rob*_*y12 64 android android-menu
我有一个PopupMenu,我知道将菜单关联到它的常用方法是使用popup.getMenuInflater().inflate(R.menu.my_menu, popup.getMenu());或类似的东西.我的问题是,我在菜单中有一系列我想要的项目,我需要能够以Java编程方式更改它们.我怎样才能做到这一点?
rob*_*y12 120
刚想通了; 对于遇到同样问题的任何人,你只需:
popup.getMenu().add(groupId, itemId, order, title);
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为每个MenuItem你想要添加.
Kha*_*jar 33
只需创建弹出菜单,注册弹出窗口将在下面显示的视图,并使用getMenu()方法添加值.别忘了打电话给show();
PopupMenu menu = new PopupMenu(this, view);
menu.getMenu().add("titleRes");
menu.show();
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Agi*_*nbu 21
试试这个:
Dynamic_PopUpMenu.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
PopupMenu menu = new PopupMenu(DialogCheckBox.this, v);
menu.getMenu().add("AGIL");
menu.getMenu().add("AGILarasan");
menu.getMenu().add("Arasan");
menu.show();
}
});
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Tar*_*ora 14
在popupmenu.xml中定义id
<?xml version="1.0" encoding="utf-8"?>
<menu xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto">
<item
android:id="@+id/slot1"
app:showAsAction="ifRoom|withText"
android:title="Movies"
android:visible="true"/>
<item
android:id="@+id/slot2"
app:showAsAction="ifRoom|withText"
android:title="Music"
android:visible="true"/>
<item
android:id="@+id/slot3"
app:showAsAction="ifRoom|withText"
android:title="Comedy"
android:visible="true"/>
</menu>
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PopupMenu popupMenu = new PopupMenu(FullMenuActivity.this, view);
popupMenu.setOnMenuItemClickListener(FullMenuActivity.this);
popupMenu.getMenu().add(1, R.id.slot1, 1, "slot1");
popupMenu.getMenu().add(1,R.id.slot2,2,"slot2");
popupMenu.getMenu().add(1,R.id.slot3,3,"slot3");
popupMenu.show();
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@Override
public boolean onMenuItemClick(MenuItem item) {
switch (item.getItemId()) {
case R.id.slot1:
SessionManager.selected_slot = item.getTitle().toString();
Toast.makeText(this, "slot1 Clicked", Toast.LENGTH_SHORT).show();
return true;
case R.id.slot2:
SessionManager.selected_slot = item.getTitle().toString();
Toast.makeText(this, "slot2 Clicked", Toast.LENGTH_SHORT).show();
return true;
case R.id.slot3:
SessionManager.selected_slot = item.getTitle().toString();
Toast.makeText(this, "slot3 Clicked", Toast.LENGTH_SHORT).show();
return true;
default:
return true;
}
}
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Abd*_*ban 11
以下是设置和处理ID的完整解决方案:
this.findViewById(R.id.hamburger_menu).setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
PopupMenu menu = new PopupMenu(getApplicationContext(), v);
menu.getMenu().add(Menu.NONE, 1, 1, "Share");
menu.getMenu().add(Menu.NONE, 2, 2, "Comment");
menu.show();
menu.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
int i = item.getItemId();
if (i == 1) {
//handle share
return true;
} else if (i == 2) {
//handle comment
return true;
} else {
return false;
}
}
});
}
});
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注意:例如,共享和注释也可以为数字1,2加上常量,以使代码更具可读性.
此外,我放,Menu.NONE因为我不关心分组菜单项.如果你想组组ID = 1,2等等...
@Voora Tarun 给出了一个很好的答案,我的答案基于它:首先,而不是使用假mune.xml资源,我认为更好的方法是创建ids文件:
ids.xml
<?xml version="1.0" encoding="utf-8"?>
<resources>
<item name="menuGroup" type="id"/>
<item name="menu1" type="id"/>
<item name="menu2" type="id"/>
<item name="menu3" type="id"/>
</resources>
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然后你可以做类似的事情:
private void showPopup(final View anchor) {
PopupMenu popupMenu = new PopupMenu(anchor.getContext(), anchor);
popupMenu.getMenu().add(R.id.menuGroup, R.id.menu1, Menu.NONE, "slot1");
popupMenu.getMenu().add(R.id.menuGroup, R.id.menu2, Menu.NONE,"slot2");
popupMenu.getMenu().add(R.id.menuGroup, R.id.menu3, Menu.NONE,"slot3");
popupMenu.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
Toast.makeText(anchor.getContext(), item.getTitle() + "clicked", Toast.LENGTH_SHORT).show();
return true;
}
});
popupMenu.show();
}
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