I don't understand why I can't insert this. I can't spot the problem. The error message is Conversion failed when converting from a character string to uniqueidentifier.
The GUIDs are the ones that I got when I did a select from some other tables.
insert into [db].[dbo].[table] (myid,friendid,time1,time2) values
( CONVERT(uniqueidentifier,'0C6A36BA-10E4-438F-BA86-0D5B68A2BB15'),
CONVERT(uniqueidentifier,'DF215E10-8BD4-4401-B2DC-99BB03135F2E'),
'2014-01-05 02:04:41.953','2014-01-05 12:04:41.953')
I use SQL Server 2012
The columns are
id uniqueidentifier,
myid uniqueidentifier,
friendid uniqueidentifier,
time1 datetime nullable,
time2 datetime nullable
This question is related to
sql-server
sql-server-2012
MSDN Documentation Here
To add a bit of context to M.Ali's Answer you can convert a string to a uniqueidentifier using the following code
SELECT CONVERT(uniqueidentifier,'DF215E10-8BD4-4401-B2DC-99BB03135F2E')
If that doesn't work check to make sure you have entered a valid GUID
DECLARE @t TABLE (ID UNIQUEIDENTIFIER DEFAULT NEWID(),myid UNIQUEIDENTIFIER
, friendid UNIQUEIDENTIFIER, time1 Datetime, time2 Datetime)
insert into @t (myid,friendid,time1,time2)
values
( CONVERT(uniqueidentifier,'0C6A36BA-10E4-438F-BA86-0D5B68A2BB15'),
CONVERT(uniqueidentifier,'DF215E10-8BD4-4401-B2DC-99BB03135F2E'),
'2014-01-05 02:04:41.953','2014-01-05 12:04:41.953')
SELECT * FROM @t
Result Set With out any errors
+------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
¦ ID ¦ myid ¦ friendid ¦ time1 ¦ time2 ¦
¦--------------------------------------+--------------------------------------+--------------------------------------+-------------------------+-------------------------¦
¦ CF628202-33F3-49CF-8828-CB2D93C69675 ¦ 0C6A36BA-10E4-438F-BA86-0D5B68A2BB15 ¦ DF215E10-8BD4-4401-B2DC-99BB03135F2E ¦ 2014-01-05 02:04:41.953 ¦ 2014-01-05 12:04:41.953 ¦
+------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
You have to check unique identifier column and you have to give a diff value to that particular field if you give the same value it will not work. It enforces uniqueness of the key.
Here is the code:
Insert into production.product
(Name,ProductNumber,MakeFlag,FinishedGoodsFlag,Color,SafetyStockLevel,ReorderPoint,StandardCost,ListPrice,Size
,SizeUnitMeasureCode,WeightUnitMeasureCode,Weight,DaysToManufacture,
ProductLine,
Class,
Style ,
ProductSubcategoryID
,ProductModelID
,SellStartDate
,SellEndDate
,DiscontinuedDate
,rowguid
,ModifiedDate
)
values ('LL lemon' ,'BC-1234',0,0,'blue',400,960,0.00,100.00,Null,Null,Null,null,1,null,null,null,null,null,'1998-06-01 00:00:00.000',null,null,'C4244F0C-ABCE-451B-A895-83C0E6D1F468','2004-03-11 10:01:36.827')
Source: Stackoverflow.com