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[英]Error Deserializing Xml to Object - xmlns='' was not expected
[英]Deserializing Xml to List<T> - xmlns='' was not expected
我在將 XML 反序列化為位置對象列表時遇到問題。
給定以下 XML:
<?xml version="1.0" encoding="utf-8"?>
<locations>
<location id="1">
<level name="3" complete="True" stars="1" firstMisson="True" secondMission="False" thridMission="False" />
</location>
<location id="2">
<level name="4" complete="True" stars="3" firstMisson="True" secondMission="True" thridMission="True" />
</location>
</locations>
以及以下課程:
[System.Serializable]
[XmlRoot(ElementName = "level")]
public class Level
{
[XmlAttribute(AttributeName = "name")]
public string Name { get; set; }
[XmlAttribute(AttributeName = "complete")]
public string Complete { get; set; }
[XmlAttribute(AttributeName = "stars")]
public string Stars { get; set; }
[XmlAttribute(AttributeName = "firstMisson")]
public string FirstMisson { get; set; }
[XmlAttribute(AttributeName = "secondMission")]
public string SecondMission { get; set; }
[XmlAttribute(AttributeName = "thridMission")]
public string ThridMission { get; set; }
}
[System.Serializable]
[XmlRoot(ElementName = "location")]
public class Location
{
[XmlElement(ElementName = "level")]
public Level Level { get; set; }
[XmlAttribute(AttributeName = "id")]
public string Id { get; set; }
}
[System.Serializable]
[XmlRoot(ElementName = "locations")]
public class Locations
{
[XmlElement(ElementName = "location")]
public Location Location { get; set; }
public List<Locations> LocDb = new List<Locations>();
}
[System.Serializable]
[XmlRoot(ElementName = "xml")]
public class Xml
{
[XmlElement(ElementName = "locations")]
public Locations Locations { get; set; }
}
反序列化方法
public List<Locations> locDB = new List<Locations>();
public static void LoadData()
{
string filepath = Application.dataPath + @"/XML/GameXMLdata.xml";
var xmlSerializer = new XmlSerializer(locDB.GetType());
var stream = File.Open(filepath, FileMode.Open);
locDB = (List<Locations>)xmlSerializer.Deserialize(stream);//locations xmlns=''> was not expected
stream.Close();
Debug.Log(locDB[1].Location.Id);
}
那么如何將 XML 反序列化為位置對象列表?
非常感謝您的幫助。
你只需要兩個類:
[XmlType("level")]
public class Level
{
[XmlAttribute("name")]
public string Name { get; set; }
[XmlAttribute("complete")]
public string Complete { get; set; }
[XmlAttribute("stars")]
public string Stars { get; set; }
[XmlAttribute("firstMisson")]
public string FirstMisson { get; set; }
[XmlAttribute("secondMission")]
public string SecondMission { get; set; }
[XmlAttribute("thridMission")]
public string ThridMission { get; set; }
}
[XmlType("location")]
public class Location
{
[XmlElement("level")]
public Level Level { get; set; }
[XmlAttribute("id")]
public string Id { get; set; }
}
重要的是將XmlType
屬性應用於類Location
而不是XmlRoot
。
現在,您可以像這樣將 XML 反序列化為List<Location>
:
var xmlSerializer = new XmlSerializer(typeof(List<Location>),
new XmlRootAttribute("locations"));
List<Location> locations;
using (var stream = File.OpenRead(filepath))
locations = (List<Location>)xmlSerializer.Deserialize(stream);
訣竅是使用XmlSerializer(Type, XmlRootAttribute)
構造函數重載指定根元素名稱(在您的情況下為“位置” XmlSerializer(Type, XmlRootAttribute)
。
我在類屬性上遇到了類似的問題,並使用不同的方法解決了它。 由於在我的場景中我無法觸及XmlSerializer
構造邏輯,這就是我所做的(使用來自 OP 的類型作為參考):
[XmlRoot("locations")]
public sealed class LocationCollection : Collection<Location>
{
}
使用自定義集合類,我可以在設計時設置根並仍然使用標准XmlSeriaizer
實例。
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