REST Assured - Generic List deserialization

Question

Let's say I have a Java Person class:

class Person {
    String name;
    String email;
}

With REST Assured, you can deserialize this JSON object

{"name":"Bob", "email":"[email protected]"}

to a Java Person instance using

Person bob = given().when().get("person/Bob/").as(Person.class);

How does one use REST Assured to deserialize this JSON array

[{"name":"Bob", "email":"[email protected]"}, 
 {"name":"Alice", "email":"[email protected]"}, 
 {"name":"Jay", "email":"[email protected]"}]

into a List<Person>? For example, this would be handy:

List<Person> persons = given().when().get("person/").as(...);

spg · Accepted Answer · 2013-07-26T13:12:25.423

77

I found a way to achieve what I wanted:

List<Person> persons = given().when().get("person/").as(Person[].class);

UPDATE: Using Rest-Assured 1.8.1, looks like cast to List is not supported anymore. You need to declare and object array like this:

Person[] persons = given().when().get("person/").as(Person[].class);

edited Jul 26 '13 at 13:12

answered Mar 22 '13 at 17:53

spg

9,309
4
36
41

did not work for me, with java.lang.NoSuchMethodError: com.fasterxml.jackson.databind.introspect.AnnotatedMethod.getParameterType(I) . Is there any prerequisites for this stuff? – Andrii Plotnikov Jul 29 '16 at 14:12
1

Its working perfectly for me. I am using https://mvnrepository.com/artifact/io.rest-assured/rest-assured/3.0.1 – Ramu Nov 24 '16 at 06:35
my version is 2.4.1. so that may be the reason. – Andrii Plotnikov Dec 13 '16 at 11:34

Christian Dräger · Answer 2 · 2018-10-26T10:26:27.597

To extract a Java List, and not an Array, from a JSON API response, you just have to remember to use jsonPath rather than as:

List<Person> persons = given()
        .when()
        .get("/person")
        .then()
        .extract()
        .body()
        // here's the magic
        .jsonPath().getList(".", Person.class);

Your json path can point to anywhere you expect to have a list of json objects in your body. in this example (and working for your question) it just points to the json root.

sidenode: rest-assured is internally using jackson for deserialization (for .jsonPath as well as .as)

Andrii Plotnikov · Answer 3 · 2018-12-11T11:41:49.980

28

for those who found out that accepted answer does not work anymore.

    List<Entity> list = new ArrayList<>();
    list = given()
            .contentType(CONTENT_TYPE)
        .when()
            .get(getRestOperationPath())
        .then()
            .extract().body().as(list.getClass());

hopefully, you understand that getRestOperationPath is returning rest operation path; and CONTENT_TYPE is placeholder for your content type (application/json for example)

upd: checked different versions, behavior differs depending on version, so you might want to try different approaches

upd2: cleaner solution was pointed by @Arigion in comments:

to use .extract().body().jsonPath().getList(".", Entity.class);

edited Dec 11 '18 at 11:41

answered Sep 20 '16 at 08:14

Andrii Plotnikov

3,022
4
17
37

2

This should be the correct answer. Clean, simple, and the most up-to-date response. – Ryan Cox Sep 27 '16 at 20:19
6

it is better to use .extract().body().jsonPath().getList(".", Entity.class); You'd get an unchecked warning otherwise. – Arigion Sep 04 '18 at 07:41
@Arigion included that into answer :) please tell me if you made Answer with that so I can link it; I didn't find it. – Andrii Plotnikov Dec 11 '18 at 12:53

score 19 · Answer 4 · answered Mar 24 '21 at 18:43

19

We can now use TypeRef much as it's possible to do it with the JsonPath library:

List<Person> persons = given().when().get("person/")
    .as(new TypeRef<List<Person>>() {});

As with https://github.com/json-path/JsonPath#what-is-returned-when - the anonymous inner class new TypeRef<List<Person>>() {} gets around type erasure and captures the type information enough that the framework can access the raw type - List in this case. The internal generic type - Person - is a safe cast that can be made under the circumstances.

answered Mar 24 '21 at 18:43

Ashley Frieze

4,993
2
29
23

2

Shorter-yet syntax also worked great for me: `.as(new TypeRef<>() {})` thanks to `import io.restassured.mapper.TypeRef;` even with old 2019 `'io.rest-assured', version: '3.3.0'` – MarkHu May 27 '21 at 00:43
@MarkHu that's nice! – Ashley Frieze May 27 '21 at 08:28

score 13 · Answer 5 · answered May 14 '13 at 15:42

13

You could also do this if you were interested in using "expect()"

expect().
 body("get(0).firstName", equalTo("Mike")).
when().
 get("person/");

This was my case

answered May 14 '13 at 15:42

allegjdm93

592
6
24

score 8 · Answer 6 · answered May 10 '18 at 11:57

If anyone's still looking. Using Java 1.8 and RestAssured 2.9 the solution is really simple and it does not throw "Unchecked Warning".

return Arrays.asList(given()
            .when()
            .get("restPath")
            .then()
            .extract()
            .response()
            .body()
            .as(Player[].class));

score 0 · Answer 7 · answered Mar 14 '18 at 05:53

0

If you are not comfortable with JsonPath, i would suggest using any java serialization/de-serialization using GSON or Jackson.

answered Mar 14 '18 at 05:53

Kshetra Mohan Prusty

301
3
8

Vikas Kumar · Answer 8 · 2021-07-05T09:59:56.570

-1

Since you want to map

[{"name":"Bob", "email":"[email protected]"}, 
 {"name":"Alice", "email":"[email protected]"}, 
 {"name":"Jay", "email":"[email protected]"}]

to java object.

Now while calling the endpoint you can directly map to the Person list as shown below

List<Person> personList = given().when().get("person/").as(Person[].class);

Note the Person class should remain same i.e. no need to do any modification for the person class.

public class Person{
 String name;
 String email;
}

edited Jul 05 '21 at 09:59

answered Feb 03 '21 at 16:07

Vikas Kumar

13
1
12

The type `PersonArray` has a _field_ `personList` which is not present in the above data. – Ashley Frieze Mar 24 '21 at 18:43
@AshleyFrieze Thanks for the comment, I have added additional wrapper. I have modified my solution – Vikas Kumar Jul 05 '21 at 10:01

score -2 · Answer 9 · answered Dec 16 '18 at 07:19

This would be helpful, works with current version of rest assured.

@Test
    public void apostUser() {
        Map<String,String> user = new HashMap<>();
        user.put("User_Id", "xyx1111");
        user.put("First_Name", "KS");
        user.put("Designation", "DR");

        given()
        .contentType("application/json")
        .body(user)
        .when().post("URI").then()
        .statusCode(200);
    }

REST Assured - Generic List deserialization

9 Answers9