CODEWITHSUNDEEP
java
Raju Sharma
Raju Sharma

Reputation: 2516

How do i filter a list of object and return that particular filtered object in Java 8

I am trying to get a filtered object through a stream api.

I have a list of VendorAddress , where a single address is a primary address. Here I am trying to filter that address and want it to return a VendorAddress object.

See below code:

 vendorAddressList.stream().filter(vendorAddressObj-> vendorAddressObj.isPrimary())

What should I do to get the filtered VendorAddress object.

Thanks in advance.

Upvotes: 1

Views: 76

Answers (3)

Seelenvirtuose
Seelenvirtuose

Reputation: 20618

Use the findAny terminal method:

Optional<VendorAddress> maybeAnAddress = vendorAddressList
        .stream()
        .filter(VendorAddress::isPrimary) // Note: I used a method reference here.
        .findAny();
VendorAddress primaryAddress = maybeAnAddress.orElse(null);

Upvotes: 3

Sergey Prokofiev
Sergey Prokofiev

Reputation: 1885

There are dozen ways to achieve that you want. For instance, you can grab first matched element like this

Optional<VendorAddress> result = vendorAddressList.stream()
    .filter(vendorAddressObj-> vendorAddressObj.isPrimary()).findFirst();

or any matched 

Optional<VendorAddress> result = vendorAddressList.stream()
    .filter(vendorAddressObj-> vendorAddressObj.isPrimary()).findAny();

In case if you're want all matched elements

List<VendorAddress> result = vendorAddressList.stream()
    .filter(vendorAddressObj-> vendorAddressObj.isPrimary()).collect(Collectors.toList());

and many more... Hope it helps!

Upvotes: 2

Jotunacorn
Jotunacorn

Reputation: 526

You can use findFirst and either get (if you want an exception when not found) or orElse(null) (if you want null when not found).

Like this

VendorAddress address =  vendorAddressList.stream()
    .filter(vendorAddressObj-> vendorAddressObj.isPrimary())
    .findFirst() //Can be changed to findAny()
    .orElse(null); //Can be changed to get()

This will return the first found instance in your list. If you are sure you only have one, or does not care that it strictly returns the same (first) element every time, you can use findAny() instead of findFirst() for better performance.

Upvotes: 1

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