Question

我想解决这个错误。我在MySQL工作，需要在公共列上连接两个派生表。两个表中的所有列都是表格别名。

Scheme是

stops (id, name)

route (num, company, pos, stop)其中

stops.id＆lt; ---＆gt; route.stop

这些表包含城市之间的公交路线，我想弄清楚从'Craiglockhart'到'Sighthill'的所有路线，这些路线只需要两条公交车（即转机点）。此外，因为route表不包含停止名称（只有ID），所以我们在派生表中使用一些连接来引用名称停止;这只是一个方便）

所以我试着制作两个派生表。其中一条路线可以吸引人们在'Craiglockhart'和任何不是'Craighartlock'的车站之间以及一个相同的第二个衍生桌而是'Sighthill'。我能够让这两个表单独工作。

但是，当我试图沿着所谓的转运站的公共列（即第一个表中的非Craighartlock站和第二个表中的非Sighthill站）加入它们时，我收到错误。

派生表1 ：这样可以正常工作并正确返回表。在这里，stop_b.name代表不是Craiglockhart的转移站，stop_a.name代表Craighill“起始”站。

SELECT * FROM 
    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Craiglockhart' AND 
                          stop_b.name <> 'Craiglockhart') AS first_route

派生表2 ：相同但不同的表别名和站限制。这里stop_b.name代表转运站（不是Sighthill），stop_a.name代表Sighthill“结束”站。

SELECT * FROM 
    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Sighthill' AND 
                          stop_b.name <> 'Sighthill') AS second_route

但是，当我尝试将它们连接到它们共同的`name_transfer'列（两者中的stop_b.name的别名）时，我收到一个错误：

SELECT * FROM
(
SELECT * FROM 
    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Craiglockhart' AND 
                          stop_b.name <> 'Craiglockhart') AS first_route
JOIN

(SELECT * FROM 
    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Sighthill' AND 
                          stop_b.name <> 'Sighthill') AS second_route)

ON (first_route.name_transfer = second_route.name_transfer)
)

我还尝试用ON替换USING (name_transfer)，因为我想要加入派生表的列在两个派生表中都被称为。

非常感谢任何帮助！

Answer 1

在FROM或JOIN子句中使用子查询时，需要围绕子查询使用括号。您可以在FROM子句中使用它，但不能在JOIN子句中使用它。当您在每个子查询周围添加不必要的SELECT * FROM SELECT时，问题似乎就出现了。

SELECT * FROM 
    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Craiglockhart' AND 
                          stop_b.name <> 'Craiglockhart') AS first_route
JOIN

    (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
            stop_b.name AS name_transfer FROM
               route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                    JOIN stops stop_a ON (a.stop=stop_a.id)
                    JOIN stops stop_b ON (b.stop=stop_b.id)
                    WHERE stop_a.name = 'Sighthill' AND 
                          stop_b.name <> 'Sighthill') AS second_route

ON first_route.name_transfer = second_route.name_transfer

Answer 2

试试这个。

SELECT * FROM
    (SELECT * FROM 
        (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
                stop_b.name AS name_transfer FROM
                   route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                        JOIN stops stop_a ON (a.stop=stop_a.id)
                        JOIN stops stop_b ON (b.stop=stop_b.id)
                        WHERE stop_a.name = 'Craiglockhart' AND 
                              stop_b.name <> 'Craiglockhart') AS first_route) AS A
    JOIN

    (SELECT * FROM 
        (SELECT a.num AS num_a, a.company AS comp_a, stop_a.name AS name_a, 
                stop_b.name AS name_transfer FROM
                   route a JOIN route b ON (a.company=b.company AND a.num=b.num)
                        JOIN stops stop_a ON (a.stop=stop_a.id)
                        JOIN stops stop_b ON (b.stop=stop_b.id)
                        WHERE stop_a.name = 'Sighthill' AND 
                              stop_b.name <> 'Sighthill') AS second_route)B
    ON (A.name_transfer = B.name_transfer)

尝试加入两个派生表时出错

2 个答案: