Question

我有一个对话框bux，当选择了textView时会打开它。该对话框包含editText。我想要做的是能够在editText中键入一个sting，当按下'Done'按钮时，它会将editText中的字符串存储到一个变量中，在该变量中可以通过原始textView查看它。根据日食，我的源代码是正确的。但是当我在我的Android手机上运行它时，它会崩溃。我知道其原因是OnClick Listner。

以下是我的源代码：

public class MainActivity extends Activity {
/** Called when the activity is first created. */

TextView showPopUpButton;//NEW
EditText getInput;//NEW

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

    showPopUpButton = (TextView) findViewById(R.id.buttonShowPopUp);//NEW
    getInput = (EditText) findViewById(R.id.editText1);//NEW

    showPopUpButton.setOnClickListener(new OnClickListener() {
        @Override
        public void onClick(View v) {
            showPopUp3();              }
        });
}


private void showPopUp3() {
    AlertDialog.Builder helpBuilder = new AlertDialog.Builder(this);
    helpBuilder.setTitle("Enter your string");

    LayoutInflater inflater = getLayoutInflater();
    View checkboxLayout = inflater.inflate(R.layout.popuplayout, null);
    helpBuilder.setView(checkboxLayout);

    helpBuilder.setPositiveButton("Done", new DialogInterface.OnClickListener() {
        TextView showPopUpButton;//NEW
        EditText getInput;//NEW
        public void onClick(DialogInterface dialog, int which) {
            // THIS IS SUPPOSED TO STORE THE VALUE OF THE EDIT-TEXT AND OUTPUT IT IN THE TEXTVIEW
                showPopUpButton = (TextView) findViewById(R.id.buttonShowPopUp);//NEW
            showPopUpButton.setText(getInput.getText());//NEW**
            }
        });

    AlertDialog helpDialog = helpBuilder.create();
    helpDialog.show();
}

}

谢谢大家，我知道这个解决方案可能很容易，我只是遗漏了一些小东西。感谢

Answer 1

我认为你的TextView在主布局中，而EditText在popuplayout.xml中。如果是这种情况，请尝试以下代码。你不能直接引用EditText，你必须从它的布局中找到它（checkboxlayout）。

private void showPopUp3() {
AlertDialog.Builder helpBuilder = new AlertDialog.Builder(this);
helpBuilder.setTitle("Enter your string");

LayoutInflater inflater = getLayoutInflater();
View checkboxLayout =inflater.inflate(R.layout.popuplayout, null);
getInput = (EditText)  checkboxLayout.findViewById(R.id.editText1);
getInput.setText(YOUR_STRING);
helpBuilder.setView(checkboxLayout);

helpBuilder.setPositiveButton("Done", new DialogInterface.OnClickListener() {


    public void onClick(DialogInterface dialog, int which) {
        // THIS IS SUPPOSED TO STORE THE VALUE OF THE EDIT-TEXT AND OUTPUT IT IN THE TEXTVIEW

        showPopUpButton.setText(getInput.getText());//NEW**
        }
    });

AlertDialog helpDialog = helpBuilder.create();
helpDialog.show();
}

Answer 2

这应该可以正常工作。如果不是这样，你需要查看logcat并发布它崩溃时的内容。

helpBuilder.setPositiveButton("Done", new DialogInterface.OnClickListener() {
    @Override
    public void onClick(DialogInterface dialog, int which) {
            showPopUpButton.setText(getInput.getText());
        }
    });

由于您已经定义了showPopUpBox和getInput，并且您已经使用findViewById找到它们，因此您无需再次执行此操作。

Android Java：Diaglog setText＆amp; getText问题

2 个答案: