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我有一個彈出窗口,顯示在最初的應用程序啓動。這裏是代碼,我用來創建彈出窗口如何製作一個彈出式窗口模式
private void loadPopup(View view, boolean loadSchool){
Log.i("Started Info","popup");
//.......
//create the popup window
int width = LinearLayout.LayoutParams.WRAP_CONTENT;
int height = LinearLayout.LayoutParams.WRAP_CONTENT;
boolean focusable = true;
popupWindow = new PopupWindow(layout, width, height, focusable);
//Show the popup window
popupWindow.showAtLocation(view, Gravity.CENTER, 0, 0);
}
這工作正常,並顯示彈出正常。但是,如果我在彈出窗口外觸摸(單擊),它將被解除。那麼我怎樣才能讓這個彈出式窗口模式化,以便用戶在他/她能夠回到其他活動之前必須做出響應?