ListView.RetrieveVirtualItem Event


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Occurs when the ListView is in virtual mode and requires a ListViewItem.

Namespace:   System.Windows.Forms
Assembly:  System.Windows.Forms (in System.Windows.Forms.dll)

public event RetrieveVirtualItemEventHandler RetrieveVirtualItem

Exception Condition

The Item property is not set to an item when the RetrieveVirtualItem event is handled.

When a ListView object is in virtual mode, it creates ListViewItem objects dynamically instead of using the Items collection. This event is raised when the object must create a ListViewItem object. A handler for this event should create the appropriate ListViewItem or retrieve it from the cache, and pass it back by way of the Item property.

For more information about handling events, see Handling and Raising Events.

The following code example demonstrates a handler for this event. In this example, listView1 needs each ListViewItem to display the square of its index. This code example is part of a larger example provided for the VirtualMode property.

//The basic VirtualMode function.  Dynamically returns a ListViewItem
//with the required properties; in this case, the square of the index.
void listView1_RetrieveVirtualItem(object sender, RetrieveVirtualItemEventArgs e)
    //Caching is not required but improves performance on large sets.
    //To leave out caching, don't connect the CacheVirtualItems event 
    //and make sure myCache is null.

    //check to see if the requested item is currently in the cache
    if (myCache != null && e.ItemIndex >= firstItem && e.ItemIndex < firstItem + myCache.Length)
        //A cache hit, so get the ListViewItem from the cache instead of making a new one.
        e.Item = myCache[e.ItemIndex - firstItem];
        //A cache miss, so create a new ListViewItem and pass it back.
        int x = e.ItemIndex * e.ItemIndex;
        e.Item = new ListViewItem(x.ToString());

.NET Framework
Available since 2.0
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