使用EF处理嵌套表

Question

Premises:

1. I have 3 tables in this order: Company -> Person -> Phone
2. A Company can have multiple Person that can have multiple Phone .
3. They all have incremental INT as PK .
4. They have an indentified relationship, so Phone have id_phone & id_company & id_person as PK .

Use Case:

1. User wants to add a Company with a Person with a Phone .

Problem:

1. How can I add a Person to a Company if I don't have the parent's PK ? (the same happens to Phone )

Answer 1

the context will track the additions made the child collections

var phone = new Phone() { //whatever };
var person = new Person() { //whatever };
person.Phones.Add(phone);

var company = new Company() { //whatever };
company.People.Add(person);

dbContext.Companies.Add(company);
dbContext.SaveChanges();

As long as you've defined the relationship, it will automatically set the correct foreign keys etc.

Here's an example for code first

public class Company
{
    [Key]
    [DatabaseGeneratedAttribute(DatabaseGeneratedOption.Identity)]
    public int Id;
    // other properties
    public virtual Collection<Person> People { get; set; }
}

public class Person
{
    [Key]
    [DatabaseGeneratedAttribute(DatabaseGeneratedOption.Identity)]
    public int Id;
    // Other properties
    [ForeignKey("Company")]
    public int CompanyId { get; set; }
    public virtual Company Company { get; set; }
    public virtual Collection<Phone> Phones { get; set; }
}

UPDATE: you can add additional children in a similar manner, you just have to fetch the entity first

var db = new DbContext();
var company = db.Companies.FirstOrDefault(x => x.Id == 1);
if(company != null)
{
    var person = new Person() { // blah };
    company.People.Add(person);
    db.SaveChanges();
}

Answer 2

Just do

company.Persons.Add(new Person
    {
        //set Person properties here but dont worry about PK - EF sorts this out for you
    });

you can even do:

company.Persons.Addcnew Person
    {
        //set Person properties here but dont worry about PK - EF sorts this out for you
        PhoneNumbers= new PhoneNumber[]
        {
            new PhoneNumber() {/*..*/ }
        }
    });