Android | Get all children elements of a ViewGroup

For those using androidx and kotlin

1 - Iterate through ViewGroup using forEach{}

If you want to iterate through all childViews, you can use predefined kotlin extension forEach in any ViewGroup. Example:

yourViewGroup.forEach{ childView -> // do something with this childView }

2 - Return a list of childViews using .children.toList in a ViewGroup

To return a list of Views, you can use function children to return a Sequence and then use toList() to transform it to a List. Example:

val yourChildViewsList: List<View> = yourViewGroup.children.toList()


(source)

If you want to get all the child views, as well as the views within children ViewGroups, you must do it recursively, since there is no provision in the API to do this out of the box.

private ArrayList<View> getAllChildren(View v) {

    if (!(v instanceof ViewGroup)) {
        ArrayList<View> viewArrayList = new ArrayList<View>();
        viewArrayList.add(v);
        return viewArrayList;
    }

    ArrayList<View> result = new ArrayList<View>();

    ViewGroup viewGroup = (ViewGroup) v;
    for (int i = 0; i < viewGroup.getChildCount(); i++) {

        View child = viewGroup.getChildAt(i);

        ArrayList<View> viewArrayList = new ArrayList<View>();
        viewArrayList.add(v);
        viewArrayList.addAll(getAllChildren(child));

        result.addAll(viewArrayList);
    }
    return result;
}

This will give you an ArrayList with all the Views in the hierarchy which you can then iterate over.

Essentially, this code call itself if it finds another ViewGroup in the hierarchy, and then returns an ArrayList to be added to the bigger ArrayList.


At the time of writing this answer, the accepted answer is flawed in that it will contains duplicates in its result.

For those who have trouble wrapping their head around recursion, here's a non-recursive alternative. You get bonus points for realizing this is also a breadth-first search alternative to the depth-first approach of the other answer.

private List<View> getAllChildrenBFS(View v) {
    List<View> visited = new ArrayList<View>();
    List<View> unvisited = new ArrayList<View>();
    unvisited.add(v);

    while (!unvisited.isEmpty()) {
        View child = unvisited.remove(0);
        visited.add(child);
        if (!(child instanceof ViewGroup)) continue;
        ViewGroup group = (ViewGroup) child;
        final int childCount = group.getChildCount();
        for (int i=0; i<childCount; i++) unvisited.add(group.getChildAt(i));
    }

    return visited;
}

A couple of quick tests (nothing formal) suggest this alternative is also faster, although that has most likely to do with the number of new ArrayList instances the other answer creates. Also, results may vary based on how vertical/horizontal the view hierarchy is.

Tags:

Android

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