Question

我有一个带有简单自定义布局的DialogFragment（TextView和EditView）。

当用户单击Dialog的Positive Button时，应用程序应将EditText字段中的用户输入保存到变量“playerName”中。无论我在尝试什么，Toast-output总是显示一个空字符串“”，或者更确切地说，当我将文本硬编码到EditText中时（如XML中所示），它始终显示该文本。

用户对键盘输入所做的更改不会在代码中得到处理，任何想法都有什么问题？

DialogFragment

@Override
public Dialog onCreateDialog(Bundle savedInstanceState) {

 builder.setPositiveButton("Save", new DialogInterface.OnClickListener() {
   public void onClick(DialogInterface dialog, int id) {
     LayoutInflater inflater = getActivity().getLayoutInflater();
     View view = inflater.inflate(R.layout.addplayer_fragment, null);

     // declare the text input field
     EditText playerNameEdit = (EditText)view.findViewById(R.id.playerNameEdit);

     //read text into String
     String playerName = playerNameEdit.getText().toString();

     // make toast with input of the edit text field
     Toast.makeText(getActivity().getApplicationContext(), playerName, Toast.LENGTH_SHORT).show();
        }

布局（addplayer_fragment.xml）

    <LinearLayout
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:padding= "10dip" >

        <TextView
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:text="Name: "/>

        <EditText
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:text="TestText"
            android:id="@+id/playerNameEdit"/>

    </LinearLayout>
  </RelativeLayout>

Answer 1

您还可以在片段

中使用以下方法

void showDialog()
{
    View view=getActivity().getLayoutInflater().inflate(R.layout.addplayer_fragment,null);
    final EditText playerNameEdit = (EditText)view.findViewById(R.id.playerNameEdit);
    new AlertDialog.Builder(getActivity())
            .setTitle("Title here")
            .setMessage("Message here")
            .setView(view)
            .setPositiveButton("Save", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialogInterface, int i) {
                    String name=playerNameEdit.getText().toString();
                    Toast.makeText(getActivity(), name, Toast.LENGTH_SHORT).show();
                }
            })
            .create().show();
}

Answer 2

试试这段代码：

@Override
public Dialog onCreateDialog(Bundle savedInstanceState) {
    // Use the Builder class for convenient dialog construction
    AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());

    builder.setTitle("Your Title");

    LayoutInflater inflater = getActivity().getLayoutInflater();
    View dialogView = inflater.inflate(R.layout.addplayer_fragment, null);
    builder.setView(dialogView)

    .setPositiveButton("Save", new DialogInterface.OnClickListener() {
            @Override
            public void onClick(DialogInterface dialog, int id) {

                EditText valueView = (EditText) dialogView.findViewById(R.id.playerNameEdit); 
                if(valueView == null) Log.d("VIEW", "NULL");
                else{
                    String playerName = playerNameEdit.getText().toString();
                    Toast.makeText(getActivity(),playerName , Toast.LENGTH_SHORT).show();
                }
            })

    .setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog, int id) {

        }
    }); 

    return builder.create();

}

Android EditText不处理键盘输入

2 个答案: