It's "@android:id/list" - NOT "@id/android:list".

25 декабря 2011 г. 12:55 пользователь KK <[email protected]>написал:

> hi all,
> I'm using FrameLayout and have defined two LinearLayouts embedded inside
> the FrameLayout. Now for each LinearLayout I've a listview . The problem
> I'm facing is picking the two listviews separately from my Java side code.
> The default listivew in XML uses the standard defined ID as 
> *android:id="@id/android:list"
> *and I'm not able to pick the right one. So I decided to give custom ids
> to each of the list view like this in Layout1:
>           <ListView
>             android:id="@+id/list1"
>             android:layout_width="fill_parent"
>             android:layout_height="fill_parent" />
>
> and the following in Layout2
>
>           <ListView
>             android:id="@+id/list2"
>             android:layout_width="fill_parent"
>             android:layout_height="fill_parent" />
>
> and this is what I'm trying to use to pick the listviews separately,
>
>       ListView listView1 = (ListView) findViewById(R.id.list1);
>       ListView listView2 = (ListView) findViewById(R.id.list2);
>
> With the above code in pace, I'm not getting any compilation errors. But
> during run time I'm getting errors with this message:
> *      java.lang.RuntimeException: Your content must have a ListView
> whose id attribute is 'android.R.id.list'*
>
> Are we not allowed to use custom IDs for listviews? If custom IDs are not
> allowed, then what is the best way to handle the above problem? Would
> appreciate if someone could point me to the right direction.
>
> BTW, I'm actually trying to define two layouts each with its own listview
> and then I'm trying to animate between two UI screens each using one layout
> and the default behavior gives only one of the listivew as picked by
> "getListView()" which happens to be the 1st one, so my 2nd UI page is not
> showing any listview.
>
>
> Thanks,
> KK
>
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