[RESOLVED] XML Serialization question

i have a class that contains a few properties and two methods, SaveToFile and LoadFromFile.

if i serialize the class in SaveToFile using serializer.Serialize(fs, this); is it possible to deserialize in the same way without creating a temporary variable to deserialize to and then setting the properties from the temporary variable?

What do you mean by a temporary variable? Deserialising creates an object of the specified type, so you just use that object.

i have a class like the following testclass:

Code:

---

class testclass
{
    string _tmp1;
    string _tmp2;

    public string Property1
    {
        get { return _tmp1; }
        set { _tmp1 = value; }
    }

    public string Property2
    {
        get { return _tmp2; }
        set { _tmp2 = value; }
    }

    void SaveToFile()
    {
        FileStream fs = new FileStream("tmp.xml", FileMode.Create, FileAccess.Write);
        XmlSerializer xmls = new XmlSerializer(typeof(testclass));
        xmls.Serialize(fs, this);
        fs.Close();
    }
}

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i would like to add a method "LoadFromFile" to the class testclass that is basically the reverse of SaveToFile.
i can't just do "this = xmls.Deserialize(fs);". is there a better way of doing this other than doing:

Code:

---

testclass tmp;
tmp = xmls.Deserialize(fs);
this._tmp1 = tmp.Property1;
this._tmp2 = tmp.Property2;

---

this method works, but is very time consuming if there is a lot of properties.

I've done a similar thing in the past and I made the Save member an public instance method that returned a bool and the Load member a public static method that returned an instance of the class. You would then do something like this to deserialise a file:

Code:

---

TestClass myTestClass = TestClass.Load(fileName);

---

great! thanks. :thumb: