Question

我有一个包含两个表，链接和巡视的数据库。游览被定义为

+----+-------------+-----+-----+--------+---------+-------+---------+
| ID | FileName    | Lat | Lon | Name   | Heading | Width | Height  |
+----+-------------+-----+-----+--------+---------+-------+---------+

和链接定义为

+-----+------------+--------+
| ID  | locationID | linkID |
+-----+------------+--------+

links.locationID是tour.ID的外键。我想要做的是获取与特定tour.name

匹配的所有链接

现在，我必须运行这两个查询并使用第一个ID作为第二个

的变量

SELECT * FROM tour WHERE tour.name = 'name'

SELECT * FROM tour
  INNER JOIN links ON tour.ID = links.linkID 
  WHERE links.locationID = ID of first query

如何将其改编为单一搜索？

我尝试了一个子查询：

SELECT * FROM tour 
  INNER JOIN links ON tour.ID = links.linkID 
  WHERE links.locationID = ( SELECT * FROM tour WHERE tour.name = 'name')

这给了我一个错误1241，说操作数应该包含一列。

我查了一些子查询的例子，但是他们都缺少我正在使用的内部连接语句。此查询需要进行哪些修改？

Answer 1

SELECT links.*, tour.*
FROM links
JOIN tour ON tour.ID = links.linkID
JOIN tour AS location ON location.ID = links.locationID -- you must create an alias, in order to refer to this table without ambiguity
WHERE location.name = 'name';

对于带子查询的版本，你可以这样做：

SELECT * FROM tour 
INNER JOIN links ON tour.ID = links.linkID 
WHERE links.locationID = (
    SELECT ID -- "operand should contain one column"
    FROM tour WHERE tour.name = 'name'
);

但更喜欢第一种选择。

构造查询以基于单个字段获取所有匹配

1 个答案: