简体   繁体   中英

SCOPE_IDENTITY() not working

When I try to get the last ID inserted on Database via SQL Server it returns __Page. Here is my code :

query = "INSERT INTO
             seekers(name, sname, lname, status, gender, dob, major, experince,
                     email, password, phone, valid, city)
             values (@name, @sname, @lname, @status ,@gender, @dob, @major,
                     @exp, @email, @password, @phone, 0, @city);
         SELECT SCOPE_IDENTITY();";

// some code here related to parameters

command = new SqlCommand(query, connection);
int id = Convert.ToInt32(command.ExecuteScalar());

If just you want to SELECT it use OUTPUT ;

INSERT INTO seekers(name,sname,lname,status,gender,dob,major,experince,email,password,phone,valid,city)
OUTPUT INSERTED.IDENTITY_COL_NAME
values(@name,@sname,@lname,@status,@gender,@dob,@major,@exp,@email,@password,@phone,0,@city);

As using ExecuteScalar . And ur first executing statement is INSERT so it returns number of records inserted.

To avoid this use SET NOCOUNT ON as shown below:

string Query = @"SET NOCOUNT ON;
INSERT INTO table1(name) values('ttt');
SELECT SCOPE_IDENTITY(); SET NOCOUNT OFF;
";

You can write a stored procedure like that;

CREATE PROCEDURE [seekers_Insert]
 @ID uniqueidentifier OUTPUT,
 @name nvarchar(50) ,
  @sname nvarchar(50),

(other input parameters)
  AS
  INSERT INTO seekers(name,sname,lname,status,gender,dob,major,experince,email,
                         password,phone,valid,city)
                  values(@name,@sname,@lname,@status,@gender,@dob,@major,@exp,@email,
                         @password,@phone,0,@city)

   SET @ID=SCOPE_IDENTITY()
    GO

and later you can take it

int ID = Convert.ToInt32(Command.ExecuteScalar());

we do this and works fine but we use return parameters for example in our stored procedure:

@RecordID int OUTPUT

... here the insert statement

SET @RecordID= SCOPE_IDENTITY();

...

SELECT @RecordID

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM