Question

我有一个XML文档如下：

<directory>
<file><monitored>0</monitored>
<xferStatus>1</xferStatus>
<name>test1.txt</name>
<size>7</size>
<created>03/31/10 11:30:02 AM</created>
<modified>03/31/10 11:30:00 AM</modified>
<tPathList><tPath>http://hwcdn.net/p2f4h2b5/cds/testing/test1.txt</tPath>
</tPathList>
<tPath>http://hwcdn.net/p2f4h2b5/cds/testing/test1.txt</tPath>
<oPathList><oPath>http://hwcdn.net/p2f4h2b5/w9m3i4q9/test1.txt</oPath>
</oPathList>
<oPath>http://hwcdn.net/p2f4h2b5/w9m3i4q9/test1.txt</oPath>
<aPath></aPath>
</file>

<file><monitored>0</monitored>
<xferStatus>1</xferStatus>
<name>GenericDAO.cs</name>
<size>1843</size>
<created>03/31/10 11:41:10 AM</created>
<modified>03/31/10 11:41:10 AM</modified>
<tPathList><tPath>http://hwcdn.net/p2f4h2b5/cds/testing/GenericDAO.cs</tPath>
</tPathList>
<tPath>http://hwcdn.net/p2f4h2b5/cds/testing/GenericDAO.cs</tPath>
<oPathList><oPath>http://hwcdn.net/p2f4h2b5/w9m3i4q9/GenericDAO.cs</oPath>
</oPathList>
<oPath>http://hwcdn.net/p2f4h2b5/w9m3i4q9/GenericDAO.cs</oPath>
<aPath></aPath>
</file>
<nEntries>2</nEntries>
</directory>

文档中有两个文件，我如何递归或迭代获取文件，大小等。？

响应采用字符串格式，并按如下方式转换为XML：

XmlTextReader textReader = new XmlTextReader(hwresponse);

Answer 1

尝试使用Linq to XML（System.Xml.Linq名称空间。）示例：

XDocument document = XDocument.Parse(xml); // xml string
var query = from file in document.Descendants("file")
            select new
                {
                    Monitored = (int)file.Element("monitored"),
                    Name = (string)file.Element("name"),
                    Size = (int)file.Element("size")
                };

foreach (var file in query)
{
    Console.WriteLine("{0}\t{1}", file.Name, file.Size);
}

递归解析XmlDOcument

1 个答案: