我们如何对 XmlNodeList 中的最后四个项目进行反向迭代

Question

此时它以相反的顺序显示项目，没有限制。XmlNodeList 可以包含许多项。我只想显示最后四个项目。如何查找或显示列表中的最后四个项目？任何人？

XmlNodeList MyTestList = MyRssDocument.SelectNodes("test/holder/item");
string Title = "";
string Link = "";

for (int i = MyTestList.Count - 1; i >= 0; i--)
{
    XmlNode MyTestDetail;
    MyTestDetail = MyTestList.Item(i).SelectSingleNode("title");
    if (MyTestDetail != null)
        Title = MyTestDetail.InnerText;
    else
        Title = "";
    MyTestDetail = MyTestList.Item(i).SelectSingleNode("link");
    if (MyTestDetail != null)
        Link = MyTestDetail.InnerText;
    else
        Link = "";
}

Answer 1

您可以替换以下行：

for (int i = MyTestList.Count - 1; i >= 0; i--)

通过这个：

for (int i = MyTestList.Count - 1; i >= (MyTestList.Count - 4) ; i--)

Answer 2

for (int i = MyTestList.Count; i >= 4; i--)
{
    XmlNode MyTestDetail;
    MyTestDetail = MyTestList.Item(i).SelectSingleNode("title");
    if (MyTestDetail != null)
        Title = MyTestDetail.InnerText;
    else
        Title = "";
    MyTestDetail = MyTestList.Item(i).SelectSingleNode("link");
    if (MyTestDetail != null)
        Link = MyTestDetail.InnerText;
    else
        Link = "";
}