我试图保存一个通用列表,并使用BinaryFormatter取回,但是我无法以保存的形式获得列表,它仅返回列表中的第一项。我认为代码尝试不覆盖文件时可能会出现错误。如果您需要更多详细信息,请告诉我,我会添加您需要的详细信息。
#region Save
/// <summary>
/// Saves the given object to the given path as a data in a generic list.
/// </summary>
protected static void Save<T>(string path, object objectToSave)
{
BinaryFormatter formatter = new BinaryFormatter();
FileStream stream;
if (!File.Exists(path))
{
stream = File.Create(path);
}
else
{
stream = File.Open(path, FileMode.Open);
}
List<T> list = new List<T>();
try
{
list = (List<T>)formatter.Deserialize(stream);
}
catch
{
}
list.Add((T)objectToSave);
formatter.Serialize(stream, list);
stream.Close();
}
#endregion
#region Load
/// <summary>
/// Loads the data from given path and returns a list of questions.
/// </summary>
protected static List<T> Load<T>(string path)
{
if (!File.Exists(path))
{
System.Windows.Forms.MessageBox.Show(path + " yolunda bir dosya bulunamadı!");
return null;
}
BinaryFormatter formatter = new BinaryFormatter();
FileStream stream = File.Open(path, FileMode.Open);
List<T> newList;
try
{
newList = (List<T>)formatter.Deserialize(stream);
}
catch
{
newList = null;
}
stream.Close();
return newList;
}
#endregion
答案 0 :(得分:0)
Okey,我刚刚发现了问题。显然,如果您在不保存数据的情况下进行了更改(我在“ list =(List)formatter.Deserialize(stream);”这一行代码中完成了此操作),然后如果尝试再次对其进行序列化,则FileStrem不能正常使用,因此您必须关闭旧流,然后重新打开它或再次打开它,或者只需键入stream = File.Open(path,FileMode.Open);再次。还是谢谢你:D