Question

我有一个规范化的数据库，我正在尝试使用JOIN和GROUP_CONCAT从多个表返回数据。

问题：行与GROUP_CONCAT重复。我不能使用DISTINCT，因为某些数据（成分mfr）确实需要重复。

这是我当前的查询和数据库结构（SQL Fiddle）：

SELECT recipe.*, 
GROUP_CONCAT(recipe_detail.ingredient_id) AS iid,  
GROUP_CONCAT(ingredient.name) AS iname, 
GROUP_CONCAT(ingredient_mfr.abbr) AS mabbr, 
GROUP_CONCAT(recipe_tag.name) AS tag
FROM  recipe
LEFT JOIN recipe_detail
    ON recipe.id = recipe_detail.recipe_id
LEFT JOIN ingredient
    ON recipe_detail.ingredient_id = ingredient.id
LEFT JOIN ingredient_mfr
    ON ingredient.mfr_id = ingredient_mfr.id
LEFT JOIN recipe_tagmap
    ON recipe.id = recipe_tagmap.recipe_id
LEFT JOIN recipe_tag
    ON recipe_tagmap.tag_id = recipe_tag.id
WHERE recipe.user_id = 1
GROUP BY recipe.id

recipe
+------------+------------+-----------+
|    id      |    name    |  user_id  |
+============+============+===========+
|     1      |  Test123   |     1     |
+------------+------------+-----------+
|     2      |  Test456   |     1     |
+------------+------------+-----------+
|     3      |  Test789   |     1     |
+------------+------------+-----------+

recipe_detail
+------------+---------------+
| recipe_id  | ingredient_id |
+============+===============+
|     1      |      193      |
+------------+---------------+
|     1      |      194      |
+------------+---------------+
|     2      |       16      |
+------------+---------------+
|     3      |      277      |
+------------+---------------+

ingredient
+------------+---------------+---------+
|     id     |      name     |  mfr_id |
+============+===============+=========+
|     16     |       Gin     |    4    |
+------------+---------------+---------+
|     193    |       Fig     |    3    |
+------------+---------------+---------+
|     194    |       Tea     |    3    |
+------------+---------------+---------+
|     277    |       Nut     |    2    |
+------------+---------------+---------+

ingredient_mfr
+------------+------------+
|    id      |    abbr    |
+============+============+
|     2      |    TFA     |
+------------+------------+
|     3      |    FA      |
+------------+------------+
|     4      |    LOR     |
+------------+------------+

recipe_tag
+------------+------------+
|    id      |    name    |
+============+============+
|     1      |    one     |
+------------+------------+
|     2      |    two     |
+------------+------------+
|     3      |    three   |
+------------+------------+
|     4      |    four    |
+------------+------------+
|     5      |    five    |
+------------+------------+
|     6      |    six     |
+------------+------------+
|     7      |    seven   |
+------------+------------+
|     8      |    eight   |
+------------+------------+
|     9      |    nine    |
+------------+------------+

recipe_tagmap
+------------+---------------+---------+
|     id     |   recipe_id   |  tag_id |
+============+===============+=========+
|     1      |       1       |    1    |
+------------+---------------+---------+
|     2      |       1       |    2    |
+------------+---------------+---------+
|     3      |       1       |    3    |
+------------+---------------+---------+
|     4      |       2       |    4    |
+------------+---------------+---------+
|     5      |       2       |    5    |
+------------+---------------+---------+
|     6      |       2       |    6    |
+------------+---------------+---------+
|     7      |       3       |    7    |
+------------+---------------+---------+
|     8      |       3       |    8    |
+------------+---------------+---------+
|     9      |       3       |    9    |
+------------+---------------+---------+

使用我当前的查询，我的结果如下所示：

+------+---------+--------------+----------- ----+---------------+------------------+
|  id  |  name   |      iid     |     iname      |    mabbr      |       tag        |
+======+=========+==============+================+===============+==================+
|   1  | Test123 | 193,193,193, | Fig, Fig, Fig, | FA, FA, FA,   | one, two, three, |
|      |         | 194,194,194  | Tea, Tea, Tea  | FA, FA, FA    | one, two, three  |
+------+---------+--------------+----------------+---------------+------------------+
|   2  | Test456 | 16,16,16     | Gin, Gin, Gin  | LOR, LOR, LOR | four, five six   |
+------+---------+--------------+----------------+---------------+------------------+
|   3  | Test789 | 277,277,277  | Nut, Nut, Nut  | TFA, TFA, TFA | seven,eight,nine |
+------+---------+--------------+----------------+---------------+------------------+

我希望我的结果如下：

+------+---------+--------------+----------- ----+---------------+------------------+
|  id  |  name   |      iid     |     iname      |    mabbr      |       tag        |
+======+=========+==============+================+===============+==================+
|   1  | Test123 |   193, 194   |    Fig, Tea    |    FA, FA     | one, two, three, |
+------+---------+--------------+----------------+---------------+------------------+
|   2  | Test456 |      16      |      Gin       |     LOR       | four, five six   |
+------+---------+--------------+----------------+---------------+------------------+
|   3  | Test789 |     277      |      Nut       |     TFA       | seven,eight,nine |
+------+---------+--------------+----------------+---------------+------------------+

如您所见，多个标签的存在会导致配料数据重复。多种成分的存在导致标签复制。我曾试图使用DISTINCT，但有时我会有多种成分，其中每一种都会返回它自己的“mabbr”，这可能与它的其他成分相同（参见预期结果的第一行）。使用DISTINCT，它只返回该“mabbr”的一个实例。

我可以对我的查询进行更改以实现我想要做的事情吗？

SQL Fiddle

Answer 1

您可以通过将tag分组提取到自己的子查询来解决此问题：

SELECT
    recipe.*,
    GROUP_CONCAT(recipe_detail.ingredient_id) AS iid,
    GROUP_CONCAT(ingredient.name) AS iname,
    GROUP_CONCAT(ingredient_mfr.abbr) AS mabbr,
    (
      SELECT GROUP_CONCAT(recipe_tag.name)
        FROM recipe_tag
          INNER JOIN recipe_tagmap
            ON recipe_tagmap.tag_id = recipe_tag.id
        WHERE recipe_tagmap.recipe_id = recipe.id
     ) AS tag

  FROM recipe
    LEFT JOIN recipe_detail
      ON recipe.id = recipe_detail.recipe_id
    LEFT JOIN ingredient
      ON recipe_detail.ingredient_id = ingredient.id
    LEFT JOIN ingredient_mfr
      ON ingredient.mfr_id = ingredient_mfr.id

  WHERE recipe.user_id = 1
  GROUP BY recipe.id

（例如fiddle）

Answer 2

在进行GROUP_CONCAT时添加唯一性将为您提供唯一的值。

SELECT recipe.*, 
GROUP_CONCAT(distinct recipe_detail.ingredient_id) AS iid,  
GROUP_CONCAT(distinct ingredient.name) AS iname, 
GROUP_CONCAT(distinct ingredient_mfr.abbr) AS mabbr, 
GROUP_CONCAT(distinct recipe_tag.name) AS tag
FROM  recipe
LEFT JOIN recipe_detail
    ON recipe.id = recipe_detail.recipe_id
LEFT JOIN ingredient
    ON recipe_detail.ingredient_id = ingredient.id
LEFT JOIN ingredient_mfr
    ON ingredient.mfr_id = ingredient_mfr.id
LEFT JOIN recipe_tagmap
    ON recipe.id = recipe_tagmap.recipe_id
LEFT JOIN recipe_tag
    ON recipe_tagmap.tag_id = recipe_tag.id
WHERE recipe.user_id = 1
GROUP BY recipe.id

SQL Fiddle

MySQL - GROUP_CONCAT返回重复数据，不能使用DISTINCT

2 个答案: