我正在尝试根据我拥有的架构构建一个 View 。但现在我完全不知道从哪里开始。我的 View 应包含有关玩家(链接)监护人、监护人的电话号码以及他们当前参加的球队的信息。如果有人能帮助我走上正确的道路，我将非常感激。请参阅附加图片的链接
谢谢!
Im hoping for this kind of result
+----------+--------------+-------------+----------+-----------+-----------+------------+
|guardianID|guardian.fName|guardianlName|Contact |playerfName|playerlName|teamName |
+----------+--------------+-------------+----------+------------+-----------+-----------+
| 1 |Dora | Kruger |0288235556|Daphne |Kruger |Taeiri Eels |
| 2 |Tonia | Wesley |0204156513|Brigitta |Wesley |Southern Mag|
| 3 |Shanene | Genovese |0211915599|Stefan |Genovese |Kaikorai |
| 4 |Jacob | Jefferson |0282316812|Yvonne |Jefferson |Pirate Skull|
| 5 |Tamika | Longstaff |0264217492|Todd |Longstaff |Zingari |
+----------+--------------+------------+-----------+-----------+-----------+------------+
我有以下数据
====PERSON TABLE====
+----------+--------------+-------------+----------+-----------+
|personID |firstName |lastName |addressID |photo |
+----------+--------------+-------------+----------+-----------+
| 1 |Dora | Kruger |1 |NULL |
| 2 |Daphne | Kruger |2 |NULL |
| 3 |Tonia | Wesley |3 |NULL |
| 4 |Brigitta | Wesley |4 |NULL |
| 5 |Shanene | Genovese |5 |NULL |
| 6 |Stefan | Genovese |6 |NULL |
| 7 |Jacob | Jefferson |7 |NULL |
| 8 |Yvibbe | Jefferson |8 |NULL |
| 9 |Tamika | Longstaff |9 |NULL |
| 10|Todd | Longstaff |10 |NULL |
+----------+--------------+------------+-----------+-----------+
====PHONE TABLE====
+----------+--------------+
|phoneID |primaryContact|
+----------+--------------+
| 1 |0288235556 |
| 2 |0204156513 |
| 3 |0211915599 |
| 4 |0282316812 |
| 5 |0264217492 |
+----------+--------------+
====GUARDIAN TABLE====
+----------+--------------+-------------+----------+
|guardianID|personID |phoneID |email |
+----------+--------------+-------------+----------+
| 1 | 1 | 1 |@ya.com |
| 2 | 3 | 2 |@yy.com |
| 3 | 5 | 3 |@ll.com |
| 4 | 7 | 4 |@pp.com |
| 5 | 9 | 5 |@no.com |
+----------+--------------+------------+-----------+
====PLAYER TABLE====
+----------+--------------+-------------+--------------+
|playerId |personID |schoolID |dob |
+----------+--------------+-------------+--------------+
| 1 | 2 | 1 |2008-04-06 |
| 2 | 4 | 2 |2011-11-19 |
| 3 | 6 | 3 |2011-01-07 |
| 4 | 8 | 4 |2006-01-01 |
| 5 | 10 | 5 |2004-04-06 |
+----------+--------------+-------------+--------------+
====FAMILY TABLE====
+----------+--------------+
|playerID |guardianID |
+----------+--------------+
| 1 |1 |
| 2 |2 |
| 3 |3 |
| 4 |4 |
| 5 |5 |
+----------+--------------+
====TEAM TABLE====
+----------+------------------+-------------+--------------+--------------+
|teamID |teamName |year |minAge |maxAge |
+----------+------------------+-------------+--------------+--------------+
| 1 | Taieri Eels | 2015-01-01 |5 |6 |
| 2 | Southern Magpies| 2014-01-01 |7 |8 |
| 3 | Kaikorai | 2016-01-01 |7 |10 |
| 4 | Pirate Skulls | 2014-01-01 |11 |12 |
| 5 | Zingari | 2013-01-01 |12 |16 |
+----------+------------------+-------------+--------------+--------------+
====TEAM ALLOCATION TABLE==== // allocating team for each player
+----------+--------------+
|teamID |playerID |
+----------+--------------+
| 2 |1 |
| 5 |2 |
| 1 |3 |
| 3 |4 |
| 4 |5 |
+----------+--------------+
更新: 监护人表用于识别玩家的监护人。我们试图使用家庭表将玩家与其监护人联系起来，因为这种关系是多对多的。然而，我们正在努力编写查询。我在想这是一个三表连接并使用别名？但我无法让它工作。
SELECT p1.firstName AS guardianFName, p1.lastName AS guardianLName,
p2.firstName AS playerFName, p2.lastName AS playerLName
from person AS p1, person as p2
join guardian on
person.personID = guardian.personID join
family on guardian.guardianID = family.guardianID
join player on
person.personID = player.personID join
family on player.playerID = family.playerID;
我尝试过一些这样的变体
DDL 和 DML 脚本位于 http://pastebin.com/euVanc2e对于那些想要帮助使用我们拥有的数据测试查询的人。
最佳答案
规范化您的架构会大大简化事情，但这最好作为一个单独的问题来处理。
创建您需要的复杂联接的最佳方法是从简单联接中构建它，并在每个步骤中进行测试以确保您走在正确的轨道上。在这种情况下，对于缺乏经验的开发人员来说，唯一真正的障碍是需要加入 person 两次；这可以通过使用表别名来解决。
让我们分解一下。
当然，第一步是将guardian加入person以获取监护人的姓名:
SELECT g.guardianId AS guardianId,
p1.firstName AS guardianFName, p1.lastName AS guardianLName
FROM guardian g
INNER JOIN person p1
ON g.personId = p1.personId;
我将 person 别名为 p1 而不仅仅是 p 的原因很快就会清楚。
接下来，我们对 player 和 person 执行相同的操作来获取玩家的姓名:
SELECT pr.playerId AS playerId,
p2.firstName AS playerFName, p2.lastName AS playerLName
FROM player pr
INNER JOIN person p2
ON pr.personId = p2.personId;
再次，我将 person 别名为 p2 而不仅仅是 p。最终查询不会显示 playerId 列；此处仅用于说明目的。
现在我们想要将监护人的名字与玩家的名字相匹配。这意味着对于结果中的每一行，我们需要从 person 读取两行。我们通过使用两个不同的别名创建两个对 person 的引用来做到这一点。我们只需通过 family 表连接前两个步骤的结果，如下所示:
SELECT g.guardianId AS guardianId,
p1.firstName AS guardianFName, p1.lastName AS guardianLName,
p2.firstName AS playerFName, p2.lastName AS playerLName
FROM guardian g
INNER JOIN person p1
ON g.personId = p1.personId
INNER JOIN family f
ON g.guardianId = f.guardianId
INNER JOIN player pr
ON f.playerId = pr.playerId
INNER JOIN person p2
ON pr.personId = p2.personId;
请注意对 person 的第二个引用，别名为 p2。这给出了对该表的第二个引用，允许结果包含来自两个引用中不同行的数据。
获取电话号码只需一步加入电话即可。
获取团队名称需要分两步加入 team_allocation，然后加入 team。您只需不从 team_allocation 返回任何列即可。
完整的查询应如下所示:
SELECT g.guardianId AS guardianId,
p1.firstName AS guardianFName, p1.lastName AS guardianLName,
ph.primaryContact AS contact,
p2.firstName AS playerFName, p2.lastName AS playerLName,
t.teamName AS teamName
FROM guardian g
INNER JOIN person p1
ON g.personId = p1.personId
INNER JOIN family f
ON g.guardianId = f.guardianId
INNER JOIN player pr
ON f.playerId = pr.playerId
INNER JOIN person p2
ON pr.personId = p2.personId
INNER JOIN phone ph
ON g.phoneId = ph.phoneId
INNER JOIN team_allocation ta
ON ta.playerId = pr.playerId
INNER JOIN team t
ON ta.teamId = t.teamId;
希望有帮助。
关于mysql - 需要在我的数据库 sql 中创建 View 方面的帮助，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37636404/