我正在尝试在xml文件中编写一些我迄今为止创建的配置文件,
输入字符串是ProfilesList(0)=“45 65 67” ProfilesList(1)=“profilename”;
public void CreateGroupXML(String GroupNameWithPath, List<String> ProfilesList)
{
ProfilesGroup.ProfilesList = ProfilesList;
XmlWriterSettings ws = new XmlWriterSettings();
ws.NewLineHandling = NewLineHandling.Entitize;
for (int i = 0; i < ProfilesList.Count; i++)
{
ProfilesList[i] += Environment.NewLine;
}
XmlSerializer serializer = new XmlSerializer(typeof(ProfilesGroup));
using (XmlWriter wr = XmlWriter.Create(GroupNameWithPath, ws))
{
serializer.Serialize(wr, ProfilesGroup);
}
}
}
在xml文件中编写的配置文件如下:
ProfilesList="45 65 67
 profilename

到目前为止一直很好,当我尝试从xml文件中读取时会出现问题 它将第一个配置文件名称拆分为3 这里是代码
public List<string> getProfilesOfGroup(string groupNameFullPath)
{
Stream stream = null;
try
{
stream = File.OpenRead(groupNameFullPath);
XmlSerializer serializer = new XmlSerializer(typeof(ProfilesGroup));
_ProfilesGroup = (ProfilesGroup)serializer.Deserialize(stream);
stream.Close();
return _ProfilesGroup.ProfilesList;
}
catch (Exception Ex)
{
log.ErrorFormat("Exception in getProfilesOfGroup: {0}", Ex.Message);
if (stream != null)
{
stream.Close();
}
return null;
}
}
the output (lets call the string ProfileList) contains :
ProfileList(0) = 45
ProfileList(1) = 65
ProfileList(2) = 67
ProfileList(3) = profilename
and i expecting the string to contain
ProfileList(0) = 45 65 67
ProfileList(1) = profilename
在这里编辑完整的xml:
?xml version =“1.0”encoding =“utf-8”?ProfilesGroup 的xmlns:的xsi = “http://www.w3.org/2001/XMLSchema-instance” xmlns:xsd =“http://www.w3.org/2001/XMLSchema”ProfilesList =“45 65 67 PROFILENAME “
和班级:
[XmlRootAttribute("VProfilesGroup", IsNullable = false, DataType = "", Namespace = "")]
public class ProfilesGroup
{
[XmlAttribute("ProfilesList")]
public List<String> ProfilesList = new List<string>();
}
答案 0 :(得分:1)
为什么不删除[XmlAttribute("ProfilesList")]
属性?您的数据将被成功序列化和反序列化。 XML将如下所示:
<VProfilesGroup xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<ProfilesList>
<string>45 65 67</string>
<string>profilename</string>
</ProfilesList>
</VProfilesGroup>
在此格式中,字符串列表明确定义为具有两个条目。这是序列化和标准化的标准方法。使用XmlSerializer
反序列化字符串数组。或者您是否有一些外部约束使您将列表声明为属性?
更新
如果必须将ProfilesList
序列化为属性而不是元素数组,则可以手动构造和解构字符串,如下所示:
[XmlRootAttribute("VProfilesGroup", IsNullable = false, DataType = "", Namespace = "")]
public class ProfilesGroup
{
static readonly char Delimiter = '\n';
[XmlIgnore]
public List<String> ProfilesList { get; set; } // Enhance the setter to throw an exception if any string contains the delimiter.
[XmlAttribute("ProfilesList")]
[DebuggerBrowsable(DebuggerBrowsableState.Never)]
public string ProfilesListText
{
get
{
return string.Join(Delimiter.ToString(), ProfilesList.ToArray());
}
set
{
ProfilesList = new List<string>(value.Split(Delimiter));
}
}
public static string CreateGroupXML(List<String> ProfilesList)
{
var group = new ProfilesGroup();
group.ProfilesList = ProfilesList;
return XmlSerializationHelper.GetXml(group);
}
public static List<string> GetProfilesOfGroup(string xml)
{
XmlSerializer serializer = new XmlSerializer(typeof(ProfilesGroup));
var group = (ProfilesGroup)serializer.Deserialize(new StringReader(xml));
return group == null ? null : group.ProfilesList;
}
public static void Test()
{
List<string> list = new List<string>(new string[] { "45 65 67", "profilename" });
var xml = CreateGroupXML(list);
var newList = GetProfilesOfGroup(xml);
bool same = list.SequenceEqual(newList);
Debug.Assert(same); // No assert.
}
}
生成的XML如下所示:
<?xml version=\"1.0\" encoding=\"utf-16\"?>\r\n<VProfilesGroup xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" ProfilesList=\"45 65 67
profilename\" />
在这种情况下,我通过序列化和反序列化为字符串而不是文件来测试代码。然后帮手:
public static class XmlSerializationHelper
{
public static string GetXml<T>(T obj, XmlSerializer serializer) where T : class
{
using (var textWriter = new StringWriter())
{
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true; // For cosmetic purposes.
settings.IndentChars = " "; // The indentation used in the test string.
using (var xmlWriter = XmlWriter.Create(textWriter, settings))
{
serializer.Serialize(xmlWriter, obj);
}
return textWriter.ToString();
}
}
public static string GetXml<T>(T obj) where T : class
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
return GetXml(obj, serializer);
}
}