tags:

views:

556

answers:

4

Hi all,

I have a simple class XmlFileHelper as follows:

public class XmlFileHelper
{
    #region Private Members

    private XmlDocument xmlDoc = new XmlDocument();
    private string xmlFilePath;

    #endregion

    #region Constructor

    public XmlFileHelper(string xmlFilePath)
    {
        this.xmlFilePath = xmlFilePath;
        xmlDoc.Load(xmlFilePath);
    }

    #endregion

    #region Public Methods

    public XmlNode SelectSingleNode(string xPathQuery)
    {
        return xmlDoc.SelectSingleNode(xPathQuery);
    }

    public string GetAttributeValueByName(XmlNode node, string attributeName)
    {
        return node.Attributes.GetNamedItem(attributeName).Value;
    }

    #endregion

    #region Public Properties

    public string XmlFilePath
    {
        get
        {
            return xmlFilePath;
        }
    }

    #endregion
}

The issue is I am getting the following error on Load:

System.IO.IOException: The process cannot access the file ''C:\CvarUAT\ReportWriterSettings.xml'' **because it is being used by another process**

this occurs when this class is used to by two running instances of a component running in parallel both attempting to load the xml file above, this is legitimate behaviour and required by the application.

I only want to read in the xml off disk once and release any reference to the file on disk and use an in memory representation from that point forward.

I would have assumed Load operates in a readonly fashion and would have no need to lock the file, what is my best way to achieve the desired result and get around this issue?

Thanks

A: 

try:

xml.Load(new StreamReader(new FileStream(path, FileMode.Open, FileAccess.Read, FileShare.Read)));
JeffreyABecker
A: 

it depends on what you need from the file,

If you need it to be threasdsafe you would need to impliment a mutex to lock the loading between instance,

If you dont really need thread safe loading (i.e. the file naver changes) you could load it via a filestream then loadf the XmlDocument form the stream

            FileStream xmlFile = new FileStream(xmlFilePath, FileMode.Open,
FileAccess.Read, FileShare.Read);
            xmlDoc.Load(xmlFile);
Pharabus
Am trying this out now, thanks
m3ntat
Excellent this did the trick!
m3ntat
A: 

If the file isn't too big to read into memory all at once:

xml.Load(new MemoryStream(File.ReadAllBytes(path)));

Michael McCloskey
+4  A: 

You can do this

using (Stream s = File.OpenRead(xmlFilePath))
{
    xmlDoc.Load(s);
}

instead of

xmlDoc.Load(xmlFilePath);
João Angelo