洗牌而初始化觀察集合

Question

我想在洗牌的ObservableCollection內容，

public static ObservableCollection<whatsnewCompleteData> whatsnewCompleteList = new ObservableCollection<whatsnewCompleteData>(); 

for (int i = 0; i < whatsnewfeed.feed.entry.Count; i++) 
{    
    whatsnewCompleteList.Add(new whatsnewCompleteData(i, content[i]); 
} 

如果是正常的名單，我可以很容易地進行洗牌！但observablecollection直接綁定到UI。我唯一能做的就是每次初始化時使用一個隨機數。 所以這裏的範圍內，

0 < whatsnewfeed.feed.entry.Count 

如何生成一個隨機數，並覆蓋從0所有值whatsnewfeed.feed.entry.count？

Answer 1

首先，你可以洗牌在就地ObservableCollection的數據，你的建議，或者你可以創建一個新的，然後覆蓋ObservableCollection。

考慮這樣的情況：

public static ObservableCollection<whatsnewCompleteData> whatsnewCompleteList = new ObservableCollection<whatsnewCompleteData>(); 

// A bindable property 
public ObservableCollection<whatsnewCompleteData> WhatsNewCompleteList 
{ 
    get 
    { 
     return whatsnewCompleteList; 
    } 
    set 
    { 
     whatsnewCompleteList = value; 
     // Your property changed notifier method 
     OnPropertyChanged("WhatsNewCompleteList"); 
    } 
} 

在你的方法：

// content should obviously be whatever object type you need to get 
// (or already have) 
public void Shuffle(List<object> content) 
{ 
    // Note: you don't have to initialize an observable collection like this. 
    // You can also create a list, shuffle it as you would normally, then call 
    // new ObservableCollection<whatsnewCompleteData(shuffledList); 
    var newWhatsnewCompleteList = new ObservableCollection<whatsnewCompleteData>(); 

    for (int i = 0; i < whatsnewfeed.feed.entry.Count; i++) 
    {    
     newWhatsnewCompleteList.Add(new whatsnewCompleteData(i, content[i]); 
    } 
    WhatsNewCompleteList = newWhatsnewCompleteList; 
} 

這將覆蓋當前目錄，也不必擔心UI嚇壞了更新綁定。

另一種選擇是簡單地創建一個新的值列表，然後在static集合上調用Clear()並將其填入新值。

希望這有助於和快樂的編碼！