Convert string to variable name

Question

I have an XML file, I have a node and I read all ChildNodes. The name of the childNode match to a variable I have to set with the value of this childNode.

In the loop, I'd like set :

myvar1 to MyValue1
myvar2 to MyValue2

The C# Code :

protected string myvar1;
protected string myvar2;

The XML content look like this :

<parameters>
 <myvar1>MyValue1</myvar1>
 <myvar2>MyValue2</myvar2>
</parameters>

C# set variables :

    foreach (var item in xmlParamInstallation.SelectNodes("parameters")[0].ChildNodes)
    {
        ??????
    }

Any idea ?

Thanks,

UPDATE 1: the value "field" in the loop is null all the time.

public class ParametersTest
{
    public string myvar1 { get; set; }
    public string myvar2 {get; set;}
}

var type = typeof(ParametersTest);
foreach (XmlNode item in xmlParamInstallation.SelectNodes("parameters")[0].ChildNodes)
{

    var field = type.GetField(item.LocalName);
    field.SetValue(field, item.InnerText);
}

Couldn't you just use XmlSerializer to do this? It is more efficient than using reflection manually. — vgru
– vgru, Commented Jan 4, 2012 at 9:39

Samich · Accepted Answer · 2012-01-04 09:02:18Z

7

You can do it using Reflection:

var type = typeof(SomeClass);
var field = type.GetField(item.Name);
field.SetValue(null, item.InnerText);

RE: UPDATE 1

var parameters = new ParametersTest();
var type = parameters.GetType();

var s = @"<parameters>
            <MyVar1>MyValue1</MyVar1>
            <MyVar2>MyValue2</MyVar2>
           </parameters>";

var xmlParamInstallation = new XmlDocument();
xmlParamInstallation.LoadXml(s);

foreach (XmlNode item in xmlParamInstallation.SelectNodes("parameters")[0].ChildNodes)
{

    var field = type.GetProperty(item.LocalName);
    field.SetValue(parameters, item.InnerText, null);
}

edited Jan 4, 2012 at 9:02

answered Jan 4, 2012 at 8:32

Samich

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1 Comment

Samich Over a year ago

First: Your xml node names doesn't match the properties names including case-sensitivity. Second: you are using properties, so you need to call GetProperty instead of GetField.

Tim M. · Accepted Answer · 2012-01-04 08:30:53Z

If you are seeking to assign variables based on the names of nodes in XML, you have at least a couple options:

Deserialize the XML structure into an object with corresponding member names
Populate the variables using reflection
Populate the variables using dynamic method calls/expression trees that know how to read the contents of the XML node into an object property.

All of these approaches suggest a more object-oriented approach to the problem then just populating a few variables, but it would be easy to create a lightweight structure with the appropriate members which is populated by reading the XML document.

You could also use a key-based collection (like a Dictionary<string, string>) to store the values if you are just looking to build a simple name/value collection from the source XML.

Guffa · Accepted Answer · 2012-01-04 08:40:46Z

3

If you put the names and values in a dictionary, you can easily get the values by name:

Dictionary<string, string> parameters =
  xmlParamInstallation.SelectNodes("parameters")[0].ChildNodes
  .ToDictionary(n => n.Name, n => n.InnerText);

myvar1 = parameters["myvar1"];
myvar2 = parameters["myvar2"];

answered Jan 4, 2012 at 8:40

Guffa

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2 Comments

TheBoubou Over a year ago

I don't want set variable one by one (if possible)

Guffa Over a year ago

Well, you can do it using reflecction, but I like to give an alternative without reflection. Besides, this way you control which properties are set, not the contents of the XML file.

Thorsten Dittmar · Accepted Answer · 2012-01-04 08:33:20Z

2

You could either do as "Default" said, or you could look into Reflection. By using the Type.GetMember(string) method you could find a member with the given name (the tag name in your XML) and set its value.

EDIT
Samich beat me, so I'll give him +1 - he's got sample code as well.

answered Jan 4, 2012 at 8:33

Thorsten Dittmar

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Comments

default · Accepted Answer · 2012-01-04 08:29:36Z

1

You can check out the XmlSerializer class

answered Jan 4, 2012 at 8:29

default

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