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执行此LINQ to XML查询的更好方法?

linq-to-xml .net c#

1

Davy8 · 技术社区 · 17 年前

<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<Root>
  <Category Name="Tasties">
    <Category Name="Pasta">
      <Category Name="Chicken">
        <Recipe Name="Chicken and Shrimp Scampi" />
        <Recipe Name="Chicken Fettucini Alfredo" />
      </Category>
      <Category Name="Beef">
        <Recipe Name="Spaghetti and Meatballs" />
        <Recipe Name="Lasagna" />
      </Category>
      <Category Name="Pork">
        <Recipe Name="Lasagna" />
      </Category>
      <Category Name="Seafood">
        <Recipe Name="Chicken and Shrimp Scampi" />
      </Category>
    </Category>
  </Category>
</Root>

var q = from chk in
            (from c in doc.Descendants("Category")
             where c.Attribute("Name").Value == "Chicken"
             select c)
        select from r in chk.Descendants("Recipe")
               select r.Attribute("Name").Value;

foreach (var recipes in q)
{
    foreach (var recipe in recipes)
    {
        Console.WriteLine("Recipe name = {0}", recipe);
    }
}

IEnumerable<IEnumerable<String>> 当我只想要一个 IEnumerable<String> .

4 回复 | 直到 17 年前

1

3

Marc Gravell 17 年前

就我个人而言,我会使用 XmlDocument SelectNodes :

foreach(XmlElement el in doc.DocumentElement.SelectNodes(
   "Category[@Name='Tasties']/Category[@Name='Pasta']/Category[@Name='Chicken']/Recipe")) {
    Console.WriteLine(el.GetAttribute("Name"));
}

var q = from c1 in doc.Root.Elements("Category")
        where c1.Attribute("Name").Value == "Tasties"
        from c2 in c1.Elements("Category")
        where c2.Attribute("Name").Value == "Pasta"
        from c3 in c2.Elements("Category")
        where c3.Attribute("Name").Value == "Chicken"
        from recipe in c3.Elements("Recipe")
        select recipe.Attribute("Name").Value;
foreach (string name in q) {
    Console.WriteLine(name);
}

    string[] categories = { "Tasties", "Pasta", "Chicken" };
    XDocument doc = XDocument.Parse(xml);
    IEnumerable<XElement> query = doc.Elements();
    foreach (string category in categories) {
        string tmp = category;
        query = query.Elements("Category")
            .Where(c => c.Attribute("Name").Value == tmp);
    }
    foreach (string name in query.Descendants("Recipe")
        .Select(r => r.Attribute("Name").Value)) {
        Console.WriteLine(name);
    }

编辑以讨论(评论)原因 Where tmp 变量:

这可能会有点复杂,但我试图公正地回答这个问题;p

foreach

class SomeWrapper {
    public string category;
    public bool AnonMethod(XElement c) {
        return c.Attribute("Name").Value == category;
    }
}
...
SomeWrapper wrapper = new SomeWrapper(); // note only 1 of these
using(var iter = categories.GetEnumerator()) {
    while(iter.MoveNext()) {
        wrapper.category = iter.Current;
        query = query.Elements("Category")
             .Where(wrapper.AnonMethod);
    }
}

哪里 category (通过谓词 AnonMethod ( 内部范围

class SecondWrapper {
    public string tmp;
    public bool AnonMethod(XElement c) {
        return c.Attribute("Name").Value == tmp;
    }
}
...
string category;
using(var iter = categories.GetEnumerator()) {
    while(iter.MoveNext()) {
        category = iter.Current;
        SecondWrapper wrapper = new SecondWrapper(); // note 1 per iteration
        wrapper.tmp = category;
        query = query.Elements("Category")
             .Where(wrapper.AnonMethod);
    }
}

2

1

CoderDennis 17 年前

var q = from t in doc.Root.Elements("Category")
        where t.Attribute("Name").Value == "Tasties"
        from p in t.Elements("Category")
        where p.Attribute("Name").Value == "Pasta"
        from c in p.Elements("Category")
        where c.Attribute("Name").Value == "Chicken"
        from r in c.Elements("Recipe")
        select r.Attribute("Name").Value;

foreach (string recipe in q)
{
    Console.WriteLine("Recipe name = {0}", recipe);
}

select IEnumerable<IEnumerable<String>>

3

1

Joel Mueller 17 年前

如果你 add a using statement 对于系统。Xml。XPath,它将向XDocument添加XPathSelectElements()扩展方法。如果您更熟悉XPath语句,这将允许您使用XPath语句选择节点。

< < 字符串 >> < 字符串 >

IEnumerable<IEnumerable<String>> foo = myLinqResults;
IEnumerable<string> bar = foo.SelectMany(x => x);

4

1

OdeToCode 17 年前

虽然有点晚,但扩展方法确实可以帮助清理看起来混乱的LINQtoXML查询。对于你的场景,你可以使用这样的代码:

var query = xml.Root
               .Category("Tasties")
               .Category("Pasta")
               .Category("Chicken")
               .Recipes();

…使用我展示的一些技巧 From LINQ To XPath And Back Again