Getting SQLCODE -530

soumyaranjan007 · New User Joined: 30 Aug 2006 Posts: 30 Location: mumbai

Hi,

we are getting SQLCODE -530. if anybody have idea in resolution. help me out.

thanks,
soumya

ragganga · New User Joined: 22 Jan 2007 Posts: 18 Location: bangalore

Hi Soumya,

SQLCODE = -530

Explanation - An UPDATE or INSERT operation attempted to place a value in a foreign key of the object table; however, this value was not equal to some value of the primary key of the parent table.When a row is inserted into a dependent table, the insert value of a foreign key must be equal to the value of the primary key of some row of the parent table in the associated relationship.When the value of the foreign key is updated, the update value of a foreign key must be equal to the value of the primary key of some row of the parent table of the associated relationship.

Response - Examine the insert or update value of the foreign key first, and then compare it with each of the primary key values of the parent table to determine the cause of the problem.

I think it is enough to solve your error

Regards
Raghu

mkk157 · Active User Joined: 17 May 2006 Posts: 310

Hi Sowmyaranjan,

I never heard about the resolution. Can u explain ur problem a bit clearly. The follwoing information about the SQLCODE -530 may help u.

-530 THE INSERT OR UPDATE VALUE OF FOREIGN KEY constraint-name IS
INVALID

Explanation: An UPDATE or INSERT operation attempted to place a value in
a foreign key of the object table; however, this value was not equal to
some value of the parent key of the parent table.

When a row is inserted into a dependent table, the insert value of a
foreign key must be equal to the value of the parent key of some row of
the parent table in the associated relationship.

When the value of the foreign key is updated, the update value of a
foreign key must be equal to the value of the parent key of some row of
the parent table of the associated relationship.

System Action: The UPDATE or INSERT statement cannot be executed. The
object table is unchanged.

Programmer Response: Examine the insert or update value of the foreign
key first, and then compare it with each of the parent key values of the
parent table to determine the cause of the problem.