Null pointer Exception - findViewById()
AndroidNullpointerexceptionOncreateFindviewbyidAndroid Problem Overview
Can anyone help me to find out what can be the issue with this program.
In the onCreate()
method the findViewById()
returns null for all ids and this causes a null pointer exception later. I can not figure out why the findViewById()
can not find the view. Any suggestions?
This is the main code:
public class MainActivity extends Activity {
ViewPager pager;
MyPagerAdapter adapter;
LinearLayout layout1, layout2, layout3;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
layout1 = (LinearLayout) findViewById(R.id.first_View);
layout2 = (LinearLayout) findViewById(R.id.second_View);
layout3 = (LinearLayout) findViewById(R.id.third_View);
adapter = new MyPagerAdapter();
pager = (ViewPager) findViewById(R.id.main_pager);
pager.setAdapter(adapter);
}
private class MyPagerAdapter extends PagerAdapter
{
@Override
public int getCount() {
return 3;
}
@Override
public Object instantiateItem(ViewGroup collection, int position) {
LinearLayout l = null;
if (position == 0 )
{
l = layout1;
}
if (position == 1)
{
l = layout2;
}
if (position == 2)
{
l = layout3;
}
collection.addView(l, position);
return l;
}
@Override
public boolean isViewFromObject(View view, Object object) {
return (view==object);
}
@Override
public void destroyItem(ViewGroup collection, int position, Object view) {
collection.removeView((View) view);
}
}
}
And the related XML files:
activity_main layout:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:background="#a4c639">
<android.support.v4.view.ViewPager
android:layout_width="match_parent"
android:layout_height="match_parent"
android:id="@+id/main_pager"/>
</LinearLayout>
activity_first layout:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:id="@+id/first_View">
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/hello_world" />
<Button
android:id="@+id/button1"
style="?android:attr/buttonStyleSmall"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Button" />
</LinearLayout>
activity_second layout:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:id="@+id/second_View">
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/hello_world" />
</LinearLayout>
And the activity_third layout:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:id="@+id/third_View">
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/hello_world" />
</LinearLayout>
Android Solutions
Solution 1 - Android
findViewById()
returns a View if it exists in the layout you provided in setContentView()
, otherwise it returns null and that's what happening to you. Note that if you don't setContentView()
, and don't have a valid view to findViewById()
on, findViewById()
will always return null until you call setContentView()
.
This also means variables in the top-level trigger an NPE, because they're called before onCreate()
, and by extension, before setContentView()
. See also the activity lifecycle
Example if you setContentView(R.layout.activity_first);
and then call findViewById(R.id.first_View);
it will return a View which is your layout.
But if you call findViewById(R.id.second_View);
before setContentView()
, it will return null
since there is not a view in your activity_first.xml
layout called @+id/second_View
.
Solution 2 - Android
Emphasis added
For those cases within an Activity class.
> Activity.findViewById(int id)
>
> Finds a view that was identified by the id
attribute from the XML that was processed in onCreate(Bundle)
.
Otherwise, such as an Fragment, Adapter, a View
from a LayoutInflater
, etc.
> View.findViewById(int id)
>
> Look for a child view with the given id
. If this view has the given id, return this view.
Either case,
> Returns
> The view if found or null
otherwise.
Now, re-check your XML files. Make sure you put the right value into setContentView
or inflater.inflate
.
In the case of an Activity, call findViewById
after setContentView
.
Then, make sure there is a View you are looking for with android:id="@+id/..."
in that layout. Make sure the +
is at @+id
, which will add the resource to the R.id
values to ensure you can find it from Java.
Solution 3 - Android
The views you're trying to get are not defined in your activity_main
layout. You need to programmatically inflate the views you're trying to add to the pager.-
@Override
public Object instantiateItem(ViewGroup collection, int position) {
LinearLayout l = null;
if (position == 0) {
l = (LinearLayout) View.inflate(this, R.layout.activity_first, null);
}
if (position == 1) {
l = (LinearLayout) View.inflate(this, R.layout.activity_second, null);
}
if (position == 2) {
l = (LinearLayout) View.inflate(this, R.layout.activity_third, null);
}
collection.addView(l, position);
return l;
}
Solution 4 - Android
Sometimes you need clean your project in Eclipse (Project - Clean..).
Solution 5 - Android
add those views to the pager adapter before accessing them.
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
adapter = new MyPagerAdapter();
pager = (ViewPager) findViewById(R.id.main_pager);
pager.setAdapter(adapter);
layout1 = (LinearLayout) findViewById(R.id.first_View);
layout2 = (LinearLayout) findViewById(R.id.second_View);
layout3 = (LinearLayout) findViewById(R.id.third_View);
}
in the pager adapter:
public Object instantiateItem(View collection, int position) {
if(position == 0){
View layout = inflater.inflate(R.layout.activity_first, null);
((ViewPager) collection).addView(layout);
return layout;
}
... and so forth.
}
from here you can access them via findViewById.
Solution 6 - Android
In my case, it was a stupid mistake on my part. I had written code in the OnCreate method but it was above the setContentView
line of code. Once I moved my code below this line the application started working fine.
setContentView(R.layout.activity_main);
Solution 7 - Android
Use the adaptor, to inflate the layout, and based on the position you can search for the view.
override fun instantiateItem(collection: ViewGroup, position: Int) : ViewGroup {
val inflater = LayoutInflater.from(mContext)
val layout = inflater.inflate(mData[position], collection, false) as ViewGroup
if(position == your_position) {
val nameOfField: TextView = layout.findViewById(R.id.item_id)
}
Solution 8 - Android
What @Warlock said above is right , you should initial LinearLayout layout1, layout2, layout3 by the right way:
LinearLayout layout1 = (LinearLayout) View.inflate(this, R.layout.first_View, null);
LinearLayout layout2 = (LinearLayout) View.inflate(this, R.layout.second_View, null);
LinearLayout layout3 = (LinearLayout) View.inflate(this, R.layout.third, null);
wish my advise help you
Solution 9 - Android
In Android, findViewById(R.id.some_id)
works when you are finding view in the layout set.
That is, if you have set a layout say:
setContentView(R.layout.my_layout);
Views can be found only in this layout (my_layout).
In your code layout1
, layout2
, layout3
all are three different layouts and they are not set to the activity.
Solution 10 - Android
The findViewById
method must find what is in the layout (which you called in the setContentView
)
Solution 11 - Android
I have gotten this error today and it was so simple to clear, that I "facepalmed".
Just try to add the UI element to your layout xml File in your res/layout-port directory!!!
Solution 12 - Android
This problem is also generated when you are having same name of many components in XML file. Even if they are different layout files you should give every element a unique name. This error occur because you are having a XML element with the same name in another layout file and android studio is trying to access that and showing this error