我试图从Joomla配方插件中导出数据，该插件已不再维护，并且在更复杂的部分出现问题。
我一直在努力研究这篇文章，其中详细介绍了一个与我尝试实现的过程类似的过程：http://greenash.net.au/thoughts/2012/05/flattening-many-to-many-fields-for-mysql-to-csv-export/
以下是我正在使用的表格：
jos_rr_recipes
+-----------+------+-------------+-----------------+---------------+------------+
| recipe_id | chef | title | introtext | portionstype | portions |
+-----------+------+-------------+-----------------+---------------+------------+
| 3 | 1 | Recipe ID 3 | Intro text ID 3 | 2 | 7-8 People |
| 6 | 1 | Recipe ID 6 | Intro text ID 6 | 2 | 3-4 People |
| 4 | 3 | Recipe ID 4 | Intro text ID 4 | 1 | 1 box |
| 5 | 3 | Recipe ID 5 | Intro text ID 5 | 1 | 10 wraps |
+-----------+------+-------------+-----------------+---------------+------------+
jos_rr_categories
+-------------+------------+-----------+-----------+
| category_id | pretty_url | title | parent_id |
+-------------+------------+-----------+-----------+
| 3 | 3-url | 3 Title | 0 |
| 198 | 198-url | 198 Title | 0 |
| 2 | 2-url | 2 Title | 0 |
| 5 | 5-url | 5 Title | 0 |
| 1 | 1-url | 1 Title | 0 |
| 169 | 169-url | 169 Title | 0 |
| 171 | 171-url | 171 Title | 0 |
| 149 | 149-url | 149 Title | 3 |
| 150 | 150--url | 150 Title | 3 |
| 151 | 151-url | 151 Title | 3 |
| 198 | 98-url | 98 Title | 3 |
| 201 | 201-url | 201 Title | 198 |
| 21 | 21--url | 21 Title | 198 |
| 100 | 100-url | 100 Title | 198 |
| 4 | 4-url | 4 Title | 169 |
| 80 | 80-url | 80 Title | 169 |
| 26 | 26-url | 26 Title | 2 |
| 213 | 213-url | 213 Title | 198 |
| 303 | 303-url | 303 Title | 171 |
| 11 | 11-url | 11 Title | 2 |
| 112 | 112-url | 112 Title | 2 |
| 231 | 231-url | 231 Title | 1 |
| 200 | 200-url | 200 Title | 1 |
| 181 | 181-url | 181 Title | 1 |
| 54 | 54-url | 54 Title | 3 |
| 195 | 195-url | 195 Title | 198 |
| 10 | 10-url | 10 Title | 198 |
| 226 | 226-url | 226 Title | 150 |
| 300 | 300-url | 300 Title | 150 |
+-------------+------------+-----------+-----------+
jos_rr_recipecategory
+-----------+-------------+
| recipe_id | category_id |
+-----------+-------------+
| 3 | 195 |
| 3 | 10 |
| 3 | 149 |
| 3 | 201 |
| 3 | 26 |
| 3 | 231 |
| 3 | 80 |
| 3 | 303 |
| 4 | 54 |
| 4 | 300 |
| 4 | 4 |
| 4 | 21 |
| 4 | 98 |
| 4 | 26 |
| 4 | 213 |
| 5 | 26 |
| 5 | 11 |
| 5 | 112 |
| 5 | 200 |
| 6 | 201 |
| 6 | 4 |
| 6 | 26 |
| 6 | 112 |
| 6 | 231 |
| 6 | 300 |
+-----------+-------------+
jos_sobi2_fields_data
+----+---------+--------------------+--------+
| id | fieldid | data_txt | itemid |
+----+---------+--------------------+--------+
| 4 | 7 | chef1@email.com | 1 |
| 5 | 13 | Description Chef 1 | 1 |
| 28 | 7 | chef3@email.com | 3 |
| 32 | 13 | description chef 3 | 3 |
+----+---------+--------------------+--------+
有些类别是多层深层的，但是没有办法将它输入到我所知道的WordPress中，所以我很乐意将它们作为group_concat放入单独的列中，输出中的每个主标头。
有父母，我将在WordPress中导入他们自己的单独分类法：
Course/Dish = 105
Meals = 199
Main Ingredient = 2
Difficulty = 5
Cuisine = 1
Equipment = 169
Rainbow = 171
这是我到目前为止所拥有的：
SELECT
recipes.recipe_id AS ID,
recipes.title AS Title,
chef.data_txt AS Chef,
recipes.introtext AS Description,
IFNULL(IF (recipes.portionstype = 1 , recipes.portions,''),'') AS Yield,
IFNULL(IF (recipes.portionstype = 2 , recipes.portions,''),'') AS Servings
FROM
jos_rr_recipes AS recipes
JOIN jos_sobi2_fields_data AS chef
ON recipes.chef = chef.itemid AND chef.fieldid = 7
GROUP BY recipes.recipe_id
现在我不知道如何将每个分类法都纳入自己的专栏。我没有成功尝试以下内容：
SELECT
recipes.recipe_id AS ID,
recipes.title AS Title,
chef.data_txt AS Chef,
recipes.introtext AS Description,
IFNULL(IF (recipes.portionstype = 1 , recipes.portions,''),'') AS Yield,
IFNULL(IF (recipes.portionstype = 2 , recipes.portions,''),'') AS Servings,
IFNULL(tax_ingr.ingredients,'') AS "Taxonomy Ingredients"
FROM (
SELECT
GROUP_CONCAT(cat.title SEPARATOR '|') AS ingredients
FROM jos_rr_recipecategory AS catlink
LEFT JOIN jos_rr_categories AS cat
ON catlink.category_id = cat.category_id AND cat.parent_id = 2
GROUP BY recipe_id
) AS tax_ingr,
jos_rr_recipes AS recipes
JOIN jos_sobi2_fields_data AS chef
ON recipes.chef = chef.itemid AND chef.fieldid = 7
GROUP BY recipes.recipe_id
这只需要很长时间来查询，并且无法提供预期结果。
3天后，我无法理解这一点。 ：（
非常感谢任何帮助或指导。
答案 0 :(得分：0)
好的，我抓住了这个并想出了一个解决方案。一旦我理解了使用子查询的概念，它实际上是相当直接的。
我头脑中的灯泡是：
每个子查询基本上都成为它在查询中使用的临时表，由它的别名引用。换句话说，我们在每个子查询中的表之间创建链接，然后在SELECT语句中使用它们的输出数据，就像它是普通表一样。
如果有人对下面的最终查询感兴趣，则由于使用不同的数据集而命名略有不同的albiet。
SELECT
recipe.recipe_id AS ID,
recipe.title AS Title,
LOWER(IFNULL(recipe.chef,'nochef@aaronportbury.com')) AS Chef,
recipe.description AS Description,
recipe.yield,
recipe.servings,
IFNULL(ingr_link.val,'') AS 'Ingredients',
IFNULL(course_link.val,'') AS 'Course',
IFNULL(meal_link.val,'') AS 'Meal',
IFNULL(dif_link.val,'') AS 'Difficulty',
IFNULL(cuis_link.val,'') AS 'Cuisine',
IFNULL(equi_link.val,'') AS 'Equipment',
FROM (
SELECT
r.recipe_id,
r.title,
c.data_txt AS chef,
r.introtext AS description,
IFNULL(IF (r.portionstype = 1 , r.portions,''),'') AS yield, # Checks to see type from Portionstype. If 1, then it's a Yield. Else leave empty
IFNULL(IF (r.portionstype = 2 , r.portions,''),'') AS servings, # Checks to see type from Portionstype. If 2, then it's a Serving. Else leave empty
FROM jos_rr_recipes AS r
LEFT JOIN jos_sobi2_fields_data AS c
ON r.chef = c.itemid AND c.fieldid = 7
) AS recipe
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val # Uses Group_Concat function to combine all Ingredients related to the recipe ID and seperates them with ;;
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id AND cat.parent_id = 2
GROUP BY rc.recipe_id
) AS ingr_link
ON recipe.recipe_id = ingr_link.rid
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id
AND (cat.parent_id = 3
OR cat.parent_id = 23
OR cat.parent_id = 26
OR cat.parent_id = 28
OR cat.parent_id = 29
OR cat.parent_id = 30
OR cat.parent_id = 59
OR cat.parent_id = 211) # Here I just found all the parent ids and created multiple OR tests to collate them.
GROUP BY rc.recipe_id
) AS course_link
ON recipe.recipe_id = course_link.rid
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id AND cat.parent_id = 198
GROUP BY rc.recipe_id
) AS meal_link
ON recipe.recipe_id = meal_link.rid
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id AND cat.parent_id = 5
GROUP BY rc.recipe_id
) AS dif_link
ON recipe.recipe_id = dif_link.rid
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id AND cat.parent_id = 1
GROUP BY rc.recipe_id
) AS cuis_link
ON recipe.recipe_id = cuis_link.rid
LEFT JOIN (
SELECT
rc.recipe_id AS rid,
IFNULL(GROUP_CONCAT(cat.title SEPARATOR ';;'),'') AS val
FROM jos_rr_recipecategory AS rc
LEFT JOIN jos_rr_categories AS cat
ON rc.category_id = cat.category_id AND cat.parent_id = 169
GROUP BY rc.recipe_id
) AS equi_link
ON recipe.recipe_id = equi_link.rid
GROUP BY recipe.recipe_id
我只是使用多个连接的子查询来将内容过滤到我需要的内容，然后针对我想要的每个类别重复。
希望这有助于其他人。