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读取行并保留xml文件中的空格

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我试图在xml文件中写一些我迄今为止创建的配置文件非常好，

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&#xA; profilename&#xA;

到目前为止这么好，问题发生在我尝试从xml文件中读取时，它将第一个配置文件名称拆分为3这里的代码

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 xmlns：xsi =“http://www.w3.org/2001/XMLSchema-instance”xmlns：xsd =“http：//www.w3 .org / 2001 / XMLSchema“ProfilesList =”45 65 67 profilename“

和 class ：

[XmlRootAttribute("VProfilesGroup", IsNullable = false, DataType = "", Namespace = "")]
public class ProfilesGroup
{
    [XmlAttribute("ProfilesList")]
    public List<String> ProfilesList = new List<string>();
}
c# xml xml-parsing

1 回答

  • 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>

    在这种格式中，字符串列表被明确定义为具有两个条目 . 这是使用 XmlSerializer 序列化和反序列化字符串数组的标准方法 . 或者您是否有一些外部约束使您将列表声明为属性？

    Update

    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看起来像：

    <?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&#xA;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 . 然后帮助者：

    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);
    }
}
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