我有一个密钥表，其中包含多个具有相同列名的数据表，以及一个用户表。我正在尝试从密钥表中选择值，将此输出与每个数据表中的一些选定值通过唯一ID（uaID）连接到已从密钥表中选择的条目（每个密钥表结果将具有相对的连接只有一个数据表范围，不是所有数据表和一些条目将返回数据表中的空结果，我们不希望这会破坏任何东西或因为null结果而被省略，然后最后将一些用户数据连接到users表中的每个结果。这将始终有结果。
让我＆＃34;画＆＃34;我的表格的基本版本，你可以看到。
keyTable
-----------------------------------------
| uaID | userID | key | appName |
|---------------------------------------|
| 1 | 7 | ABC01 | Physics |
| 2 | 9 | DEF20 | Geometry |
| 3 | 12 | XJG14 | Biology |
| 4 | 19 | DAF09 | Chemistry |
| 5 | 27 | KYT78 | Algebra |
| 6 | 29 | PLF43 | Statistics|
| 7 | 34 | COG89 | Geology |
| 8 | 45 | HYL72 | Art |
| 9 | 48 | TSK45 | History |
| 10 | 53 | BBS94 | GeoChem |
| 11 | 59 | DOD10 | BioChem |
| 12 | 27 | HKV62 | Music |
-----------------------------------------
dataTable01
-----------------------------------------------
| uaID | sector | subSector | topic |
|---------------------------------------------|
| 2 | circle | volumn | measure |
| 7 | triangle | hypotenuse |pythagoras |
| 8 | square | | |
| 11 | triangle | hypotenuse |pythagoras |
-----------------------------------------------
dataTable02
---------------------
| uaID | topic |
|-------------------|
| 1 | door |
| 3 | window |
| 9 | porch |
| 12 | |
---------------------
dataTable03
-----------------------------------------------
| uaID | sector | subSector | topic |
|---------------------------------------------|
| 4 | cat | feline | kitty |
| 5 | dog | canine | rover |
| 6 | kangaroo | marsupial | jack |
| 10 | bunny | leporidae | peter |
-----------------------------------------------
users
------------------------------------------------------------------------
| userID | Title | firstName | lastName | email |
|----------------------------------------------------------------------|
| 7 | Dr | Melissa | Smith | mel@email.com |
| 9 | Mr | Bob | Andrews | bob@email.com |
| 12 | Miss | Clare | Greco | clare@email.com |
| 19 | Mr | Dan | Fonseca | dan@email.com |
| 27 | Mr | Matt | Jones | matt@email.com |
| 29 | Mr | Chris | Nimmo | chris@email.com |
| 34 | Mrs | Lisa | Araujo | lisa@email.com |
| 45 | Miss | Raquel | Bailey | raquel@email.com |
| 48 | Dr | Steven | Dowd | steven@email.com |
| 53 | Prof | Roger | Hesp | roger@email.com |
| 59 | Prof | Sally | Bryce | sally@email.com |
| 65 | Mrs | Elena | Eraway | elena@email.com |
------------------------------------------------------------------------
这就是我想要达到的最终结果：
-------------------------------------------------------------------------------------------------------------------------------
| uaID | key | appName | sector | subSector | topic | title | firstName | lastName | email |
|-----------------------------------------------------------------------------------------------------------------------------|
| 1 | ABC01 | Physics | | | door | Dr | Melissa | Smith | mel@email.com |
| 2 | DEF20 | Geometry | circle | volumn | measure | Mr | Bob | Andrews | bob@email.com |
| 3 | XJG14 | Biology | | | window | Miss | Clare | Greco | clare@email.com |
| 4 | DAF09 | Chemistry | cat | feline | kitty | Mr | Dan | Fonseca | dan@email.com |
| 5 | KYT78 | Algebra | dog | canine | rover | Mr | Matt | Jones | matt@email.com |
| 6 | PLF43 | Statistics| kangaroo | marsupial | jack | Mr | Chris | Nimmo | chris@email.com |
| 7 | COG89 | Geology | triangle | hypotenuse |pythagoras | Mrs | Lisa | Araujo | lisa@email.com |
| 8 | HYL72 | Art | square | | | Miss | Raquel | Bailey | raquel@email.com |
| 9 | TSK45 | History | | | porch | Dr | Steven | Dowd | steven@email.com |
| 10 | BBS94 | GeoChem | bunny | leporidae | peter | Prof | Roger | Hesp | roger@email.com |
| 11 | DOD10 | BioChem | triangle | hypotenuse |pythagoras | Prof | Sally | Bryce | sally@email.com |
| 12 | HKV62 | Music | | | | Mr | Matt | Jones | matt@email.com |
-------------------------------------------------------------------------------------------------------------------------------
我试图通过执行：
来实现这一目标$sql = "SELECT keyTable.uaID, keyTable.userID, keyTable.key,
keyTable.appName, dataTable01.sector, dataTable01.subSector,
dataTable01.topic, dataTable02.topic, dataTable03.sector,
dataTable03.subSector, dataTable03.topic, users.title,
users.firstName, users.lastName, users.email
FROM keyTable
LEFT OUTER JOIN dataTable01 ON keyTable.uaID = dataTable01.uaID
LEFT OUTER JOIN dataTable02 ON keyTable.uaID = dataTable02.uaID
LEFT OUTER JOIN dataTable03 ON keyTable.uaID = dataTable03.uaID
LEFT OUTER JOIN users ON keyTable.userID = users.userID";
我获得了所有keyTable数据。我将所有用户数据放在它应该加入的所有地方。我也获得了所有dataTable03数据，但我没有从结果中显示dataTable01或dataTable02获得任何数据。如果我省略对dataTable03的调用，则会显示dataTable02显示的所有相关数据，但不会显示来自dataTable01的数据。对users表的调用结束时始终显示正常。很明显，数据表中匹配的字段名称存在问题。我没有得到任何错误，并且该过程完成，只是缺少所提到的数据。我试过了不同的JOINS - INNER JOIN，OUTER JOIN，LEFT OUTER JOIN。显然必须有一种方法来实现这一点，但似乎无法在网上找到任何关于这个特定问题的参考。有人可以告诉我，我做错了吗？
答案 0 :(得分：2)
加入后，您可以使用COALESCE从表中获取具有匹配行的非空值。
$sql = "SELECT k.uaID, k.userID, k.key, k.appName,
COALESCE(d1.sector, d3.sector, '') AS sector,
COALESCE(d1.subSector, d3.subSector, '') AS subSector,
COALESCE(d1.topic, d2.topic, d3.topic, '') AS topic,
users.title, users.firstName, users.lastName, users.email
FROM keyTable AS k
LEFT OUTER JOIN dataTable01 AS d1 ON k.uaID = d1.uaID
LEFT OUTER JOIN dataTable02 AS d2 ON k.uaID = d2.uaID
LEFT OUTER JOIN dataTable03 AS d3 ON k.uaID = d3.uaID
LEFT OUTER JOIN users ON k.userID = users.userID
ORDER BY k.uaID";
将datatablesNN表中的数据合并到同一列的另一种方法是使用UNION。
SELECT k.uaID, k.userID, k.key, k.appName, IFNULL(d.sector, '') AS sector, IFNULL(d.subSector, '') AS subSector, IFNULL(d.topic, '') AS topic,
u.title, u.firstName, u.lastName, u.email
FROM keyTable AS k
LEFT OUTER JOIN (
SELECT uaID, sector, subSector, topic
FROM dataTable01
UNION
SELECT uaID, NULL, NULL, topic
FROM datatable02
UNION
SELECT uaID, sector, subSector, topic
FROM datatable03) AS d
ON k.uaID = d.uaID
LEFT JOIN users AS u ON u.userID = k.userID
ORDER BY k.uaID
答案 1 :(得分：0)
你必须使用别名
这里的类似问题和解决方案：
php-mysql-how-to-resolve-ambiguous-column-names-in-join-operation
select * from common inner join (
(select link from table1)
union
(select link from table2)
) as unionT
on unionT.link = common.link