Question

我有这个mysql设置：

*words* table
+----+-----------+
| id |   word    |
+----+-----------+
| 1  |  curse1   |
| 2  |  curse2   |
| 3  | niceWord1 |

*Category* table
+----+-----------+
| id |    name   |
+----+-----------+
| 1  | niceWords |
| 2  | badWords  |

*link* table
+----+--------+-------+
| id | wordID | catID |
+----+-----------+----+
| 1  |    1   |   2   |
| 2  |    2   |   2   |
| 3  |    3   |   1   |

我想创建一个MySQL SELECT，我说我想用链接表选择所有坏词所以我会得到类似这样的输出：

+----------+------------+---------------+
| words.id | words.word | Category.name |
+----------+------------+---------------+
|    1     |  curse1    |   badWords    |
|    2     |  curse2    |   badWords    |

我想我需要使用某种类型的JOIN，我不太擅长SQL。

使用相同的Pastebin：http://pastebin.com/ML684RRA

Answer 1

select words.id, words.word, Category.name
from words
join link on link.wordid = words.id
join category on category.id = link.catid
where category.name = 'badWords'

Answer 2

只需通过相关ID加入表格，然后通过使用WHERE坏词ID = 2来过滤掉好词...我还在表名中添加了别名，因为这是一个很好的做法。

SELECT
    w.word,
    c.name,
    c.id
FROM words w
JOIN link l on l.wordID = w.id
JOIN Category c on c.id = l.catID
WHERE c.id = 2

如何从具有其他表的条件的表中进行选择

2 个答案: