Optional one-to-one mapping in Hibernate

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旧时难觅i
旧时难觅i 2021-02-18 23:05

How do I create an optional one-to-one mapping in the hibernate hbm file? For example, suppose that I have a User and a last_visited_page table. The user may or may not have a l

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  • 2021-02-18 23:07

    Just spend most of the day today trying to do a similar thing, finally found the following solution (just in-case this might be useful for other people)

    @OneToOne
    @JoinColumn(name="ClassA_Id", referencedColumnName="ClassB_Id", nullable=true)
    

    Hope that might help save somebody some time

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  • 2021-02-18 23:13

    I was having a simliar problem, but using annotations. Google brought me here, so if anyone else finds themselves in the same sitatuions, this worked for me:

    http://opensource.atlassian.com/projects/hibernate/browse/ANN-725

    If you're using annotations, you can use the @NotFound(action=NotFoundAction.IGNORE) annotation so that you don't get an exception. Just make sure your code checks for nulls because it's now might not be there ;-)

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  • 2021-02-18 23:14

    Had the same issue, resolved with @OneToOne(optional = true) on the User class (hibernate 5.2.17.Final)

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  • 2021-02-18 23:18

    To my knowledge, Hibernate doesn't support optional one-to-one (see HHH-2007) so you'll have to use a fake many-to-one with not-null="false" instead.

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  • 2021-02-18 23:22

    If a user has at most one last_visited page, then there are two cases :

    (a) some given user has no last_visited page, in which case there will not be any tuple for this user in the last_visited_page table, (b) some given user has exactly one last_visited page, in which case there will be exactly one tuple for this user in the last_visited_page table.

    That should make it obvious that userid is a candidate key in your last-visited-page table.

    And that should make it obvious that you should declare that key to the DBMS.

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