4)
    product TEXT, -- (6)
    feature TEXT -- (3)
);
CREATE TABLE feature_text (
    year INTEGER, -- (4)
    product_group TEXT,
     -- (2) usually matches the first two digits of product
     -- but is sometimes empty / NULL or blank ' '
    feature TEXT, -- (3)
    language INTEGER, -- (2)
    description1 TEXT -- up to 40
);

INSERT INTO product_features (year, product, feature) VALUES
    (2024, 'B12XBE', 'A1C'), (2024, 'B12XBE', '7B0'),
    (2024, 'B12XBE', 'DUL'), (2024, 'UTB2EF', 'A1C'),
    (2024, 'UTB2EF', '7B0'), (2024, 'UTB2EF', 'DUL'),
    (2024, 'X9Y8Z7', 'DUH'), (2024, 'X9Y8Z7', '7B0');

INSERT INTO feature_text
 (year, product_group, feature, language, description1)
VALUES
  (2024, 'B1', 'A1C', 1, 'B1 hardened steel'),
  (2024, 'B1', '7B0', 1, 'B1 diamond tip'),
  (2024, 'B1', 'DSP', 1, 'display 1 inch'),
  (2024, 'B1', 'DSP', 1, 'Anzeige 1,5 cm'),
  (2024, 'UT', 'A1C', 1, 'UT hardened steel'),
  (2024, 'UT', '7B0', 1, 'UT diamond tip'),
  (2024, 'UT', 'DSP', 1, 'display 1,5 inch'),
  (2024, 'UT', 'DSP', 1, 'Anzeige 2,25 cm'),
  (2024, ' ', 'DUL', 1, '10mm for light duty'),
  (2024, 'X9', '7B0', 1, 'X9 diamond tip'),
  (2024, ' ', 'DUH', 1, '13mm for heavy duty'),
  (2024, 'B1', 'A1C', 2, 'B1 gehärteter Stahl'),
  (2024, 'B1', '7B0', 2, 'B1 Diamant Spitze'),
  (2024, 'UT', 'A1C', 2, 'UT gehärteter Stahl'),
  (2024, 'UT', '7B0', 2, 'UT Diamant Spitze'),
  (2024, ' ', 'DUL', 2, '10mm für leichte Aufgaben'),
  (2024, 'X9', '7B0', 2, 'X9 Diamant Spitze'),
  (2024, ' ', 'DUH', 2, '13mm für schwere Aufgaben'),
  (2024, NULL, 'DSP', 1, 'display'),
  (2024, NULL, 'DSP', 1, 'Anzeige');

我的做法：

WITH prft_en as (
SELECT
pf.year, pf.product, pf.feature,
COALESCE(ft1.product_group, ft2.product_group) AS product_group,
COALESCE(ft1.description1, ft2.description1) AS feature_text_en,
COALESCE(ft1.language, ft2.language) AS language_en
  FROM product_features pf
    LEFT JOIN feature_text  ft1
                 ON pf.year = ft1.year
                 AND pf.feature = ft1.feature
                 AND substr(pf.product, 1,2) = ft1.product_group
    LEFT JOIN feature_text ft2
                 ON pf.year = ft2.year
                 AND pf.feature = ft2.feature
                 AND ft2.product_group is ' ' OR ft2.product_group is NULL
WHERE ft1.language = 1 OR ft2.language = 1
), prft_de AS (
SELECT
pf.year, pf.product, pf.feature,
COALESCE(ft1.product_group, ft2.product_group) AS product_group,
COALESCE(ft1.description1, ft2.description1) AS feature_text_de,
COALESCE(ft1.language, ft2.language) AS language_de
  FROM product_features pf
    LEFT JOIN feature_text  ft1
                 ON pf.year = ft1.year
                 AND pf.feature = ft1.feature
                 AND substr(pf.product, 1,2) = ft1.product_group
    LEFT JOIN feature_text ft2
                 ON pf.year = ft2.year
                 AND pf.feature = ft2.feature
                 AND ft2.product_group is ' ' OR ft2.product_group is NULL
WHERE ft1.language = 2 OR ft2.language = 2
)

查询CTE：

SELECT prft_en.year, prft_en.product,
       prft_en.feature, prft_en.product_group,
       prft_en.feature_text_en, prft_de.feature_text_de
FROM prft_en
     LEFT JOIN prft_de
     ON prft_en.year = prft_de.year
     AND prft_en.feature = prft_de.feature
     AND SUBSTR(prft_en.product, 1,2) = prft_en.product_group

Answer 1

我认为整个 CTE 的东西确实不是你想要的。我认为您想要每个产品一行，所以从那开始

select
    f.*,
    substring(f.product, 1, 2) as product_group
from product_features f

然后对于每个产品，您想要获取相应的描述（如果存在），这可以简单地在英语子查询列中完成：

    (
        select description1
        from feature_text t
        where f.year = t.year
            and (product_group = t.product_group or t.product_group = ' ' or t.product_group is null)
            and f.feature = t.feature
            and language = 1
    ) as feature_text_en

对于德国人来说：

    (
        select description1
        from feature_text t
        where f.year = t.year
            and (product_group = t.product_group or t.product_group = ' ' or t.product_group is null)
            and f.feature = t.feature
            and language = 2
    ) as feature_text_de

这是 两者都在 sqlime 上 并生成此结果表：

年	产品	特点	产品组	feature_text_en	feature_text_de
2024	B12XBE	A1C	B1	B1硬化偷窃	B1 斯塔尔
2024	B12XBE	7B0	B1	B1钻石尖头	B1 钻石斯皮策
2024	B12XBE	DUL	B1	10mm 适用于轻型	10mm 轻量级
2024	UTB2EF	A1C	UT	UT硬化窃取	UT gehärteter Stahl
2024	UTB2EF	7B0	UT	UT金刚石刀尖	UT 钻石斯皮茨
2024	UTB2EF	DUL	UT	10mm 适用于轻型	10mm 轻量级
2024	X9Y8Z7	呃	X9	13mm 适用于重型	13mm für schwere Aufgaben
2024	X9Y8Z7	7B0	X9	X9钻石尖头	X9 钻石斯皮策

Answer 2

我想您所需要的只是按复杂条件进行左连接。像这样：

select 
    product_features.year,
    product_features.product,
    product_features.feature,
    substr(product_features.product, 1, 2) AS product_group,
    text_en.description1 AS en,
    text_de.description1 AS de
FROM product_features
LEFT JOIN feature_text AS text_en
ON text_en.year = product_features.year
AND text_en.feature = product_features.feature
AND (substr(product_features.product, 1, 2) = text_en.product_group OR text_en.product_group IS NULL OR TRIM(text_en.product_group) = '')
AND text_en.language = 1
LEFT JOIN feature_text AS text_de
ON text_de.year = product_features.year
AND text_de.feature = product_features.feature
AND (substr(product_features.product, 1, 2) = text_de.product_group OR text_de.product_group IS NULL OR TRIM(text_de.product_group) = '')
AND text_de.language = 2

请注意，如果为单个产品和产品组提供不同的描述，您无论如何都会得到重复的内容。如果您需要解决此类重复，那么每个 LEFT JOIN 都需要我制作 2 次（对于产品和组），然后应使用 COALESCE 进行适当的描述。

结果

编辑：示例如何避免后一种重复

select 
    product_features.year,
    product_features.product,
    product_features.feature,
    substr(product_features.product, 1, 2) AS product_group,
    COALESCE(text_en.description1, text_en_gen.description1) AS en,
    COALESCE(text_de.description1, text_de_gen.description1) AS de
FROM product_features
LEFT JOIN feature_text AS text_en
ON text_en.year = product_features.year
AND text_en.feature = product_features.feature
AND (substr(product_features.product, 1, 2) = text_en.product_group)
AND text_en.language = 1
LEFT JOIN feature_text AS text_en_gen
ON text_en_gen.year = product_features.year
AND text_en_gen.feature = product_features.feature
AND (text_en_gen.product_group IS NULL OR TRIM(text_en_gen.product_group) = '')
AND text_en_gen.language = 1
LEFT JOIN feature_text AS text_de
ON text_de.year = product_features.year
AND text_de.feature = product_features.feature
AND (substr(product_features.product, 1, 2) = text_de.product_group)
AND text_de.language = 2
LEFT JOIN feature_text AS text_de_gen
ON text_de_gen.year = product_features.year
AND text_de_gen.feature = product_features.feature
AND (text_de_gen.product_group IS NULL OR TRIM(text_de_gen.product_group) = '')
AND text_de_gen.language = 2