2

I'm not sure my question is even worded correctly but here goes.

I have a table called Contacts that has FK references to tables Address, Email, Phone (these have 1 to many with Contacts). I need to create a query that will pull all the data and has a column called Contact Method that shows which sub table that row came from.

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

I need the resulting table to look like this:

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

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

Alternatively it could list all the combined columns in a row if that's simpler, I can work with that as well. i.e.

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

I looked at PIVOT, which is neat but doesn't seem to solve my problem by itself. Do I need to combine it with COALESCE?

Thanks for any help.

EDIT

My data, on table Contact would look like this:

ID | AddressID | PhoneID | EmailID

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

The proposed solution works except that I get 3 rows per ID. Make sense?

4

2 に答える 2

7

and句を使用してデータのピボットを解除し、結果を取得できます。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)

SQL Fiddle with Demoを参照してください。

これにより、次の結果が得られます。

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

2 番目の結果が必要な場合は、複数の結合を使用して取得できます。

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';

SQL Fiddle with Demoを参照してください。結果は次のとおりです。

| 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 |

編集＃1、変更に基づいて、クエリを次のように変更できます。

3 つのvalue列を持つ最初の列では、句を追加して値WHEREを除外することができます。null

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

SQL Fiddle with Demoを参照してください。結果は次のとおりです。

 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 |

結果を 1 行で表示する場合は、次のUNPIVOT関数を使用します。

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';

SQL Fiddle with Demoを参照してください。このクエリの結果は次のとおりです。

| 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) |
于 2013-03-14T19:47:23.723 に答える
0

2 番目の列のレイアウトに到達する方がはるかに簡単です。そのためには、参加するだけです：

SELECT *
FROM dbo.Contact c
JOIN dbo.Address a
ON c.AddressID = a.ID
JOIN dbo. Email e
ON c. EmailID = e.ID
JOIN dbo. Phone p
ON c. PhoneID = p.ID

を使用SELECT *しましたが、すべての列が必要なわけではないため、実際にはすべての列をリストする必要があります。各子テーブルに必ずしも行があるとは限らない場合はLEFT OUTER JOIN、単に の代わりに使用する必要がありますJOIN。

JOIN の詳細については、このシリーズをチェックしてください: http://sqlity.net/en/1146/a-join-a-day-introduction/

複数の行が必要な場合は、これを使用できます。

SELECT *
FROM dbo.Contact c
CROSS JOIN (VALUES('Address','Email','Phone'))X(ContactMethod)
LEFT JOIN dbo.Address a
ON c.AddressID = a.ID
AND X.ContactMethod = 'Address'
LEFT JOIN dbo. Email e
ON c. EmailID = e.ID
AND X.ContactMethod = 'Email'
LEFT JOIN dbo. Phone p
ON c. PhoneID = p.ID
AND X.ContactMethod = 'Phone'

「拡張」バージョンを使用する利点は、データ型の非互換性に対処する必要がないことです。

于 2013-03-14T19:40:53.913 に答える