在ArrayList中搜索特定对象

Question

I have a class called Person . It has the following attributes ; It has 2 attributes, ID , and Telephone . 1 person can have many telephones, so you may see people with multiple ID's below.

public ArrayList<Person> all(){

    p = new ArrayList<Person>();
    p.add(new Person(1,266763));
    p.add(new Person(1, 358643));
    p.add(new Person(2, 4667763));

    return p; 
}

There's another class called PersonDB . and it will have a method called, findPersonWithTheTelephoneNumber(int telephone) .

public void findPersonWithTheTelephoneNumber(int telephone) {
   Person pp = new Person();
   ArrayList<Person> personList = pp.all();

   // Now i want to find the Person object that will match the telephone number of these list of personList.


}

The personList, has 3-4 Person objects. I need to search the PersonArrayList and find the object that will match the Person object. How can i get this done ?

Note: i tried personList.contains() . But this doesn't work.

Answer 1

//...
Person foundPerson = null;
for (Person p : personList){
    if (p.getTelephone() == telephone){
         foundPerson = p; //or simply return it from there
         break;
    }
}

For implementing hashCode and equals you can observe this tutorial .

Answer 2

First of all, why are you not having a List<Integer> for storing all your telephoneNumbers for a particular person. That way, you won't have to create a separate Person instance for each telephoneNumber for the same personId , which simply makes no sense.

You can change your attributes of Person class to: -

private int id;
private List<Integer> telephoneNumbers = new ArrayList<Integer>();

And then have a list of Person, as you are having.

To find a Person with a particular telephoneNumber, you need to iterate through your List<Person> .

for (Person person: personList) {
    if (person.getTelephoneNumbers().contains(telephone)) {
        return person;
    }
}

Answer 3

i tried personList.contains()

Make sure you override Object.equals() and Object.hashCode() for Person class. But you have to have equality check on telephone number assuming telephone number unique. This would not be a solution but a workaround. Use bellum's answer. Mark it as correct answer.

Answer 4

您需要迭代数组，逐个检查人员的电话号码，当您到达那个时，您需要将其分配给变量。

Answer 5

Use a for loop to iterate through the list. In the body of the loop, check if the telephone number of the person in the list is the telephone number you're looking for. If yes, then return the person.

See The for Statement in Oracle's Java Tutorials.

Answer 6

Many solutions, define two persons as equal by their phone number but what if two persons that live in the same house and / or have the same phone number are added to the list? Which one is the correct one?.

Before rushing to find the person, you have to define a way to determine if two persons are, indeed, equal without ambiguous results. Unless you restrict the creation of persons based on that very phone number by making it unique (you do not clarify this in your question, so I assume there is no such restriction), the result of the search is undefined.

You're using an ArrayList so you can't guarantee even a result by insertion order.

I suggest you base the equality test in a person's ID instead of its phone. To prevent the modification of id 's just define a getter for it and do not define a setId method at all. Then, you can redefine equals (and hashcode if you feel like it) based on id .