在Kotlin中创建Spring的ParameterizedTypeReference实例 [英] Create instance of Spring´s ParameterizedTypeReference in Kotlin

问题描述

我正在尝试学习Kotlin，并测试它在Spring Boot中如何工作.我的应用程序使用mongo数据库存储数据，并且我有一个Jersey资源来检索数据.我正在使用spring-boot-test和RestTestTemplate对其进行测试.

I am trying to learn Kotlin, and test how it works with spring boot. My application is using a mongo database to store data and I have a Jersey resource for retrieving data. I am testing it using spring-boot-test and RestTestTemplate.

RestTestTemplate具有采用ParameterizedTypeReference的exchange方法.此类具有受保护的构造函数.因此，我从Kotlin设法使用它的唯一方法是:

The RestTestTemplate has an exchange method which takes a ParameterizedTypeReference. This class has a protected constructor. So the only way I managed to use it from Kotlin was like this:

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

这是我的测试方法:

@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}]")
    }
}

这是正确的(或唯一的)方法吗?还是有更好的方法?

Is this the correct (or only) way to do this, or is there a better way?

如果有帮助，这里是整个测试的链接:

If it helps, here is a link for the whole test: testclass on github

推荐答案

虽然使用对象表达式的答案是正确的，并且直接等效于您在Java中的处理方式，但是在需要大量代码的情况下，通过类型化的类型参数可以简化它ParameterizedTypeReference s:

While the answer using object expression is correct and the direct equivalent of the way you do it in Java, reified type parameters allow you to simplify it if you need many ParameterizedTypeReferences:

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

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

当编译器看到typeReference<SomeType>调用时，它将被定义替换，因此结果与编写object : ParameterizedTypeReference<SomeType>() {}相同.

When the compiler sees a typeReference<SomeType> call, it's replaced by the definition, so the result is the same as if you wrote object : ParameterizedTypeReference<SomeType>() {}.

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