我正在尝试合并 4 个表，以获取服装项目的相应数据，在为每个项目选择时类似这样..
Array(
[0] => array ([item] => 1, [article] => 10, [layer1] => 6, [layer2] => 8, [layer3] => 9, [layer4] => 10),
[1] => array ([item] => 2, [article] => 5, [layer1] => 3, [layer2] => 4, [layer3] => 5, [layer4] => 0/null),
[2] => array ([item] => 3, [article] => 7, [layer1] => 7, [layer2] => 0/null etc),
[3] => array ([item] => 4, [article] => 1, [layer1] => 1, [layer2] => 2, [layer3] => 0/null etc))
我不确定如何嵌套和连接这些表以获得正确的数据，但是(不幸的是)所有 4 个表都是存储各种排序和预览数据所必需的，最远的表在树下(article_layers，和层)需要多行连接。骨架表如下所示。
items table
item_id item_article item_name
1 10 red dress
2 5 green polo
3 7 jeans
4 1 black leather jacket
5 10 black dress
articles table
article_id article_name
1 jacket
5 shirt
7 pants
10 dress
article_layers table
id article_id layer_id
1 1 1
2 1 2
3 5 3
4 5 4
5 5 5
6 7 7
7 10 6
8 10 8
9 10 9
10 10 10
layers table
layer_id layer_name
1 jacket_right_sleeve
2 jacket_left_sleeve
3 shirt_right_sleeve
4 shirt_left_sleeve
5 shirt_torso
6 dress_torso
7 pants
8 dress_left_sleeve
9 dress_right_sleeve
10 dress_skirt
我试过了
SELECT items.*, articles.*, article_layers.*, layers.*,
(SELECT * FROM layers where article_layers.layer_id = layers.layer_id) as layer1,
FROM items
JOIN articles ON items.article_id = articles.article_id
JOIN article_layers ON articles.article_id = article_layers.article_id
和许多类似的查询，但我找不到神奇的公式来获取我需要的数据。感谢您的帮助。
最佳答案
试试这个:
SELECT *
FROM items i
INNER JOIN articles a ON i.article_id = a.article_id
INNER JOIN article_layers al ON a.article_id = al.article_id
INNER JOIN layers l ON al.layer_id = l.layer_id
关于mysql - 使用连接嵌套子查询，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21087132/