Question

我试图将存储在.resx文件中的一些XML数据用作单元测试的测试数据。

我有这个代码，它接受测试数据，将其转换为流，并尝试将其反序列化为学生集合，我可以使用我的模拟dbset。

var stream = TestData.Students.ToStream();
var reader = new StreamReader(stream);
var serializer = new XmlSerializer(typeof(Collection<Student>));
_students = (Collection<Student>)serializer.Deserialize(reader);

学生班

public class Student
{
    public int Id { get; set; }
    public string LastName { get; set; }
    public string FirstName { get; set; }
    public char MiddleInitial { get; set; }
    public DateTime? EnrollmentDate { get; set; }

    [XmlIgnore]
    public virtual ICollection<Enrollment> Enrollments { get; set; }
}

在最后一行失败，出现此异常：

{"<Students xmlns=''> was not expected."}

我尝试将Serializable属性添加到Student类，尝试向其添加XMLRoot属性，也在Google上搜索了一段时间，但无法找到处理资源文件的任何内容。< / p>

另外，是否有更容易/更好的方法来获取测试数据？

修改的如果它有助于我关注this文章。

编辑2 因此深入到内部异常的堆栈跟踪中，我找到了这一行：

at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderCollection1.Read4_ArrayOfStudent()

所以我将XML的根元素从Students更改为ArrayOfStudent，瞧，它正确反序列化。

这是资源文件中的XML数据。

<Students>
  <Student>
    <Id>1</Id>
    <LastName>Alexander</LastName>
    <FirstName>Carson</FirstName>
    <EnrollmentDate>2010-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>2</Id>
    <LastName>Alonso</LastName>
    <FirstName>Meredith</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>3</Id>
    <LastName>Anand</LastName>
    <FirstName>Arturo</FirstName>
    <EnrollmentDate>2013-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>4</Id>
    <LastName>Barzdukas</LastName>
    <FirstName>Gytis</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>5</Id>
    <LastName>Li</LastName>
    <FirstName>Yan</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>6</Id>
    <LastName>Justice</LastName>
    <FirstName>Peggy</FirstName>
    <EnrollmentDate>2011-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>7</Id>
    <LastName>Norman</LastName>
    <FirstName>Laura</FirstName>
    <EnrollmentDate>2013-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>8</Id>
    <LastName>Olivetto</LastName>
    <FirstName>Nino</FirstName>
    <EnrollmentDate>2005-08-11T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>9</Id>
    <LastName>Alexander</LastName>
    <FirstName>Carson</FirstName>
    <EnrollmentDate>2010-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>10</Id>
    <LastName>Alonso</LastName>
    <FirstName>Meredith</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>11</Id>
    <LastName>Anand</LastName>
    <FirstName>Arturo</FirstName>
    <EnrollmentDate>2013-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>12</Id>
    <LastName>Barzdukas</LastName>
    <FirstName>Gytis</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>13</Id>
    <LastName>Li</LastName>
    <FirstName>Yan</FirstName>
    <EnrollmentDate>2012-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>14</Id>
    <LastName>Justice</LastName>
    <FirstName>Peggy</FirstName>
    <EnrollmentDate>2011-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>15</Id>
    <LastName>Norman</LastName>
    <FirstName>Laura</FirstName>
    <EnrollmentDate>2013-09-01T00:00:00</EnrollmentDate>
  </Student>
  <Student>
    <Id>16</Id>
    <LastName>Olivetto</LastName>
    <FirstName>Nino</FirstName>
    <EnrollmentDate>2005-08-11T00:00:00</EnrollmentDate>
  </Student>
</Students>

Answer 1

您可以使用此代码从xml文件中获取数据

using (FileStream sr = new FileStream(filePath, FileMode.Open, FileAccess.Read))
                {
                    XmlSerializer serializer = new XmlSerializer(typeof(Students));
                    return serializer.Deserialize(sr) as Students;
                }

对于 Xmlns ，您无法使用属性来添加应添加的名称空间

XmlSerializerNamespaces np = new XmlSerializerNamespaces();

然后将新命名空间添加到 np ，添加此 np 以序列化或反序列化为参数

Answer 2

尝试此代码（为部分xml添加根元素）

var xmlString = File.ReadAllText(@"C:\YourDirectory\Students.xml");

var rootAttribute = new XmlRootAttribute { ElementName = "Students", IsNullable = true };
XmlSerializer serializer = new XmlSerializer(typeof(Student[]), rootAttribute);
StringReader reader = new StringReader(xmlString);
Student[] students = (Student[])serializer.Deserialize(reader);

我只是在这里阅读文件中的全文，你可以使用一个流，赢了。

Answer 3

在反序列化时，序列化程序正在寻找ArrayOf {ClassName}的根元素，因此将根元素Students更改为ArrayOfStudent就可以了。

使用存储在资源文件中的XML作为单元测试的模拟数据

3 个答案: