Why does this throw a NullReferenceException?

Question

private void alterNodeValue(string xmlFile, string parent, string node, string newVal)
{
    XDocument xml = XDocument.Load(this.dir + xmlFile);

    if (xml.Element(parent).Element(node).Value != null)
    {
        xml.Element(parent).Element(node).Value = newVal;
    }
    else
    {
        xml.Element(parent).Add(new XElement(node, newVal)); 
    }

    xml.Save(dir + xmlFile); 
}  

Why does this throw

System.NullReferenceException was unhandled by user code

on this line

if (xml.Element(parent).Element(node).Value != null)

?

I'm guessing it's because the XML node doesn't exist, but that's what the != null is suppose to check for. How do I fix that?

I've tried several things and they ALL throw the same exception at some point during the not null check.

Thanks for any help.

Answer 1

Either xml.Element(parent) or the Element(node) you are trying to access from the return value of xml.Element(parent) is null .

Restructuring like this will enable to you see which one it is:

var myElement = xml.Element(parent);
if (xmyElement != null)
{
    var myNode = myElement.Element(node);
    if(myNode != null)
    {
      myNode.Value = newVal;
    }
}

---

Update:

From your comment it looks like you want to do this:

if (xml.Element(parent).Element(node) != null)  // <--- No .Value
{
    xml.Element(parent).Element(node).Value = newVal;
}
else
{
    xml.Element(parent).Add(new XElement(node, newVal)); 
}

Answer 2

几乎可以肯定，因为它返回null：

xml.Element(parent)

Answer 3

You need to check if:

Element(node) != null

Before you call .Value. If Element(node) == null, then the call to .Value will throw a null reference exception.

Dan

Answer 4

Try changing your if statement to this:

if (xml.Element(parent).Element(node) != null)

If the node in the parent element is null, you cannot access a member of a null object.

Answer 5

At least:

private void alterNodeValue(string xmlFile, string parent, string node, string newVal)
{
    XDocument xml = XDocument.Load(this.dir + xmlFile);
    XElement parent = xml.Element(parent).Element(node);
    if (parent  != null)
    {
        parent.Value = newVal;
    }
    else
    {
        xml.Element(parent).Add(new XElement(node, newVal)); 
    }    
    xml.Save(dir + xmlFile); 
}  

Better:

private void alterNodeValue(string xmlFile, string parent, string node, string newVal)
{
    string path = System.IO.Path.Combine(dir, xmlFile);
    XDocument xml = XDocument.Load(path );
    XElement parent = xml.Element(parent).Element(node);
    if (parent != null)
    {
        XElement node = parent.Element(parent);
        if (node != null)
        {
            node.Value = newVal;
        }
        else
        {
            // no node
        }
    }
    else
    {
        // no parent
    }    
    xml.Save(path); 
}  

Answer 6

        if (xml.Element(parent) != null)
        {
            var myNode = xml.Element(parent).Element(node);
            if (node != null)
                myNode.Value = newVal;
        }
        else
        {
            xml.Element(parent).Add(new XElement(node, newVal));
        }