Custom xml attribute for @layout link
I would like to create my own view with my own xml attributes. I would like to specify the header layout to be inflated in my custom xml view, something like this:
<com.example.MyCustomWidget
android:layout_width="match_parent"
android:layout_height="match_parent"
app:headerLayout="@layout/my_header"
/>
Is this possible and how to get the layout resource from TypedArray
?
So in the end, I would like to do something like this:
class MyCustomWidget extends FrameLayout {
public ProfileTabLayout(Context context, AttributeSet attrs, int defStyle) {
super(context, attrs, defStyle);
TypedArray a = getContext().obtainStyledAttributes(attrs, R.styleable.ProfileTabLayout);
int headerLayout = a.getLayout(R.styleable.MyCustomView_headerLayout, 0); // There is no such method
a.recycle();
LayoutInflater.from(context)
.inflate(headerLayout, this, true);
}
}
<?xml version="1.0" encoding="utf-8"?>
<resources>
<declare-styleable name="MyCustomWidget">
<attr name="headerLayout" format="reference" />
</declare-styleable>
</resources>
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1 answer
You must first create your own field. To do this add the code belowres/values/attrs.xml
<declare-styleable name="MyCustomView">
<attr name="headerLayout" format="reference" />
</declare-styleable>
Then in your custom view, you can get that value in the constructor
public MyCustomView(Context context, AttributeSet attrs, int defStyle) {
super(context, attrs, defStyle);
...
TypedArray a = context.getTheme().obtainStyledAttributes(
attrs,
R.styleable.MyCustomView,
defStyle, 0
);
try {
int headerLayout = a.getResourceId(R.styleable.MyCustomView_headerLayout, 0);
} finally {
a.recycle();
}
...
}
Here you can inflate headerLayout
withLayoutInflater
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