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如何将 XmlNodeList 拆分为两个较小的 XmlNodeList，其中一个的大小为 N，另一个的总大小为 N？

请参阅下面的示例以及我尝试使用的地方：

    public static void Main(string[] args)
    {
        XmlDocument someDoc = new XmlDocument();
        someDoc.LoadXml(@"<bananas>
                            <banana tasty='yes'></banana>
                            <banana tasty='very'></banana>
                            <banana tasty='amazing'></banana>
                            <banana tasty='mind-blowing'></banana>
                            <banana tasty='disgusting'></banana>
                          </bananas>");
        XmlNodeList bananaNodeList = someDoc.SelectNodes("//banana");
        eatSomeBananas(bananaNodeList, 2);
    }

    /** Splits a XmlNodeList into two XmlNodeList, first one is a slice from 0 to numberOfBananas-1, and the other slice is from numberOfBananas and onwards
    */
    public static void eatSomeBananas(XmlNodeList subBananaNodeList, int numberOfBananas)
    {
        XmlNodeList bananasToEat = subBananaNodeList.Cast<XmlNode>().Take(numberOfBananas) as XmlNodeList; //Error down-casting - null!
        if (bananasToEat == null)
            Console.WriteLine("Error! Did not work");
        /*else 
            doSomethingHere(bananasToEat); */
        XmlNodeList remainingBananas = subBananaNodeList.Cast<XmlNode>().Skip(numberOfBananas) as XmlNodeList; //Error down-casting - null!
        eatSomeBananas(remainingBananas, numberOfBananas);
    }

我试图将 aXmlNodeList转换为 a IEnumerable<XmlNode>（因为前者继承自后者） - 我相信这应该是一个向上转换。我不应该可以将其重新转换为XmlNodeList之后吗？但如果不是，为什么不呢？

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2 回答 2

2

之后我不应该可以将它转换回 XmlNodeList 吗？但如果不是，为什么不呢？

不，因为从返回的值Skip不是XmlNodeList. 它只被IEnumerable<XmlNode>声明为. 将以完全相同的方式工作。SkipSkipXmlNodeListTake

就我个人而言，我会完全避免使用旧的 XML API，而只使用 LINQ to XML。这与 LINQ 很自然 - 通常是一个更好的 XML API，IMO。

请注意，您不必使用它-您可以将整个代码更改为使用IEnumerable<XmlNode>，而不是XmlNodeList：

public static void eatSomeBananas(IEnumerable<XmlNode> subBananaNodeList, int numberOfBananas)
{
    IEnumerable<XmlNode> bananasToEat = subBananaNodeList.Take(numberOfBananas);
    IEnumerable<XmlNode> remainingBananas = subBananaNodeList.Skip(numberOfBananas);
    // Added condition to avoid infinite recursion
    if (remainingBananas.Any())
    {
        eatSomeBananas(remainingBananas, numberOfBananas);
    }
}

然后你可以在调用Cast方法时调用一次：

eatSomeBananas(bananaNodeList.Cast<XmlNode>(), 2);

不过，这是 LINQ to XML 版本，我更喜欢：

using System.Collections.Generic;
using System.Linq;
using System.Xml.Linq;

public class Test
{
    public static void Main(string[] args)
    {
        XDocument someDoc = XDocument.Parse(
            @"<bananas>
                <banana tasty='yes'></banana>
                <banana tasty='very'></banana>
                <banana tasty='amazing'></banana>
                <banana tasty='mind-blowing'></banana>
                <banana tasty='disgusting'></banana>
              </bananas>");
        IEnumerable<XElement> bananas = someDoc.Descendants("banana");
        EatSomeBananas(bananas, 2);
    }

    public static void EatSomeBananas(IEnumerable<XElement> bananas, int numberOfBananas)
    {
        var bananasToEat = bananas.Take(numberOfBananas);
        Console.WriteLine("Eating some bananas"); 
        foreach (var element in bananasToEat)
        {
            var tasty = element.Attribute("tasty").Value;
            Console.WriteLine($"Tasty: {tasty}");
        }
        Console.WriteLine("Eaten the bananas");
        var remainingBananas = bananas.Skip(numberOfBananas);
        if (remainingBananas.Any())
        {
            EatSomeBananas(remainingBananas, numberOfBananas);
        }
    }
}

请注意，对于生产实现，我会避免递归并可能定期实现结果 - 否则它将每次从头开始迭代，跳过负载然后采取一些。

于 2018-05-08T22:30:34.533 回答
1

我刚刚转换为 IQueryable 并使用它，比 XmlNodeList 更容易使用。

    public static void Main(string[] args)
    {
        XmlDocument someDoc = new XmlDocument();
        someDoc.LoadXml(@"<bananas>
                        <banana tasty='yes'></banana>
                        <banana tasty='very'></banana>
                        <banana tasty='amazing'></banana>
                        <banana tasty='mind-blowing'></banana>
                        <banana tasty='disgusting'></banana>
                      </bananas>");
        XmlNodeList bananaNodeList = someDoc.SelectNodes("//banana");

        var allBananas = bananaNodeList.Cast<XmlNode>().AsQueryable();

        eatSomeBananas(allBananas, 2);
    }

    public static void eatSomeBananas(IQueryable<XmlNode> subBananas, int numberOfBananas)
    {
        var bananasToEat = subBananas.Take(numberOfBananas);
        var remainingBananas = subBananas.Skip(numberOfBananas);

        Console.WriteLine(string.Format("Bananas to eat: {0}", bananasToEat.Count()));
        Console.WriteLine(string.Format("Remaining bananas: {0}", remainingBananas.Count()));
        if (!bananasToEat.Any())
            Console.WriteLine("Error! Did not work (not enough bananas!)");
        else 
            eatSomeBananas(remainingBananas, numberOfBananas);
    }

输出：

Bananas to eat: 2
Remaining bananas: 3
Bananas to eat: 2
Remaining bananas: 1
Bananas to eat: 1
Remaining bananas: 0
Bananas to eat: 0
Remaining bananas: 0
Error! Did not work (not enough bananas!)
于 2018-05-08T22:14:36.903 回答