CODEWITHSUNDEEP
c#
SharePointDummy
SharePointDummy

Reputation: 357

How to declare a variable with a variable name?

How can I declare a variable inside a for loop

foreach (System.Xml.XmlNode xmlnode in node)
{
    string AMSListName = xmlnode.Attributes["Title"].Value.ToString();
    /* the below assignment should be variable for each iteration */
    XmlNode ndViewFields + AMSListName = xmlDoc.CreateNode(XmlNodeType.Element,
                                                           "ViewFields", "");
}

How do I achieve this? I want for each value in the for loop to have the xmlnode to have a different name. Is this possible at all?

Upvotes: 0

Views: 277

Answers (2)

Tim Schmelter
Tim Schmelter

Reputation: 460108

Use a collection:

List<XmlNode> nodes = new List<XmlNode>();
foreach (System.Xml.XmlNode xmlnode in node)
{
    string AMSListName = xmlnode.Attributes["Title"].Value.ToString();
    nodes.Add(xmlDoc.CreateNode(XmlNodeType.Element, "ViewFields", ""));
}

You can access this list via index or in a loop:

foreach(var node in nodes)
{
    // ...
}

another approach If the name is an identifier, use a Dictionary:

Dictionary<string, System.Xml.XmlNode> nodeNames = new Dictionary<string, System.Xml.XmlNode>();
foreach (System.Xml.XmlNode xmlnode in node)
{
    string AMSListName = xmlnode.Attributes["Title"].Value.ToString();
    nodeNames[AMSListName] = xmlDoc.CreateNode(XmlNodeType.Element, "ViewFields", "");
}

This will replace an already available node with a given name, otherwise it'll add it.

You can access it via name:

XmlNode node
if(nodeNames.TryGetValue("Some Name", out node)
{
    // ..
};

Upvotes: 5

Scott Emick
Scott Emick

Reputation: 125

I would suggest using something like a dictionary or hastable. Because even if you did create dynamic variables, how would you reference them later? I am not sure this is possible.

            Hashtable ViewFields = new Hashtable();

        foreach (System.Xml.XmlNode xmlnode in node)
        {
            string AMSListName = xmlnode.Attributes["Title"].Value.ToString();
            XmlNode nd = xmlDoc.CreateNode(XmlNodeType.Element, "ViewFields", "");
            ViewFields.Add(AMSListName,nd);
        }

Upvotes: 0

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