Question

Android 按钮弹出提示框问题

private void showDialog(String title,int msg){
AlertDialog.Builder builder= new AlertDialog.Builder(this);
builder.setIcon(android.R.drawable.ic_dialog_info);
builder.setTitle(title);
builder.setMessage(msg);
builder.setPositiveButton("确认", null);
builder.create().show();
}
public void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);
setContentView(R.layout.main);

button1=(Button)findViewById(R.id.button1);
button1.setOnClickListener(new Button.OnClickListener(){

public void onClick(View v) {
// TODO Auto-generated method stub
showDialog("结果",1);
}
});
}
这样写为什么在虚拟器里点击button后 运行不了呢？求高手指点

Answer 1

你改成这样试试：
public void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);
setContentView(R.layout.main);

button1=(Button)findViewById(R.id.button1);
button1.setOnClickListener(new Button.OnClickListener(){

public void onClick(View v) {
// TODO Auto-generated method stub
AlertDialog.Builder builder= new AlertDialog.Builder(this);
builder.setIcon(android.R.drawable.ic_dialog_info);
builder.setTitle(title);
builder.setMessage(msg);
builder.setPositiveButton("确认", null);
builder.create().show();
}
});
}