我有以下格式的xml文件:
<TestDataFile name="Register">
<TestSuite name="Positive">
<TestCase>
<StoryName>Register user</StoryName>
<ScenarioName>Positive</ScenarioName>
<TestCaseName>Register new user</TestCaseName>
<ResponseCode>200</ResponseCode>
<UserDetails>
<DateOfBirth></DateOfBirth>
</UserDetails>
</TestCase>
<TestCase>
<StoryName>Register user</StoryName>
<ScenarioName>Positive</ScenarioName>
<TestCaseName>Register new user</TestCaseName>
<ResponseCode>200</ResponseCode>
<UserDetails>
<DateOfBirth></DateOfBirth>
</UserDetails>
</TestCase>
</TestSuite>
<TestSuite name="Negative">
<TestCase>
<StoryName>Register user</StoryName>
<ScenarioName>Positive</ScenarioName>
<TestCaseName>Register new user</TestCaseName>
<ResponseCode>200</ResponseCode>
<UserDetails>
<DateOfBirth></DateOfBirth>
</UserDetails>
</TestCase>
<TestCase>
<StoryName>Register user</StoryName>
<ScenarioName>Positive</ScenarioName>
<TestCaseName>Register new user</TestCaseName>
<ResponseCode>200</ResponseCode>
<UserDetails>
<DateOfBirth></DateOfBirth>
</UserDetails>
</TestCase>
</TestSuite>
</TestDataFile>
以此格式包含许多xml文件。我想读取所有xmls并使用LINQ存储在List:List中。尝试使用以下代码实现它但不起作用:
foreach (string file in files)
{
TestDataFile testData = new TestDataFile();
var doc = XDocument.Load(file);
var result = doc.Descendants("TestDataFile")
.Select(x => new TestDataFile
{
TestSuites = new List<TestSuite>
(from ts in doc.Descendants("TestSuite")
select new TestSuite
{
TestCases = new List<TestCase>(from test in doc.Descendants("TestCase")
select new TestCase
{
StoryName = x.Element("StoryName").Value
})
})
});
}
有人可以帮帮我吗?
答案 0 :(得分:2)
试试这个: -
var result = XDocument.Load(@"Path\Data.xml").Root.Descendants("TestCase")
.Select(x => new TestCase
{
StoryName = x.Element("StoryName").Value,
ScenarioName = x.Element("ScenarioName").Value,
TestCaseName = x.Element("TestCaseName").Value,
ResponseCode = Convert.ToInt32(x.Element("ResponseCode").Value),
userDetails = x.Descendants("UserDetails")
.Select(z => new UserDetail
{
DateOfBirth = !String.IsNullOrEmpty(z.Element("DateOfBirth").Value) ?
Convert.ToDateTime(z.Element("DateOfBirth").Value) : DateTime.MinValue }).FirstOrDefault()
}).ToList();
我使用了以下类型: -
public class TestCase
{
public string StoryName { get; set; }
public string ScenarioName { get; set; }
public string TestCaseName { get; set; }
public int ResponseCode { get; set; }
public UserDetail userDetails { get; set; }
}
public class UserDetail
{
public DateTime DateOfBirth { get; set; }
}
修改强>
我认为您的UserDetail在TestCase类型中不是List
,因此根据该假设更新了代码。
答案 1 :(得分:0)
您可以使用以下代码序列化和反序列化任何对象。 在你的情况下,TestDataFile的对象应该是obj。在TestDataFile中,您应该声明TestSuite的属性,并在其中声明TestCase的Type List属性。
获取XML文件这是我的代码:
TextReader Treader = null;
Treader = new StreamReader(fullFileName);
ContactsModel = (ContactsDBModel)OSINFO.FromXml<ContactsDBModel>(Treader);
助手的内容如下:
public static string ToXml(object obj)
{
XmlSerializer s = new XmlSerializer(obj.GetType());
using (StringWriterWithEncoding writer = new StringWriterWithEncoding(Encoding.UTF8))
{
s.Serialize(writer, obj);
return writer.ToString();
}
}
public static object FromXml<T>(this TextReader data)
{
XmlSerializer s = new XmlSerializer(typeof(T));
object obj = s.Deserialize(data);
return (T)obj;
}
我为UTF8编码添加了一个自定义类StringWriterWithEncoding。
public class StringWriterWithEncoding : StringWriter
{
public StringWriterWithEncoding(StringBuilder sb, Encoding encoding)
: base(sb)
{
this.m_Encoding = encoding;
}
public StringWriterWithEncoding(Encoding encoding)
: base()
{
this.m_Encoding = encoding;
}
private readonly Encoding m_Encoding;
public override Encoding Encoding
{
get
{
return this.m_Encoding;
}
}
}