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在Kotlin中创建Springs ParameterizeTypeReference的实例

spring-test kotlin generics spring-boot

thomas77 · 技术社区 · 8 年前

spring-boot-test RestTestTemplate .

RestTestTemplate测试模板 有一个 exchange 方法需要 ParameterizedTypeReference

class ListOfPeople : ParameterizedTypeReference<List<Person>>()

以下是我的测试方法:

@Test
fun `get list of people`() {
    // create testdata
    datastore.save(Person(firstname = "test1", lastname = "lastname1"))
    datastore.save(Person(firstname = "test2", lastname = "lastname2"))
    datastore.save(Person(firstname = "test3", lastname = "lastname2"))
    datastore.save(Person(firstname = "test4", lastname = "lastname2"))

    val requestEntity = RequestEntity<Any>(HttpMethod.GET, URI.create("/person"))

    // create typereference for response de-serialization
    class ListOfPeople : ParameterizedTypeReference<List<Person>>() // can this be done inline in the exchange method?
    val responseEntity : ResponseEntity<List<Person>> = restTemplate.exchange(requestEntity, ListOfPeople())

    assertNotNull(responseEntity)
    assertEquals(200, responseEntity.statusCodeValue)
    assertTrue( responseEntity.body.size >= 4 )

    responseEntity.body.forEach { person ->
        println("Found person: [${person.firstname} ${person.lastname}] " +
                ", born [${person.birthdate}]")
    }
}

如果有帮助的话,下面是整个测试的链接: testclass on github

2 回复 | 直到 8 年前

Alexey Romanov 8 年前

虽然使用对象表达式的答案是正确的,并且与在Java中的方法直接等效,但是如果需要很多参数,具体化的类型参数允许您简化它 ParameterizedTypeReference 学生:

inline fun <reified T> typeReference() = object : ParameterizedTypeReference<T>() {}

// called as
restTemplate.exchange(requestEntity, typeReference<List<Person>>())

当编译器看到 typeReference<SomeType> object : ParameterizedTypeReference<SomeType>() {} .

thomas77 8 年前

感谢JB Nizet给我指出了正确的文档。

val responseEntity : ResponseEntity<List<Person>> =
            restTemplate.exchange(requestEntity, 
            object: ParameterizedTypeReference<List<Person>> () {})

如果我读对了,这就叫对象表达式。

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