Question

最近我一直试图用ExpandableListView重新组织一些菜单，并让它工作，但硬编码。现在我正在寻找一种从strings.xml获取值的方法，并遇到了一些问题。

这是我的ATM。

这是标题。

@Override
protected void onCreate(Bundle savedInstanceState)
{
  super.onCreate(savedInstanceState);
  setContentView(R.layout.activity_main);

  headersList = new ArrayList<>();
    String[] parentHeaders = getResources().getStringArray(R.array.headers);
    List<String> parentHeaders = new ArrayList<String>(Arrays.asList(headers));
    headersList.addAll(parentHeaders);

    HashMap<String, List<String>> allChildItems = returnGroupedChildItems();
    expandableListView = (ExpandableListView)findViewById(R.id.expandableListView);
    ExpandableListViewAdapter expandableListViewAdapter = new ExpandableListViewAdapter(getApplicationContext(), headersList, allChildItems);
    expandableListView.setAdapter(expandableListViewAdapter);
}

而不是那样，我想用我的strings-arrays中的以下片段填充它：

<string-array name = "headers">
  <item>@string/parent1</item>
  <item>@string/parent2</item>
  <item>@string/parent3</item>
</string-array>

最后，我为孩子们准备了这个：

private HashMap<String, List<String>> returnGroupedChildItems()
{
  HashMap<String, List<String>> childList = new HashMap<String, List<String>>();

  List<String> parent1 = new ArrayList<String>();
    String[] child1 = getResources().getStringArray(R.array.parent1);
    List<String> childHeaders1 = new ArrayList<String>(Arrays.asList(child1));
    parent1.addAll(childHeaders1);

  List<String> parent2 = new ArrayList<String>();
    String[] child2 = getResources().getStringArray(R.array.parent2);
    List<String> childHeaders2 = new ArrayList<String>(Arrays.asList(child2));
    parent1.addAll(childHeaders2);

  List<String> parent3 = new ArrayList<String>();
    String[] child3 = getResources().getStringArray(R.array.parent3);
    List<String> childHeaders3 = new ArrayList<String>(Arrays.asList(child3));
    parent3.addAll(childHeaders3);

  childList.put(headersList.get(0), parent1);
  childList.put(headersList.get(1), parent2);
  childList.put(headersList.get(2), parent3);

  return childList;
}

同样，我有3个其他字符串数组，包含所有这些值。

<string-array name = "parent1">
  <item>@string/child1_1</item>
  <item>@string/child1_2</item>
  <item>@string/child1_3</item>
  <item>@string/child1_4</item>
</string-array>

<string-array name = "parent2">
  <item>@string/child2_1</item>
  <item>@string/child2_2</item>
  <item>@string/child2_3</item>
  <item>@string/child2_4</item>
</string-array>

<string-array name = "parent3">
  <item>@string/child3_1</item>
  <item>@string/child3_2</item>
  <item>@string/child3_3</item>
  <item>@string/child3_4</item>
</string-array>

尝试使用getResources().getStringArray(R.array.parent1)，但它不起作用。

这也是我的适配器，如果有帮助的话。

import android.content.Context;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.BaseExpandableListAdapter;
import android.widget.TextView;
import java.util.HashMap;
import java.util.List;


public class ExpandableListViewAdapter extends BaseExpandableListAdapter
{

  private Context context;
  private List<String> parentDataSource;
  private HashMap<String, List<String>> childDataSource;

  public ExpandableListViewAdapter(Context context, List<String> childParent, HashMap<String, List<String>> child)
  {
    this.context = context;
    this.parentDataSource = childParent;
    this.childDataSource = child;
  }

  @Override
  public int getGroupCount()
  {
    return this.parentDataSource.size();
  }

  @Override
  public int getChildrenCount(int groupPosition)
  {
    return this.childDataSource.get(this.parentDataSource.get(groupPosition)).size();
  }

  @Override
  public Object getGroup(int groupPosition)
  {
    return parentDataSource.get(groupPosition);
  }

  @Override
  public Object getChild(int groupPosition, int childPosition)
  {
    return this.childDataSource.get(parentDataSource.get(groupPosition)).get(childPosition);
  }

  @Override
  public long getGroupId(int groupPosition)
  {
    return groupPosition;
  }

  @Override
  public long getChildId(int groupPosition, int childPosition)
  {
    return childPosition;
  }

  @Override
  public boolean hasStableIds()
  {
    return false;
  }

  @Override
  public View getGroupView(int groupPosition, boolean isExpanded, View convertView, ViewGroup parent)
  {
    View view = convertView;
    if(view == null)
    {
      LayoutInflater inflater = (LayoutInflater)this.context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
      view = inflater.inflate(R.layout.parent_layout, parent, false);
    }
    String parentHeader = (String)getGroup(groupPosition);
    TextView parentItem = (TextView)view.findViewById(R.id.parent_layout);
    parentItem.setText(parentHeader);
    return view;
  }

  @Override
  public View getChildView(int groupPosition, int childPosition, boolean isLastChild, View convertView, ViewGroup parent)
  {
    View view = convertView;
    if(view == null)
    {
      LayoutInflater inflater = (LayoutInflater)this.context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
      view = inflater.inflate(R.layout.child_layout, parent, false);
    }
    String childName = (String)getChild(groupPosition, childPosition);
    TextView childItem = (TextView)view.findViewById(R.id.child_layout);
    childItem.setText(childName);
    return view;
  }

  @Override
  public boolean isChildSelectable(int groupPosition, int childPosition)
  {
    return true;
  }
}

最后，设法把它放在一起。我可以直接从字符串填充我的列表。所以，我的最后一个问题是，如何创建一个循环来检查我的父数组，获取它的名称并使用相应的数组填充每个子项？

编辑3：我的循环上有一些变化。它仍然是硬编码的，但它不需要额外的变量（虽然使用了很多 ifs ）。

int size = headersList.size();

for (int i = 1; i <= size; i++)
  {
  List<String> childList = new ArrayList<>();
  if (i == 1)
    {
      String[] child = getResources().getStringArray(R.array.parent1);
      List<String> childHeaders = new ArrayList<>(Arrays.asList(child));
      childList.addAll(childHeaders);
    } else if (i == 2)
    {
      String[] child = getResources().getStringArray(R.array.parent2);
      List<String> childHeaders = new ArrayList<>(Arrays.asList(child));
      childList.addAll(childHeaders);
    } else if (i == 3)
    {
      String[] child = getResources().getStringArray(R.array.parent3);
      List<String> childHeaders = new ArrayList<>(Arrays.asList(child));
      childList.addAll(childHeaders);
    }

    childContent.put(headersList.get(i - 1), childList);
  }

正如我所说，这是有效的，并使用期望的值填充我的列表，但我想在没有所有条件的情况下执行它（我只需要动态地将每个父＃array传递给我的 child String ）。

所以，我的最后一个问题是，有没有办法在循环的每次迭代中检索headersList.get（i-1），并在getStringArray（）中使用该名称？

Answer 1

getResources().getStringArray(R.array.parent1)将返回String []而不是List<String>。要将String []转换为列表，请使用以下

String [] parent1 = getResources().getStringArray(R.array.parent1);

List<String> list = new ArrayList<String>(Arrays.asList(parent1));

Answer 2

您可以尝试重载构造函数

public ExpandableListViewAdapter(Context context, List<String> childParent, HashMap<String, List<String>> child)
{
    this.context = context;
    this.parentDataSource = childParent;
    this.childDataSource = child;
}

public ExpandableListViewAdapter(Context context, String[] childParent, HashMap<String, String[]> child) 
{
    this.context = context;
    this.parentDataSource = Arrays.asList(childParent);
    this.childDataSource = new HashMap<>();
    for (Entry<String, String[]> entry : child.entrySet()) {
        childDataSource.put(entry.getKey(), Arrays.asList(entry.getValue());
    } 
}

并回答

但是我想在没有所有条件的情况下这样做（我只需要动态地将每个父＃array传递给我的子String）。

所以，我的最后一个问题是，有没有办法在循环的每次迭代中检索headersList.get（i-1），并在getStringArray（）中使用该名称？

是的，getResources().getIdentifier

@Override
protected void onCreate(Bundle savedInstanceState)
{
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    final Resources res = getResources();

    // Build Hashmap 
    Map<String, String[]> allChildItems = new HashMap<>();
    String[] headersList = res.getStringArray(R.array.headers);
    for (String header : headersList) {
        int resId = res.getIdentifier(header, "array", getPackageName());
        allChildItems.put(header, res.getStringArray(resId));
    }

    // Build adapter
    ExpandableListViewAdapter expandableListViewAdapter = new ExpandableListViewAdapter(MainActivity.this, headersList, allChildItems);
    expandableListView.setAdapter(expandableListViewAdapter);
}

（代码未经测试）

strings.xml中的ExpandableListView

2 个答案: