INSERT Record Through A View

Apr 20, 2004

Is is possible to insert a record through a view. If so, how?

USE Northwind

GO

CREATE TABLE tbForms (
FormID INT IDENTITY (1,1) NOT NULL,
Form varchar (100) NOT NULL
)

GO

ALTER TABLE tbForms
ADD CONSTRAINT tbForms_pk PRIMARY KEY (FormID)
GO

CREATE TABLE tbDoubleTeeForms (
fkFormID INT NOT NULL,
Form varchar(100) NOT NULL,
Width FLOAT,
Height FLOAT,
Flange FLOAT,
Leg FLOAT,
LegCount INT
)

GO

ALTER TABLE tbDoubleTeeForms
ADD CONSTRAINT tbDoubleTeeForms_pk PRIMARY KEY (fkFormID)
GO

ALTER TABLE tbDoubleTeeForms
ADD CONSTRAINT tbDoubleTeeForms_fk FOREIGN KEY (fkFormID)
REFERENCES tbForms (FormID)
GO

CREATE TABLE tbFlatPanelForms (
fkFormID INT NOT NULL,
Form varchar(100) NOT NULL,
Width FLOAT,
HEIGHT FLOAT
)

GO

ALTER TABLE tbFlatPanelForms
ADD CONSTRAINT tbFlatPanelForms_pk PRIMARY KEY (fkFormID)
GO

ALTER TABLE tbFlatPanelForms
ADD CONSTRAINT tbFlatPanelForms_fk FOREIGN KEY (fkFormID)
REFERENCES tbForms (FormID)
GO

CREATE VIEW MyProducts AS
SELECT fkFormID, Form FROM tbDoubleTeeForms UNION ALL
SELECT fkFormID, FOrm FROM tbFlatPanelForms

GO

-- How can I insert a new record, the pk of the forms table is identity.
-- Can this be done?
INSERT INTO MyProducts (Form)
VALUES ('My First Entry')
GO

SELECT * FROM MyProducts
GO

DROP VIEW MyProducts
GO

DROP TABLE tbFlatPanelForms
GO

DROP TABLE tbDoubleTeeForms
GO

DROP TABLE tbForms
GO

Mike B

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Exception --->  System.Data.SqlClient.SqlException:
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ALTER procedure sp_Cigna_build_export
as

Declare CignaExport Cursor
For
Select
exprectype,
expssn,
expfiller,
expbody

From cigna_employee_sort

Declare
@RecordType char(2),
@SSN char(11),
@Filler char(20),
@Body char(867),
@Counter int,
@CountLength int,
@RecordCount char(9)

select @counter = 0

Open CignaExport
Fetch Next from CignaExport

Into
@RecordType,
@SSN,
@Filler,
@Body

While @@fetch_status = 0

Begin
select @counter = @counter + 1
insert into cigna_export_table
(exportfield)
select
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Fetch next from CignaExport
Into
@RecordType,
@SSN,
@Filler,
@Body
End


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select @counter = @counter + 2
select @RecordCount = '000000000'
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<%
Dim dbConn
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