刷新UI以反映添加到列表中的项目

Question

While my UI is displayed, data is being passed in the back end and added to a List<string> that I would in turn like to display on my UI.

I've seen several examples using background workers however I don't have access to the actual component due to how I layout my User Controls and programmatically build them.

Question: How can I run this method repeatedly behind my UI without locking up my UI in a loop?

public void UpdatePanel()
{
    foreach (var item in list)
    {
        AddMethod(item);
    }
}

Answer 1

Instead of using a loop or time intervals to monitor a list, as an option when possible, you can use a BindingList<T> or ObservableCollection<T> and receive notification when list changes.

Then you can update user interface in the event handler which you attaced to ListChanged event of BindingList<T> or CollectionChanged event of ObservableCOllection<T> .

Example

Here is an example based on ObservableCollection<string> .

ObservableCollection<string> list;
private void Form1_Load(object sender, EventArgs e)
{
    list = new ObservableCollection<string>();
    list.CollectionChanged += list_CollectionChanged;
    list.Add("Item 1");
    list.Add("Item 2");
    list.RemoveAt(0);
    list[0] = "New Item";
}
void list_CollectionChanged(object sender, NotifyCollectionChangedEventArgs e)
{
    if (e.Action == NotifyCollectionChangedAction.Add)
    {
        var items = string.Join(",", e.NewItems.Cast<String>());
        MessageBox.Show(string.Format("'{0}' Added", items));
    }
    else if (e.Action == NotifyCollectionChangedAction.Remove)
    {
        var items = string.Join(",", e.OldItems.Cast<String>());
        MessageBox.Show(string.Format("'{0}' Removed", items));
    }
    else if (e.Action == NotifyCollectionChangedAction.Replace)
    {
        var oldItems = string.Join(",", e.OldItems.Cast<String>());
        var newItems = string.Join(",", e.NewItems.Cast<String>());
        MessageBox.Show(string.Format("'{0}' replaced by '{1}'", oldItems, newItems));
    }
    else
    {
        MessageBox.Show("Reset or Move");
    }
}

Answer 2

You can use Task, Async and await, where is some code which insert an element in a listbox each second without blocking UI.

In your case, you have to return the data from backend asynchronously.

    public async void LoadItemsAsync()
    {
        for (int i = 0; i < 10; i++)
            listBox1.Items.Add(await GetItem());
    }

    public Task<string> GetItem()
    {
        return Task<string>.Factory.StartNew(() =>
        {
            Thread.Sleep(1000);
            return "Item";
        });
    }