Question

我在Android中使用Scala，我想将scala.collection.mutable.ArrayBuffer转换为Java.util.List。但它没有成功，有人可以帮助我吗？

代码：

package com.ssj.DoubanPipi.activity.fragment

import android.app.Fragment
import android.view.{View, LayoutInflater, ViewGroup}
import android.os.Bundle
import com.ssj.DoubanPipi.R
import android.widget.{ListView, SimpleAdapter}
import scala.collection.mutable.ArrayBuffer
import scala.collection.JavaConversions._

/**
 * Created by shenshijun on 14-5-31.
 */
class MainFragment extends Fragment {
  override def onCreateView(inflater: LayoutInflater, container: ViewGroup, savedInstanceState: Bundle) = {
    val rootView = inflater.inflate(R.layout.main_fragment, container, false)
    val listView = (rootView findViewById R.id.fragment_list).asInstanceOf[ListView]
    val adapter = new SimpleAdapter(getActivity, MainFragment.menu_list,
      R.layout.main_fragment_list, Array("text", "image"), Array(R.id.action, R.id.action_image))
    listView setAdapter adapter
    rootView
  }

  def onSign(view: View) {
  }
}

object MainFragment {
  val menu_list_text = Array(R.string.login, R.string.exit)
  val menu_list_image = Array(R.drawable.douban, R.drawable.exit)
  val menu_list = new ArrayBuffer[Map[String, Any]]()
  val len = menu_list_image.length
  for (index <- 0 until len) {
    val map = Map("text" -> menu_list_text(index), "image" -> menu_list_image(index))
    menu_list += map
  }
}

，错误消息如下：

type mismatch;
 found   : scala.collection.mutable.ArrayBuffer[scala.collection.immutable.Map[String,Any]]
 required: java.util.List[_ <: java.util.Map[String, _]]
    val adapter = new SimpleAdapter(getActivity, MainFragment.menu_list,
                                                              ^

Answer 1

在menu_list上调用.toList（）可以完成这项工作。

    MainFragment.menu_list.toList

编辑：

如果您还需要将scala Map转换为java.util.Map，则此代码段应该有效：

~~menu_list.map（_。asInstanceOf [java.util.Map [String，Any]]）~~

val javaMenuList: java.util.List[java.util.Map[String, Any]] = menu_list.map { map =>
  val javaMap: java.util.Map[String, Any] = map
  javaMap
}

为什么在Java List出现时，不会为缓冲区发生JavaConversions._？

1 个答案: