根据T-SQL中的源表填充列数据

Question

我不确定我的问题是否措辞正确但是这里也是如此.

我有一个名为Contacts的表,其中包含表格地址,电子邮件,电话的FK引用(这些表格中包含1对多个联系人).我需要创建一个将提取所有数据的查询,并有一个名为Contact Method的列,显示该行来自哪个子表.

Contact: ID, AddressID, EmailID, PhoneID
Address: ID, Line1, City, State
Email :  ID, EAddress
Phone :  ID, Number, Extension

我需要结果表看起来像这样:

ContactMethod | ID | [Value1] | [Value2] | [Value3]

Address         2      N5980    Onalaska     WI
Email           8     myEmail@
Phone           5     555-5555    1234

或者它可以列出一行中的所有组合列,如果这更简单,我也可以使用它.即

ContactMethad | ID | Line1 | City | State | ID | EAddress | ID | Number | Extension

我看着PIVOT,这很整洁,但似乎并没有解决我的问题.我需要将它与COALESCE结合使用吗？

谢谢你的帮助.

---

编辑

我的数据,表格上的联系人看起来像这样:

ID | AddressID | PhoneID | EmailID

1      3           null      null
2     null         null      7
3     null          5        null
4     4            null      null
5     null         6         null

建议的解决方案有效,除了我每个ID获得3行.合理？

Answer 1

您可以使用和子句取消数据以获取结果:CROSS APPLYVALUES

select d.ContactMethod, d.id, d.Value1, d.Value2, d.Value3
from contacts c
left join address a
  on c.addressid = a.id
left join email e
  on c.emailid = e.id
left join phone p
  on c.phoneid = p.id
cross apply
(
  values
    ('Address', c.addressid, a.Line1, a.City, a.State),
    ('Email', c.emailid, e.eAddress, '', ''),
    ('Phone', c.phoneid, p.number, cast(p.extension as varchar(10)), '')
) d (ContactMethod, id, Value1, Value2, Value3)

请参阅SQL Fiddle with Demo.

这给出了结果:

| CONTACTMETHOD | ID |   VALUE1 |   VALUE2 | VALUE3 |
-----------------------------------------------------
|       Address |  2 |    N5980 | Onalaska |     WI |
|         Email |  8 | myEmail@ |          |        |
|         Phone |  5 | 555-5555 |     1234 |        |

如果您想要第二个结果,那么您可以使用多个连接来获取它:

select cm.ContactMethod,
  a.id addressid,
  a.line1,
  a.city,
  a.state,
  e.id emailid,  
  e.eaddress,
  p.id phoneid,
  p.number,
  p.extension
from contacts c
cross join
(
  VALUES ('Address'),('Email'),('Phone')
) cm (ContactMethod)
left join address a
  on c.addressid = a.id
  and cm.ContactMethod = 'Address'
left join email e
  on c.emailid = e.id
  and cm.ContactMethod = 'Email'
left join phone p
  on c.phoneid = p.id
  and cm.ContactMethod = 'Phone';

请参阅SQL Fiddle with Demo.结果是:

| CONTACTMETHOD | ADDRESSID |  LINE1 |     CITY |  STATE | EMAILID | EADDRESS | PHONEID |   NUMBER | EXTENSION |
----------------------------------------------------------------------------------------------------------------
|       Address |         2 |  N5980 | Onalaska |     WI |  (null) |   (null) |  (null) |   (null) |    (null) |
|         Email |    (null) | (null) |   (null) | (null) |       8 | myEmail@ |  (null) |   (null) |    (null) |
|         Phone |    (null) | (null) |   (null) | (null) |  (null) |   (null) |       5 | 555-5555 |      1234 |

编辑#1,根据您的更改,您可以将查询更改为以下内容.

第一个有三value列,然后你可以添加一个WHERE子句来过滤掉任何null值:

select c.ID, ContactMethod, Value1, Value2, Value3
from contacts c
left join address a
  on c.addressid = a.id
left join email e
  on c.emailid = e.id
left join phone p
  on c.phoneid = p.id
cross apply
(
  values
    ('Address', c.addressid, a.Line1, a.City, a.State),
    ('Email', c.emailid, e.eAddress, null, null),
    ('Phone', c.phoneid, p.number, cast(p.extension as varchar(10)), null)
) d (ContactMethod, id, Value1, Value2, Value3)
where value1 is not null
  or value2 is not null
  or value3 is not null

请参阅SQL Fiddle with Demo.结果是:

 ID | CONTACTMETHOD |            VALUE1 |    VALUE2 | VALUE3 |
---------------------------------------------------------------
|  1 |       Address |             N5980 |  Onalaska |     WI |
|  2 |         Email |          myEmail@ |    (null) | (null) |
|  3 |         Phone |          555-5555 |      1234 | (null) |
|  4 |       Address | 1417 Saint Andrew | La Crosse |     WI |

如果您想将结果放在一行中,那么您将需要使用该UNPIVOT函数:

select *
from
(
  select id,
    case col 
      when 'addressid' then 'address'
      when 'emailid' then 'email'
      when 'phoneid' then 'phone' end ContactMethod,
    contact_id
  from contacts
  unpivot
  (
    contact_id
    for col in (addressid, emailid, phoneid)
  ) unpiv
) c
left join address a
  on c.contact_id = a.id
  and c.ContactMethod = 'Address'
left join email e
  on c.contact_id = e.id
  and c.ContactMethod = 'Email'
left join phone p
  on c.contact_id = p.id
  and c.ContactMethod = 'Phone';

请参阅SQL Fiddle with Demo.此查询的结果是:

| ID | CONTACTMETHOD | CONTACT_ID |             LINE1 |      CITY |  STATE | EADDRESS |   NUMBER | EXTENSION |
--------------------------------------------------------------------------------------------------------------
|  1 |       address |          2 |             N5980 |  Onalaska |     WI |   (null) |   (null) |    (null) |
|  2 |         email |          8 |            (null) |    (null) | (null) | myEmail@ |   (null) |    (null) |
|  3 |         phone |          5 |            (null) |    (null) | (null) |   (null) | 555-5555 |      1234 |
|  4 |       address |          3 | 1417 Saint Andrew | La Crosse |     WI |   (null) |   (null) |    (null) |