我已经在android中创建了一个按钮,当点击它时会显示弹出窗口..但代码不能正常工作..它没有错误但没有显示弹出窗口...请帮助..我的代码是 公共类MainActivity扩展了Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
final RelativeLayout objrl = (RelativeLayout) findViewById(R.id.myrl);
final Button objButton = (Button) findViewById(R.id.mybutton);
objButton.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
PopupWindow objPopupWindow = new PopupWindow(objrl, LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT, true);
objPopupWindow.setAnimationStyle(R.drawable.background2);
objPopupWindow.showAtLocation(objButton, Gravity.CENTER_HORIZONTAL, 10, 10);
}
});
}
答案 0 :(得分:0)
你试过这个:objPopupWindow.showAsDropDown(popupButton, 0, 0);
或尝试此http://rajeshandroiddeveloper.blogspot.in/2013/07/android-popupwindow-example-in-listview.html
答案 1 :(得分:0)
PopupWindow popupWindowDogs = popupWindowDogs();
called below function where they want ::-
public PopupWindow popupWindowDogs() {
// initialize a pop up window type
PopupWindow popupWindow = new PopupWindow(this);
// the drop down list is a list view
ListView listViewDogs = new ListView(this);
// set our adapter and pass our pop up window contents
listViewDogs.setAdapter(dogsAdapter(popUpContents));
// set the item click listener
listViewDogs.setOnItemClickListener(new DogsDropdownOnItemClickListener());
// some other visual settings
popupWindow.setFocusable(true);
popupWindow.setWidth(250);
popupWindow.setHeight(WindowManager.LayoutParams.WRAP_CONTENT);
// set the list view as pop up window content
popupWindow.setContentView(listViewDogs);
return popupWindow;
}
答案 2 :(得分:0)
我在你的代码中发现了一些奇怪的东西
setAnimationStyle
方法。在我看来,如果你指定一个有效的动画风格和一个内容视图,它应该出现。
答案 3 :(得分:0)
我认为你在 OnClickListener 中错过了这段代码
objPopupWindow.setContentView(objrl);