我正在尝试为我的Visual C#程序实现“打开/保存”功能。我需要保存的对象是列表列表。更具体地说,它是List<List<Components>>
,其中Components
是我的自定义类。我设法导出它,我用XML阅读器读取我的变量,但是当我将它导入程序时,我得到了错误的变量。例如,导出之前的列表容量为2.导入后,它变为4.任何帮助?
如果您可以建议另一种“打开/保存”功能,请随意。我只需要将自定义类列表的列表存储并恢复到文件中。
public void open_click(object send, EventArgs e)
{
XmlSerializer xml = new XmlSerializer(typeof(List<List<Components>>));
OpenFileDialog oDialog = new OpenFileDialog();
oDialog.Filter = "XML|*.xml";
if (oDialog.ShowDialog() == DialogResult.OK)
{
Variables.compBuffer = new List<List<Components>>();
using (FileStream s = File.OpenRead(oDialog.FileName))
using (StreamReader sw = new StreamReader(s))
{
Variables.compBuffer = (List<List<Components>>)xml.Deserialize(s);
}
}
}
//
public void saveAs_click(object send, EventArgs e)
{
XmlSerializer xml = new XmlSerializer(typeof(List<List<Components>>));
SaveFileDialog sDialog = new SaveFileDialog();
sDialog.FileName = "myLadder";
sDialog.Filter = "XML|*.xml";
sDialog.OverwritePrompt = true;
if (sDialog.ShowDialog() == DialogResult.OK)
{
using (FileStream s = File.OpenWrite(sDialog.FileName))
using (StreamWriter sw = new StreamWriter(s))
{
xml.Serialize(s, Variables.compBuffer);
}
}
}
//
public class Components
{
//Private variables
private string _type = "empty";
private string _name = "";
private int _time = 0; //If it's a "timer" (in ms)
private int _index = -1;
private string _comment = "";
private bool _output = false;
public Components() { }
public string Type
{
get { return this._type; }
set { this._type = value; }
}
public string Name
{
get { return this._name; }
set { this._name = value; }
}
public int Time
{
get { return this._time; }
set { this._time = value; }
}
public int Index
{
get { return this._index; }
set { this._index = value; }
}
public string Comment
{
get { return this._comment; }
set { this._comment = value; }
}
public bool Output
{
get { return this._output; }
set { this._output = value; }
}
public void reset()
{
_type = "empty";
_name = "";
_time = 0; //If it's a "timer" (in ms)
_index = -1;
_comment = "";
_output = false;
}
}
}
答案 0 :(得分:0)
由于您没有查看特定于xml的解决方案,我只是将其作为二进制表示形式序列化到文件中,然后将其还原:
using System.IO;
[Serializable]
public class Components
{
...
}
var components = new List<Components>();
string pathToFile = @"c:\dev\components.bin";
SerializeFile(pathToFile, components);
var fetchComponents = DeserializeFile(pathToFile);
private void SerializeFile(string file, IList<Components> data)
{
using (Stream stream = File.Open(file, FileMode.Create))
{
var formatter = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
formatter.Serialize(stream, data);
}
}
private IList<Components> DeserializeFile(string file)
{
using (Stream stream = File.Open(file, FileMode.Create))
{
var formatter = new System.Runtime.Serialization.Formatters.Binary.BinaryFormatter();
return (List<Components>)formatter.Deserialize(stream);
}
}
此解决方案的缺点是您无法查看文件并浏览数据。进一步的改进是使用泛型,以便您可以序列化/反序列化任何类型的数据。
不要忘记使用属性[Serializable]