Question

我已经构建了一个辅助类来存储各种AlertDialog类型。我认为这将有用，以便我可以在我的代码中的任何地方调用它们。不幸的是，我在new AlertDialog.Builder()处收到了错误。它说Cannot resolve constructor `Builder()。我怎样才能让它发挥作用？

public class AlertDialogHelper {

    public void showAboutDialog() {
        AlertDialog.Builder builder = new AlertDialog.Builder();
        builder.setTitle(R.string.about);
        builder.setMessage("A weather app made by Martin Erlic")
                .setCancelable(false)
                .setPositiveButton("Ok", new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int id) {
                        // Ok
                    }
                });
        AlertDialog alert = builder.create();
        alert.show();
    }

}

在我的活动中：

private void showAboutAlertDialog() {
        AlertDialogHelper alertDialogHelper = new AlertDialogHelper();
        alertDialogHelper.showAboutDialog();
    }

Answer 1

你应该在构造函数中传递一个Context，如下所示：

AlertDialog.Builder builder = new AlertDialog.Builder(context);

来自您的活动：

alertDialogHelper.showAboutDialog(this);

现在：

public void showAboutDialog(Context context) {
        AlertDialog.Builder builder = new AlertDialog.Builder(context);
        ....

Answer 2

我是这样做的：

public class AlertDialogHelper {

    public static Dialog CreateDialog(Context mContext) {

        AlertDialog.Builder builder = new AlertDialog.Builder(mContext);
        builder.setTitle(R.string.about);
        builder.setMessage("A weather app made by Martin Erlic")
                .setCancelable(false)
                .setPositiveButton("Ok", new DialogInterface.OnClickListener() {
                    public void onClick(DialogInterface dialog, int id) {
                        // Ok
                    }
                });
        return builder.create();
    }
}

在我的活动中：

AlertDialogHelper.CreateDialog(this).show();

AlertDialog.Builder助手类无法解析构造函数

2 个答案: