How to create connection between two tables in SQL Server 2008? - sql-server

Question

I have database that has Student table with information about each student in the system. Each record in that table has unique identifier that was created with NEWID() in SQL 2008. Then I have three other tables that use the same ID to link the student and the record. I'm wondering if I need to set any kind of property that will link these tables/records. Here is example of my Student table:
st_id -> Auto increment id
st_studentGUID -> Primary key
st_firstName
st_lastName
st_dob
st_gender
st_uid
st_udt
st_utime
And here is example of my other three tables:
Table 1 Table 2 Table 3
tb1_id -> auto increment tb2_id -> auto increment tb3_id -> auto increment
tb1_studentGUID tb2_studentGUID tb3_studentGUID
I have tried to create foreign key for each Table 1,2 and 3 on studentGUID but I got an error:
The ALTER TABLE statement conflicted with the FOREIGN KEY constraint "FK_Table1_Student". The conflict occurred in database "testDB", table "Student", column 'st_studentGUID'.
After I have done some research seems like this error occurred because Table 1 doesn't have all st_studentGUID. This make sense because student record might exist in Table 3 but not in Table 1. I'm wondering if there is any other sort of relation that I can set between the tables? Should I maybe use the Indexes on studentGUID fields? This is the reason why I'm asking this, in one of my update queries I use auto incremented id in where clause and for some reason I received an error that looks like this:
ErrorCode 1205
Message [Macromedia][SQLServer JDBC Driver][SQLServer]Transaction (Process ID 111) was deadlocked on lock | communication buffer resources with another process and has been chosen as the deadlock victim. Rerun the transaction.
SQLState 40001
and error was pointing to the line that is end of the update query that looks like this:
<cfquery name="updateTable1" datasource="testDB">
UPDATE Table1
SET
tb1_testdt = <cfqueryparam value="#FORM.frm_testdt#" cfsqltype="cf_sql_date" maxlength="10" null="#yesNoFormat(!len(FORM.frmhs_testdtgr))#" />
FROM Table1 AS Table
LEFT OUTER JOIN studentLocked AS Locked
ON Locked.lk_studentGUID = Table.tb1_studentGUID
WHERE tb1_id = <cfqueryparam value="#FORM.frm_id#" cfsqltype="cf_sql_integer" />
AND lk_active = '1'
AND lk_staffID = <cfqueryparam value="#trim(appStaff)#" cfsqltype="cf_sql_char" maxlength="10" />
</cfquery>
Query above is checking if studentGUID exists in Lock table, reason why I'm doing this is because we do not want two different users updating the same record at the same time. Query updates single record and should be pretty fast. I'm confused whit the error message above deadlock victim. Is there something in my database design that is off or my update query is causing this error? I couldn't find anything on the web that could help me fix this issue. if anyone has experience with this kind of problem please let me know.

Answer 1

If all StudentGUIDs exist in the Student table, then the correct design is to have the StudentGUID in Table1, Table2 and Table3 be a Foreign Key to the Student table.

Answer 2

Now that you've sorted out your foreign key issue with the missing record from Student, try setting up your FKs on all of your tables, then re-try your update query.
If the deadlock issue persists and is reproducible, submit a new question for that, including the schema for the table you're updating and the values you're trying to pass.

Answer 3

Here's a script that demonstrates you can create the relationships you described provided all the keys are present in the students table:
CREATE TABLE students (st_id AS NEWID(),
st_studentGUID UNIQUEIDENTIFIER PRIMARY KEY)
INSERT INTO students (st_studentGUID) VALUES ('3B9F59DD-BF0A-4A09-BF4E-A396E2978B24')
INSERT INTO students (st_studentGUID) VALUES ('7CC5FF67-DAB8-426A-B9F7-E9F041718B6B')
INSERT INTO students (st_studentGUID) VALUES ('84B80D3E-44C4-4291-857D-B6CA6552369D')
CREATE TABLE t1 (tb1_id AS NEWID(), tb1_studentGUID UNIQUEIDENTIFIER)
ALTER TABLE t1
ADD CONSTRAINT FK_t1_tbl_studentGUID FOREIGN KEY (tb1_studentGUID) REFERENCES students(st_studentGUID)
INSERT INTO t1 (tb1_studentGUID) VALUES ('3B9F59DD-BF0A-4A09-BF4E-A396E2978B24')
CREATE TABLE t2 (tb2_id AS NEWID(), tb2_studentGUID UNIQUEIDENTIFIER)
ALTER TABLE t2
ADD CONSTRAINT FK_t2_tbl_studentGUID FOREIGN KEY (tb2_studentGUID) REFERENCES students(st_studentGUID)
INSERT INTO t2 (tb2_studentGUID) VALUES ('7CC5FF67-DAB8-426A-B9F7-E9F041718B6B')
CREATE TABLE t3 (tb3_id AS NEWID(), tb3_studentGUID UNIQUEIDENTIFIER)
ALTER TABLE t3
ADD CONSTRAINT FK_t3_tbl_studentGUID FOREIGN KEY (tb3_studentGUID) REFERENCES students(st_studentGUID)
INSERT INTO t3 (tb3_studentGUID) VALUES ('84B80D3E-44C4-4291-857D-B6CA6552369D')
Now create a 4th table, add a UUID that isn't in the students table, and try to create the foreign key constraint:
CREATE TABLE t4 (tb4_id AS NEWID(), tb4_studentGUID UNIQUEIDENTIFIER)
INSERT INTO t4 (tb4_studentGUID) VALUES ('897925F1-BE92-44EB-82C3-88E1C33C7792')
ALTER TABLE t4
ADD CONSTRAINT FK_t4_tbl_studentGUID FOREIGN KEY (tb4_studentGUID) REFERENCES students(st_studentGUID)
Which reproduces your error message:
The ALTER TABLE statement conflicted with the FOREIGN KEY constraint "FK_t4_tbl_studentGUID". The conflict occurred in database "IconERP", table "dbo.students", column 'st_studentGUID'.