Question

假设我有下一个xml：

<Report>
<File id="1">
    <Variables>
            <Variable id="1" name="integer"> 1 </Variable>
            <Variable id="1" name="string"> x </Variable>
    </Variables>
</File>
<File id="2">
    <Variables>
            <Variable id="2" name="integer"> 1 </Variable>
            <Variable id="2" name="string"> x </Variable>
    </Variables>
</File>
<File id="3">
    <Variables>
            <Variable id="3" name="integer"> 1 </Variable>
            <Variable id="3" name="string"> y </Variable>
    </Variables>
</File>

如何获取变量字符串为“x”的所有文件？

注意linq到xml vb或c＃但是vb是首选

Answer 1

假设您已将XML加载到XDocument实例，您可以执行以下操作：

var files = from f in xDoc.Root.Elements("File")
            where f.Element("Variables")
                   .Elements("Variable")
                   .Any(v => (string)v.Attribute == "string" &&
                             (string)v == "x")
            select f;

或使用等效的基于方法的查询：

var files = xDoc.Root.Elements("File")
                     .Where(f => f.Element("Variables")
                                  .Elements("Variable")
                                  .Any(v => (string)v.Attribute == "string" &&
                                            (string)v == "x"))

Answer 2

给出以下xml：

Dim xml = <Report>
                <File id="1">
                    <Variables>
                            <Variable id="1" name="integer">1</Variable>
                            <Variable id="1" name="string">x</Variable>
                    </Variables>
                </File>
                <File id="2">
                    <Variables>
                            <Variable id="2" name="integer">1</Variable>
                            <Variable id="2" name="string">x</Variable>
                    </Variables>
                </File>
                <File id="3">
                    <Variables>
                            <Variable id="3" name="integer">1</Variable>
                            <Variable id="3" name="string">y</Variable>
                    </Variables>
                </File>
                </Report>

你可以使用Linq：

Dim result = xml.<File> _
                .Where(Function(file) file.<Variables>.<Variable> _
                  .Any(Function(variable) variable.@name = "string" AndAlso 
                                          variable.Value = "x"))

或更短，使用XPath-Expression查询节点：

Dim doc = New XmlDocument() ' Use XmlDocument instead of XElement '
doc.LoadXml(xml.ToString())
Dim result = doc.SelectNodes("/Report/File[Variables/Variable[@name='string' and text()='x']]")

Linq到xml，当子节点满足一定条件时

2 个答案: