Php 将两个子表与一个不重复的父表联接 问题

Php 将两个子表与一个不重复的父表联接 问题,php,mysql,sorting,join,Php,Mysql,Sorting,Join,我有三张桌子:人、电话和电子邮件。每个人都有一个唯一的ID,每个人可以有多个号码或多封电子邮件 它看起来是这样的: +---------+----------+ | ID | Name | +---------+----------+ | 5000003 | Amy | | 5000004 | George | | 5000005 | John | | 5000008 | Steven |

我有三张桌子:人、电话和电子邮件。每个人都有一个唯一的ID,每个人可以有多个号码或多封电子邮件

它看起来是这样的:

    +---------+----------+
    | ID      | Name     |
    +---------+----------+
    | 5000003 | Amy      |
    | 5000004 | George   |
    | 5000005 | John     |
    | 5000008 | Steven   |
    | 8000009 | Ashley   |
    +---------+----------+

    +---------+-----------------+
    | ID      | Number          |
    +---------+-----------------+
    | 5000005 | 5551234         |
    | 5000005 | 5154324         |
    | 5000008 | 2487312         |
    | 8000009 | 7134584         |
    | 5000008 | 8451384         |
    +---------+-----------------+

+---------+------------------------------+
| ID      | Email                        |
+---------+------------------------------+
| 5000005 | Smithley@goodmail.com.com    |
| 5000005 | Smithley.j@gmail.com         |
| 5000008 | Smithley@gmail.com           |
| 5000008 | tech@goodmail.com            |
| 5000008 | feler@campus.uni.com         |
| 8000009 | Ashley.hill86@gmail.com      |
| 5000004 | georgestanko@hotmail.com     |
+---------+------------------------------+
SELECT People.Name, People.ID, Phones.Number, Emails.Email
    FROM People 
    LEFT OUTER JOIN Phones ON People.ID = Phones.ID
    LEFT OUTER JOIN Emails ON People.ID = Emails.ID 
    ORDER BY Name, ID, Number, Email;

+----------+---------+-----------------+------------------------------+
| Name     | ID      | Number          | Email                        |
+----------+---------+-----------------+------------------------------+
| Steven   | 5000008 | 8451384         | feler@campus.uni.com         |
| Steven   | 5000008 | 8451384         | Smithley@gmail.com           |
| Steven   | 5000008 | 8451384         | tech@goodmail.com            |
| Steven   | 5000008 | 24887312        | feler@campus.uni.com         |
| Steven   | 5000008 | 24887312        | Smithley@gmail.com           |
| Steven   | 5000008 | 24887312        | tech@goodmail.com            |
| John     | 5000005 | 5551234         | Smithley@goodmail.com        |
| John     | 5000005 | 5551234         | Smithley.j@gmail.com         |
| John     | 5000005 | 5154324         | Smithley@goodmail.com        |
| John     | 5000005 | 5154324         | Smithley.j@gmail.com         |
| George   | 5000004 | NULL            | georgestanko@hotmail.com     |
| Ashley   | 8000009 | 7134584         | Ashley.hill86@gmail.com      |
| Amy      | 5000003 | NULL            | NULL                         |
+----------+---------+-----------------+------------------------------+
我正在尝试将它们连接在一起,而不重复。当我尝试只加入与人的电子邮件或只加入与人的电话时,效果非常好。

SELECT People.Name, People.ID, Phones.Number
    FROM People 
    LEFT OUTER JOIN Phones ON People.ID=Phones.ID 
    ORDER BY Name, ID, Number;
+----------+---------+-----------------+
| Name     | ID      | Number          |
+----------+---------+-----------------+
| Steven   | 5000008 | 8451384         |
| Steven   | 5000008 | 24887312        |
| John     | 5000005 | 5551234         |
| John     | 5000005 | 5154324         |
| George   | 5000004 | NULL            |
| Ashley   | 8000009 | 7134584         |
| Amy      | 5000003 | NULL            |
+----------+---------+-----------------+

SELECT People.Name, People.ID, Emails.Email
    FROM People 
    LEFT OUTER JOIN Emails ON People.ID=Emails.ID 
    ORDER BY Name, ID, Email;
+----------+---------+------------------------------+
| Name     | ID      | Email                        |
+----------+---------+------------------------------+
| Steven   | 5000008 | Smithley@gmail.com           |
| Steven   | 5000008 | tech@goodmail.com            |
| Steven   | 5000008 | feler@campus.uni.com         |
| John     | 5000005 | Smithley@goodmail.com.com    |
| John     | 5000005 | Smithley.j@gmail.com         |
| George   | 5000004 | georgestanko@hotmail.com     |
| Ashley   | 8000009 | Ashley.hill86@gmail.com      |
| Amy      | 5000003 | NULL                         |
+----------+---------+------------------------------+
然而,当我尝试加入电子邮件和电话时,我得到了这样一个信息:

    +---------+----------+
    | ID      | Name     |
    +---------+----------+
    | 5000003 | Amy      |
    | 5000004 | George   |
    | 5000005 | John     |
    | 5000008 | Steven   |
    | 8000009 | Ashley   |
    +---------+----------+

    +---------+-----------------+
    | ID      | Number          |
    +---------+-----------------+
    | 5000005 | 5551234         |
    | 5000005 | 5154324         |
    | 5000008 | 2487312         |
    | 8000009 | 7134584         |
    | 5000008 | 8451384         |
    +---------+-----------------+

+---------+------------------------------+
| ID      | Email                        |
+---------+------------------------------+
| 5000005 | Smithley@goodmail.com.com    |
| 5000005 | Smithley.j@gmail.com         |
| 5000008 | Smithley@gmail.com           |
| 5000008 | tech@goodmail.com            |
| 5000008 | feler@campus.uni.com         |
| 8000009 | Ashley.hill86@gmail.com      |
| 5000004 | georgestanko@hotmail.com     |
+---------+------------------------------+
SELECT People.Name, People.ID, Phones.Number, Emails.Email
    FROM People 
    LEFT OUTER JOIN Phones ON People.ID = Phones.ID
    LEFT OUTER JOIN Emails ON People.ID = Emails.ID 
    ORDER BY Name, ID, Number, Email;

+----------+---------+-----------------+------------------------------+
| Name     | ID      | Number          | Email                        |
+----------+---------+-----------------+------------------------------+
| Steven   | 5000008 | 8451384         | feler@campus.uni.com         |
| Steven   | 5000008 | 8451384         | Smithley@gmail.com           |
| Steven   | 5000008 | 8451384         | tech@goodmail.com            |
| Steven   | 5000008 | 24887312        | feler@campus.uni.com         |
| Steven   | 5000008 | 24887312        | Smithley@gmail.com           |
| Steven   | 5000008 | 24887312        | tech@goodmail.com            |
| John     | 5000005 | 5551234         | Smithley@goodmail.com        |
| John     | 5000005 | 5551234         | Smithley.j@gmail.com         |
| John     | 5000005 | 5154324         | Smithley@goodmail.com        |
| John     | 5000005 | 5154324         | Smithley.j@gmail.com         |
| George   | 5000004 | NULL            | georgestanko@hotmail.com     |
| Ashley   | 8000009 | 7134584         | Ashley.hill86@gmail.com      |
| Amy      | 5000003 | NULL            | NULL                         |
+----------+---------+-----------------+------------------------------+
发生的情况是-如果一个人有两个号码,他的所有电子邮件都会显示两次(它们无法排序!这意味着@last无法删除它们)

我想要的是: 总之,玩@last,我想最终得到这样的东西,但是如果我不按正确的方式排列订单列,@last将不起作用-这似乎是一个大问题..在电子邮件列中排序。因为从上面的例子可以看出:

史蒂文有2个电话号码和3封电子邮件。带有数字的连接电子邮件发生在每封电子邮件中-因此无法排序的重复值(排序方式对它们不起作用)

现在有人告诉我,最好将电子邮件和数字放在分开的表格中,因为一个人可以有很多电子邮件。那么,如果这是一件很常见的事情,有什么不简单的解决方案呢

我也很乐意使用PHP解决方案

我现在知道该怎么做了,这让我很满意,但并没有那么漂亮。 如果我与GROUP_CONTACT这样做,结果会令人满意,但它看起来并不漂亮:我不能在它旁边放一个“Email type=work”

   SELECT People.Ime,  
    GROUP_CONCAT(DISTINCT Phones.Number),  
    GROUP_CONCAT(DISTINCT Emails.Email)  
    FROM People 
    LEFT OUTER JOIN Phones ON People.ID=Phones.ID
    LEFT OUTER JOIN Emails ON People.ID=Emails.ID
    GROUP BY Name;
+----------+----------------------------------------------+---------------------------------------------------------------------+
| Name     | GROUP_CONCAT(DISTINCT Phones.Number)         | GROUP_CONCAT(DISTINCT Emails.Email)                                 |
+----------+----------------------------------------------+---------------------------------------------------------------------+
| Steven   | 8451384,24887312                             | Smithley@gmail.com,tech@goodmail.com,feler@campus.uni.com           |
| John     | 5551234,5154324                              | Smithley@goodmail.com,Smithley.j@gmail.com                          |
| George   | NULL                                         | georgestanko@hotmail.com                                            |
| Ashley   | 7134584                                      | Ashley.hill86@gmail.com                                             |
| Amy      | NULL                                         | NULL                                                                |
+----------+----------------------------------------------+---------------------------------------------------------------------+

你想要的其实根本不是你想要的,如果这有道理的话。。。实际上,您无法通过编程方式对数据库输出执行任何操作,您需要对其执行一些操作(除非您只是直接在数据库上运行查询)

既然你说“我也很乐意使用PHP解决方案”。。。您真正想要的是类似于PHP“用户”对象的东西,类似于(当然这都是假设的):


然后,您的数据库抽象对象只需要执行一些非常简单的查询来读取用户、用户电子邮件和电话号码表,并将结果作为数组返回,然后您可以将这些数组直接放入PHP对象中。例如:

<?php

/**
 * this implements a database connection object as a private class member
 */
class DBUser {

  private $_conn;

  /* constructor other functionality */

  /**
   * method to pass an SQL query to the database and return an array of results
   */
  public function readTelNos($iUserId) {
    return $this->_conn->read("SELECT `number` from `tel` WHERE `user_id` = " . (int) $iUserId);
  }

}
?>

这会将您的问题分解为更小、更易于处理的问题,并将它们封装在漂亮的PHP对象中,然后您可以实际使用这些对象


您的用户对象将有一个列表,如果电话号码和电子邮件可以很容易地通过
$oUser->getEmails()等方式检索$oUser->getEmail('work')检索它们

如果我理解正确,您还需要将其他字段与电子邮件关联(例如:键入)

使用group_concat,您可以将电子邮件类型添加到结果中,在我的示例中是“;”将邮件与类型分开“,”将结果分开

SELECT u.name,  
    GROUP_CONCAT( distinct p.phone),  
    GROUP_CONCAT( distinct 'mail=',e.mail,';type=' ,type)  
    FROM people u
    LEFT OUTER JOIN phone p ON u.id=p.id
    LEFT OUTER JOIN mail e ON u.id=e.id
    GROUP BY u.id

首先创建临时表名temp_tb,然后插入记录。然后使用Row_Number()进行分区,并多次删除列出的所有记录。就像被打了一样

    Insert into temp_tb SELECT People.Name, People.ID, Phones.Number, Emails.Email
    FROM People 
    LEFT OUTER JOIN Phones ON People.ID = Phones.ID
   LEFT OUTER JOIN Emails ON People.ID = Emails.ID 
   ORDER BY Name, ID, Number, Email;
   SELECT * from temp_tb 
    With A as
  (
 select People.Name, People.ID, Phones.Number, Emails.Email ROW_NUMBER() 
  OVER (Partition by ID ORDER BY Fee_temp.Name) As Number from temp_tb 
 )
 DELETE FROM A WHERE Number>1
 SELECT * FROM temp_tb 

希望能对您有所帮助。

您的表达有问题。不要在SQL中解决这个问题。这不是它的目的。+1用于详细说明和显示您面临的问题所需的基本背景信息。:)如果MySQL有一种简单的方法来实现带有窗口功能的行号,这将是非常容易的。@bluefeet:使用一个变量
从aTable中选择@rownum:=rownum+1,(选择@rownum:=0)r
@juergend right,但MySQL在使用Thanke you时确实存在问题。我在学习php之前先学习数据库。(我知道C语言编程,所以我很擅长让某些东西工作,但还不知道如何让它看起来漂亮)。不客气,但如果这解决了你的问题,请记住标记我的答案为正确;)