[英]Read last n elements of an XML file with Linq XDocument
I am using the follow Code to read an XML file: 我正在使用以下代码读取XML文件:
XDocument doc = XDocument.Load(xmlPath);
foreach (var node in doc.Descendants("LogInfo"))
{
ListViewItem item = new ListViewItem(new string[]
{
node.Element("MailBox").Value,
node.Element("LastRun").Value,
});
listViewHome.Items.Add(item);
}
How can I change this Code to receive just the last "n" number of elements from that XML file? 如何更改此代码以仅接收该XML文件中最后“ n”个元素? Thanks in advance.
提前致谢。
You could do this: 您可以这样做:
var nodes = doc.Descendants("LogInfo");
foreach (var node in nodes.Skip(Items.Count() - n))
{
...
}
Or this: 或这个:
var nodes = doc.Descendants("LogInfo");
foreach (var node in nodes.Reverse().Take(n).Reverse())
{
...
}
If you're feeling adventurous, you could also write your own extension method which should be more efficient than either of these. 如果您喜欢冒险,还可以编写自己的扩展方法 ,该方法应该比这两种方法都有效。 Here's my quick and dirty solution:
这是我的快速而肮脏的解决方案:
public static IEnumerable<T> TakeFromEnd<T>(this IEnumerable<T> items, int n)
{
var arry = new T[n];
int i = 0;
foreach(var x in items)
{
arry[i++ % n] = x;
}
if (i < n)
{
n = i;
i = 0;
}
for (int j = 0; j < n; j++)
{
yield return arry[(i + j) % n];
}
}
var nodes = doc.Descendants("LogInfo");
foreach (var node in nodes.TakeFromEnd(n))
{
...
}
XDocument doc = XDocument.Load(xmlPath);
var logs = doc.Descendants("LogInfo");
var logsCount = logs.Count();
foreach (var node in logs.Skip(logsCount - n).Take(n))
{
ListViewItem item = new ListViewItem(new string[]
{
node.Element("MailBox").Value,
node.Element("LastRun").Value,
});
listViewHome.Items.Add(item);
}
And here is XPath solution, which will enumerate xml once 这是XPath解决方案,它将一次枚举xml
var xpath = String.Format("//LogInfo[position()>last()-{0}]", n);
foreach (var log in doc.XPathSelectElements(xpath))
{
ListViewItem item = new ListViewItem(new string[]
{
(string)log.Element("MailBox"),
(string)log.Element("LastRun")
});
listViewHome.Items.Add(item);
}
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