Question

我正在使用类别的嵌套集。我理解查找深度和叶节点等的基本原理。

我正在制作的网站是多语言的，因此需要根据用户选择的语言显示类别名称。网址是不变的。

简化表格如下......

分类

+----+-----+-----+-------------------------+
| id | lft | rgt | url                     |
+----+-----+-----+-------------------------+
|  1 |   1 |  12 | top_level               |
|  2 |   2 |   5 | foo                     |
|  3 |   3 |   4 | foo_sub_cat             |
|  4 |   6 |  11 | bar                     |
|  5 |   7 |   8 | bar_sub_cat_1           |
|  6 |   9 |  10 | bar_sub_cat_2           |
+----+-----+-----+-------------------------+

Category_Info：

+-------------+---------+----------------------------+
| category_id | lang_id | name                       |
+-------------+---------+----------------------------+
|          1  |       1 | One cat to rule them all   |
|          2  |       1 | Foo Cat                    |
|          3  |       1 | Subcategory of Foo         |
|          4  |       1 | Bar Cat                    |
|          5  |       1 | Bar SubCat                 |
|          6  |       1 | Another Bar SubCat         |
+-------------+---------+----------------------------+

我运行的查询是这样的......

SELECT node.*, 
       category_info.name, 
       ( Count(parent.url) - 1 ) AS depth 
FROM   categories AS node, 
       categories AS parent 
       JOIN category_info 
         ON parent.id = category_info.category_id 
WHERE  node.lft BETWEEN parent.lft AND parent.rgt 
       AND category_info.lang_id = 1 
GROUP  BY node.url 
ORDER  BY node.lft

在上面的示例中，返回的是每个结果的名称One cat to rule them all。这是你所期望的，因为我匹配了parent.id。

JOIN正在杀死我。如果我尝试JOIN node.id = category_info.category_id，那么我会收到错误告诉我找不到node.id，就像我使用JOIN categories.id = category_info.category_id一样。

我知道我必须亲近，但我真的无法理解。

Answer 1

你能试试吗？

SELECT node.*, category_info.name, (COUNT(parent.url) - 1) AS depth
FROM categories AS node INNER JOIN
                        ^^^^^^^^^^^ <= this is added
    categories AS parent
JOIN category_info ON node.id = category_info.category_id
WHERE node.lft BETWEEN parent.lft AND parent.rgt AND category_info.lang_id = 1
GROUP BY node.url
ORDER BY node.lft

如果缺少JOIN，则无法在ON子句中引用该列，但可以在WHERE部分

中引用该列

这是我尝试过的。

mysql> SELECT * FROM t1 tab1 INNER JOIN t1 tab2 INNER JOIN t2 tab3 ON tab1.a = tab3.a;
Empty set (0.00 sec)

mysql> SELECT * FROM t1 tab1,  t1 tab2 INNER JOIN t2 tab3 ON tab1.a = tab3.a;
ERROR 1054 (42S22): Unknown column 'tab1.a' in 'on clause

将MySQL嵌套集连接到另一个表

1 个答案: