Question

我正在通过此代码更新xml文件，

public static void UpdateDesignCfg(string ChildName, string[,] AttribWithValue)
{        
    try
    {
        XmlDocument doc = new XmlDocument();
        doc.Load("DesignCfg.xml");
        XmlElement formData = (XmlElement)doc.SelectSingleNode("//" + ChildName);
        if (formData != null)
        {
            for (int i = 0; i < AttribWithValue.GetLength(0); i++)
            {
                formData.SetAttribute(AttribWithValue[i, 0], AttribWithValue[i, 1]);                    
            }
        }
        doc.Save("DesignCfg.xml");                                 
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.Message);
    }
}

但我经常会收到此错误（并非每次都有）

the process cannot access the file because it is being used by another process

所以，有没有办法在每次更改后“释放”文件？

Answer 1

<强>更新

该文件可从其他地方访问，但未关闭。在其他地方使用相同的Load方法。

Answer 2

我解决了我的问题，非常感谢@Ulugbek Umirov的帮助，问题出在我的ReadXmlFile方法上，就像这样：

public static Color GetColor(string ChildName, string Attribute) 
{
 Color clr = new Color(); string v = ""; 
 XmlTextReader reader = new XmlTextReader("DesignCfg.xml"); 
 XmlDocument doc = new XmlDocument(); 
 XmlNode node = doc.ReadNode(reader); 
foreach (XmlNode chldNode in node.ChildNodes) 
{ 
   if (chldNode.Name == ChildName) 
      v = chldNode.Attributes["" + Attribute + ""].Value; 
} 
clr = System.Drawing.ColorTranslator.FromHtml(v);
return clr; 
}

而新的是：

public static Color GetColor(string ChildName, string Attribute)
{
  Color clr = new Color();
  string v = "";
  XmlDocument doc = new XmlDocument();
  doc.Load("DesignCfg.xml");
  XmlElement formData = (XmlElement)doc.SelectSingleNode("//" + ChildName);
  if (formData != null)
     v = formData.GetAttribute(Attribute);    
  clr = System.Drawing.ColorTranslator.FromHtml(v);
  return clr;
}

谢谢大家的帮助:)

更新xml文件中的元素值

2 个答案: