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android - 单击后如何将新项目添加到弹出菜单

转载 作者:太空狗 更新时间:2023-10-29 14:23:42 25 4
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我有一个按钮:

<Button
android:id="@+id/bot_button1"
android:layout_width="120dp"
android:layout_height="fill_parent"
android:layout_weight="1"
android:textSize="24sp"
android:text="@string/bot_button1_tx" />

这是我的主要 Activity :

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
activity = this;

setContentView(R.layout.activity_main);

button1 = (Button) findViewById(R.id.bot_button1);
button1.setOnClickListener(onClickListener);

Log.d(className, "onCreate");

private OnClickListener onClickListener = new OnClickListener() {
@Override
public void onClick(final View v) {
switch(v.getId()){
case R.id.bot_button1:
showPopupMenu(v, 1);
break;
}
}
};

private void showPopupMenu(final View v, Integer i){
PopupMenu popupMenu = new PopupMenu(MainActivity.this, v);
switch (i) {
case 1:
popupMenu.getMenuInflater().inflate(R.menu.menu1, popupMenu.getMenu());
break;
}

popupMenu.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {

new AlertDialog.Builder(activity).
setTitle("TITLE").
setMessage("MESSAGE").
setAdapter(adapter, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
// TODO: user specific action
dialog.dismiss();
}
}).create().show();
return true;
}
});
popupMenu.show();
}

选择弹出窗口的项目时,如何显示更多项目(单击)?理想情况:PopupMenu 在选择后保持可见,并且新的 PopupMenu 出现在所选项目的旁边。或弹出菜单展开以显示子项。

我试图引入第二个 PopupMenu onMenuItemClick(),但它只替换了第一个 PopupMenu。

最佳答案

我使用了一个新的弹出菜单来解决这个问题

public OnMenuItemClickListener listener = new OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {

switch (item.getItemId()) {
case R.id.menu2_3_3:
popupMenu2.getMenuInflater().inflate(R.menu.menu2_3_3, popupMenu2.getMenu());
popupMenu2.show();
popupMenu2.setOnMenuItemClickListener(listener2);

关于android - 单击后如何将新项目添加到弹出菜单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14382614/

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