many-to-many

Say, for example, I had two entities: Article and Tag (like in a typical blog). Each article can have many tags, and each tag can be used by many articles, so it is a classical m:n relationship. I need to specify an owning side with JPA. But which side should be the owning side? An article doesn't depend on a certain tag and vice versa. Is there a rule of thumb for determining which side should be the owning side? Every bidirectional relationship requires an owning side in JPA. In the particular case of ManyToMany : @JoinTable is specified on the owning side of the relationship. the owning

The situation: So, I have a basic many-to-many relationship in SQLAlchemy using an association table . For example, a person can attend many parties, and a party can have many persons as guests: class Person(Base): __tablename__ = 'person' id = Column(Integer, primary_key=True) name = db.Column(db.String(50)) class SexyParty(Base): __tablename__ = 'sexy_party' id = Column(Integer, primary_key=True) guests = relationship('Person', secondary='guest_association', lazy='dynamic', backref='parties') guest_association = Table( 'guest_association', Column('user_id', Integer(), ForeignKey('person.id')

I am having some trouble with a many to many relationship. I have Users and Assets . I would like to be able to assign users to an asset on the asset page. The code below displays a list of users when creating/editing an asset, however changes made to the user checkboxes do not save, while the rest of the data is persisted. If I add an entry to users_assets through the mysql client, these changes are shown in the asset list. User class User extends BaseUser { /** * @ORM\ManyToMany(targetEntity="Asset", inversedBy="users") */ private $assets; } Asset class Asset { /** * @ORM\ManyToMany

I've two model classes: public class Candidate { public int Id { get; set; } public string Name { get; set; } public ICollection<Job> Jobs { get; set; } } public class Job { public int Id { get; set; } public string Name { get; set; } public ICollection<Candidate> Candidates { get; set; } } My DbContext name is JobsContext. The above code generates me 3 tables Candidates, Jobs & CandidatesJobs(autogenerated by EF) Now I've records in Jobs table : Id = 1, Name = "Sales" : Id = 2, Name = "Engineer". I want to associate a new Candidate which I'll be inserting into Candidates table with the 2

I'm working on a back office of a restaurant's website. When I add a dish , I can add ingredients in two ways. In my form template, I manually added a text input field. I applied on this field the autocomplete method of jQuery UI that allows: Select existing ingredients (previously added) Add new ingredients However, when I submit the form, each ingredients are inserted in the database (normal behaviour you will tell me ). For the ingredients that do not exist it is good, but I don't want to insert again the ingredients already inserted. Then I thought about Doctrine events , like prePersist()

I have 3 tables, foo, foo2bar, and bar. foo2bar is a many to many map between foo and bar. Here are the contents. select * from foo +------+ | fid | +------+ | 1 | | 2 | | 3 | | 4 | +------+ select * from foo2bar +------+------+ | fid | bid | +------+------+ | 1 | 1 | | 1 | 2 | | 2 | 1 | | 2 | 3 | | 4 | 4 | +------+------+ select * from bar +------+-------+------+ | bid | value | zid | +------+-------+------+ | 1 | 2 | 10 | | 2 | 4 | 20 | | 3 | 8 | 30 | | 4 | 42 | 30 | +------+-------+------+ What I want to request is, "Give me a list of all the fid and values with zid of 30" I expect an

Wow, it's hard to find a simple explanation to this topic. A simple many-to-many relationship. Three tables, tableA, tableB and a junction tableA_B. I know how to set up the relationship, with keys and all, but I get a little confused when time comes to perform INSERT, UPDATE and DELETE queries.... Basically, what I am looking for is an example that shows: How to get all records in TableA, based on an ID in TableB How to get all records in TableB, based on an ID in TableA 3 How to INSERT in either TableA or TableB, and then make the appropriate INSERT in the junction table to make the