构造函数的参数太多

Question

正在尝试学习一些 Scala。

我的项目中有以下类：

package com.fluentaws

class AwsProvider(val accountId: String, val accountSecret: String) {

 def AwsAccount = new AwsAccount(accountId, accountSecret)

}

class AwsAccount(val accountId : String, val accountSecret : String) {

}

以及以下测试：

package com.fluentaws

import org.scalatest._

class AwsProvider extends FunSuite {

  test("When providing AwsProvider with AWS Credentials we can retrieve an AwsAccount with the same values") {

    val awsAccountId = "abc"
    val awsAccountSecret = "secret"

    val awsProvider = new AwsProvider(awsAccountId, awsAccountSecret)

    val awsAccount = awsProvider.AwsAccount

    assert(awsAccount.accountId == awsAccountId)
    assert(awsAccount.accountSecret == awsAccountSecret)
  }

}

当我的测试套件运行时，出现编译时错误：

too many arguments for constructor AwsProvider: ()com.fluentaws.AwsProvider [error] val awsProvider = new AwsProvider(awsAccountId, awsAccountSecret) [error]

从错误消息来看，它似乎看到了一个参数为零的构造函数？

谁能看出我做错了什么？

Answer 1

这是一个典型的菜鸟错误。我修复了我的 test-class' 名称，因为使用相同的名称会影响原始名称，因此我实际上是在测试我的测试 class:

package com.fluentaws

import org.scalatest._

class AwsProviderTestSuite extends FunSuite {

  test("When providing AwsProvider with AWS Credentials we can retrieve an AwsAccount with the same values") {

    val awsAccountId = "abc"
    val awsAccountSecret = "secret"

    val awsProvider = new AwsProvider(awsAccountId, awsAccountSecret)

    val awsAccount = awsProvider.AwsAccount

    assert(awsAccount.accountId == awsAccountId)
    assert(awsAccount.accountSecret == awsAccountSecret)
  }

}

现在它过去了。

构造函数的参数太多

Too many arguments for constructor

tdd

constructor

scala

classname