Android：ListView条件滚动

Question

I have a 2 pane layout. On the left is a ListView , on the rights some content linked with the list. The content is scrollable. If content moves forward, I want the list to go forward only if this item is not yet displayed. This prevents that the list has to update on every content change. I also want the list to go ahead one full page. Is there something like

listView.isThisElementCurrentlyShown(int nr)

and

listView.moveAheadOnePage(int direction)

?

Thanks in advance.

Answer 1

Here is an example of isThisElementCurrentlyShown(Element e) :

    public boolean isThisElementCurrentlyShown(Element e){
                ListView lv = getListView();
                int start = lv.getFirstVisiblePosition();
                for(int i=start, j=lv.getLastVisiblePosition();i<=j;i++){
                    if(e==lv.getItemAtPosition(i)){
                        return true;
                    }
                }
                return false;
}

This will tell you whether the element is visible or not.

For the moveAheadOnePage you should be able to use listview's built in functions getFirstVisiblePosition() and getLastVisiblePosition() to calculate the number of rows visible on that device, then advance the listview ahead that many positions. Something like this pseudocode (you will have to write this one):

public void moveAheadOnePage(int direction){//direction: 0-up, 1-down
    int numVisibleRows = getLastVisiblePosition() - getFirstVisiblePosition();
    this.setSelected(currentSelection + numVisibleRows) // +/- depending on direction
}

Mind you these are both expensive calls to make while scrolling through a listview, so you will definitely have to implement wisely~

EDIT: Updated code to include return false