不显示警报对话框
not showing alertdialog box
DialogInterface.OnClickListener clickListener= new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
switch (which)
{
case BUTTON_POSITIVE :
udb.signout();
break;
case BUTTON_NEGATIVE:
finish();
break;
}
}
};
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
builder.setTitle("Notification");
builder.setMessage("You are already logged in.\nDo you want to signout and login with different account?");
builder.setPositiveButton("Yes",clickListener);
builder.setNegativeButton("No",clickListener);
builder.show();
这是我显示弹出对话框的代码..但我遇到了问题
"builder.show()" 行。而且我不明白我做错了什么。请。我会很感激任何帮助
首先调用 builder.create() 。您不能显示构建器本身。
您必须 create
您的 AlertDialog.Builder
对话框然后显示它...删除 builder.show();
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
builder.setTitle("Notification");
builder.setMessage("You are already logged in.\nDo you want to signout and login with different account?");
builder.setPositiveButton("Yes",clickListener);
builder.setNegativeButton("No",clickListener);
AlertDialog alert = builder.create();
alert.show();
您需要使用 AlertDialog.Builder 对象创建 AlertDialog 对象并显示对话框。例如,
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
/*All your dialog code*/
//Create AlertDialog object
AlertDialog alertDialog = builder.create ();
//show the AlertDialog using show() method
alertDialog.show();
首先,当我在我的设备上亲自测试时,
builder.show();
应该和
效果一样
AlertDialog dialog = builder.create();
dialog.show();
不管是android.support.v7.app.AlertDialog还是android.app.AlertDialog。
就我而言,我发现导致问题的原因是 AlertDialog.Builder 的初始化方式。
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
从 Dialog documentation 开始,您需要将 Activity 传递给此构造函数,以下内容将起作用:
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
或者,
AlertDialog.Builder builder = new AlertDialog.Builder(<YourActivity>.this);
DialogInterface.OnClickListener clickListener= new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
switch (which)
{
case BUTTON_POSITIVE :
udb.signout();
break;
case BUTTON_NEGATIVE:
finish();
break;
}
}
};
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
builder.setTitle("Notification");
builder.setMessage("You are already logged in.\nDo you want to signout and login with different account?");
builder.setPositiveButton("Yes",clickListener);
builder.setNegativeButton("No",clickListener);
builder.show();
这是我显示弹出对话框的代码..但我遇到了问题 "builder.show()" 行。而且我不明白我做错了什么。请。我会很感激任何帮助
首先调用 builder.create() 。您不能显示构建器本身。
您必须 create
您的 AlertDialog.Builder
对话框然后显示它...删除 builder.show();
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
builder.setTitle("Notification");
builder.setMessage("You are already logged in.\nDo you want to signout and login with different account?");
builder.setPositiveButton("Yes",clickListener);
builder.setNegativeButton("No",clickListener);
AlertDialog alert = builder.create();
alert.show();
您需要使用 AlertDialog.Builder 对象创建 AlertDialog 对象并显示对话框。例如,
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
/*All your dialog code*/
//Create AlertDialog object
AlertDialog alertDialog = builder.create ();
//show the AlertDialog using show() method
alertDialog.show();
首先,当我在我的设备上亲自测试时,
builder.show();
应该和
效果一样AlertDialog dialog = builder.create();
dialog.show();
不管是android.support.v7.app.AlertDialog还是android.app.AlertDialog。
就我而言,我发现导致问题的原因是 AlertDialog.Builder 的初始化方式。
AlertDialog.Builder builder = new AlertDialog.Builder(getApplicationContext());
从 Dialog documentation 开始,您需要将 Activity 传递给此构造函数,以下内容将起作用:
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
或者,
AlertDialog.Builder builder = new AlertDialog.Builder(<YourActivity>.this);