带有规范的 kotlin 单元测试

问题描述

我需要获取带有规范的查询结果。为了做到这一点，我使用了 JpaSpecificationExecutor 的 List<T> findAll(@Nullable Specification<T> spec) 方法。问题是我不能在测试中做同样的事情，因为它有几个具有相同参数的方法。 这是我的方法：

fun getFaqList(category: FaqCategory?,subcategory: FaqCategory?,searchText: String?): List<FaqEntity> {

    val spec = Specification.where(FaqSpecification.categoryEquals(category))
        ?.and(FaqSpecification.subcategoryEquals(subcategory))
        ?.and(
            stringFieldContains("title",searchText)
                ?.or(stringFieldContains("description",searchText))
        )

    return faqRepository.findAll(spec)
}

以及我尝试运行的测试：

@MockK
private lateinit var faqRepository: FaqRepository

@InjectMockKs
private lateinit var faqService: FaqService

companion object {
    val FAQ_CATEGORY_ENTITY = FaqCategoryEntity(
        id = AGRICULTURE
    )

    val FAQ_SUBCATEGORY_ENTITY = FaqCategoryEntity(
        id = AGRICULTURE_GENERAL
    )

    val FAQ_ENTITY = FaqEntity(
        id = FAQ_ID,title = "title",description = "description",category = FAQ_CATEGORY_ENTITY,subcategory = FAQ_SUBCATEGORY_ENTITY
    )
}

@Test
fun `getFaqList - should return faq list`() {
    val faqList = listOf(FAQ_ENTITY)

    every { faqRepository.findAll(any()) } returns faqList

    val response = faqService.getFaqList(AGRICULTURE,AGRICULTURE_GENERAL,FAQ_SEARCH_TEXT)

    assertThat(response).isEqualTo(faqList)
}

我收到错误：

Overload resolution ambiguity. All these functions match.
public abstract fun <S : FaqEntity!> findAll(example: Example<TypeVariable(S)!>): 
(Mutable)List<TypeVariable(S)!> defined in kz.btsd.backkotlin.faq.FaqRepository
public abstract fun findAll(pageable: Pageable): Page<FaqEntity!> defined in 
kz.btsd.backkotlin.faq.FaqRepository
public abstract fun findAll(sort: Sort): (Mutable)List<FaqEntity!> defined in 
kz.btsd.backkotlin.faq.FaqRepository
public abstract fun findAll(spec: Specification<FaqEntity!>?): (Mutable)List<FaqEntity!> 
defined in kz.btsd.backkotlin.faq.FaqRepository

我应该在 findAll() 参数中写什么以便 spring 理解：

faqRepository.findAll(any())

解决方法

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