C# 根据后代节点的属性选择节点的最佳 LINQ-to-XML 查询？
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Best LINQ-to-XML query to select nodes based on properties of descendant nodes?
提问by Gayle
I have the following XML document:
我有以下 XML 文档：
<?xml version="1.0" encoding="UTF-8"?>
<FamilyTree>
<Parent name="Ken">
<Child name="Lorna">
<Grandchild name="Andrew"/>
<Grandchild name="Brian"/>
</Child>
<Child name="Mike">
<Grandchild name="Ann"/>
<Grandchild name="Beth"/>
</Child>
</Parent>
<Parent name="Norma">
<Child name="Owen">
<Grandchild name="Charles"/>
</Child>
<Child name="Peter">
<Grandchild name="Charlotte"/>
</Child>
</Parent>
<Parent name="Quinn">
<Child name="Robert">
<Grandchild name="Debbie"/>
<Grandchild name="Eric"/>
</Child>
<Child name="Susan">
<Grandchild name="Frank"/>
</Child>
</Parent>
<Parent name="Tom">
<Child name="Ursula">
<Grandchild name="George"/>
<Grandchild name="Harriet"/>
</Child>
<Child name="Victor">
<Grandchild name="Ian"/>
<Grandchild name="Juliet"/>
</Child>
</Parent>
</FamilyTree>
I'm trying to select all the "Parents" with a Child who has at least two children ("Grandchild") of his/her own. Note that I'm notlooking for "Parents" with at least two "Grandchild[ren]".
我试图选择所有“父母”，孩子至少有两个孩子（“孙子”）。请注意，我不是在寻找至少有两个“孙[ren]”的“父母”。
The following LINQ query works, but I've a feeling it's not the most elegant.
以下 LINQ 查询有效，但我感觉它不是最优雅的。
IEnumerable<XElement> parents = (from c in familyTreeElement.Descendants("Child")
where c.Elements().Count() > 1
select c.Parent).Distinct();
Is there a better way to specify this?
有没有更好的方来指定这一点？
采纳答案by Marc Gravell
Ahh the edit (2 grand-children) helps ;-p
啊，编辑（2 个孙子）有帮助；-p
While XDocumentis useful, at times I miss XPath/XQuery. With XmlDocumentyou could just use doc.DocumentElement.SelectNodes("Parent[Child/Grandchild[2]]").
虽然XDocument很有用，但有时我会想念 XPath/XQuery。有了XmlDocument你就可以使用doc.DocumentElement.SelectNodes("Parent[Child/Grandchild[2]]").
回答by mqp
I don't know the "SQL-like" syntax enough to guarantee that I'll get the syntax right if I write it that way, but you want to use .Any()instead of .Count(), and if you select in a different manner, you don't need the Distinct()at the end. Try this:
我不太了解“类似 SQL”的语，无保证如果我这样写，我会得到正确的语，但是您想使用.Any()代替.Count()，如果您以不同的方式进行选择，则不会不需要Distinct()最后。尝试这个：
IEnumerable<XElement> parents =
familyTreeElement.Elements("Parent").Where(
parent => parent.Elements("Child").Any(
child => child.Elements().Count() >= 2));
EDIT:If you want to ensure that there are at least 2, you do pretty much have to use .Count().
编辑：如果你想确保至少有 2 个，你几乎必须使用.Count().
回答by Jon Skeet
Hmmm... I'm finding it hard to get my head round it exactly :)
嗯......我发现很难完全理解它:)
Normally to find out if there are anyelements, I'd use Any- but you want to see if there are at least twoelements. We still don't need to use Countthough - because there being at least two elements is the same as skipping an element and seeing if there are still any. So...
通常要找出是否有任何元素，我会使用Any- 但您想看看是否至少有两个元素。我们仍然不需要使用Count- 因为至少有两个元素与跳过一个元素并查看是否还有任何元素相同。所以...
var parents = familyTreeElement.Elements("Parent")
.Where(parent => parent.Elements("Child").Any(
child => child.Elements("Grandchild").Skip(1).Any()));
I think that works - and actually it doesn't read toobadly:
我认为这是有效的 - 实际上它读起来并不太糟糕：
For each parent, see whether anyof there children has any(grand)children after ignoring the first (grand)child.
对于每一个家长，看有无任何有孩子有任何忽略第一（大）后的孩子（大）的儿童。
I suspect using XPath (as per Marc's answer) would be the most readable option though.
我怀疑使用 XPath（根据 Marc 的回答）将是最易读的选项。
