Unity XML Get list values from xml list to c# list

Question

I need to retrieve list values from attributes in a file with this XML:

<LISTS>
<LIST list="213 82 14 2 3 4 18 4 1 5 5 3 6 5 2"/>
<LIST list="2 4 5 1 4 0 0 0 2 3 5 532 7 10 0"/>
<LIST list="10 511 4 8 2 6 4 721 2 5 7 10 10 4"/>
</LISTS>

Using a stream reader I put the XML in a string inside a list. So it looks like this:

 <LISTS><LIST list="213 82 14 2 3 4 18 4 1 5 5 3 6 5 2"/><LIST list="2 4 5 1 4 0 0 0 2 3 5 532 7 10 0"/><LIST list="10 511 4 8 2 6 4 721 2 5 7 10 10 4"/></LISTS>

Then I create the XML document

Xmldoc = new XmlDocument();
Xmldoc.LoadXml(XmlLineList[0]) //i get an obj reference not set to instance of obj ex here. Even tho i can see in the inspector the xml line is there.

If I use

Xmldoc.Load(XmlLineList[0]) // Illegal characters in Path

And that is as far as I have gone. I have been reading documentation all night but cant find a good answer or a case that resembles mine. Need to pass the elements of the xml list into a c# list. Every list in a different one, so three lists.

Answer 1

Your XML basically consists of a sequence <LIST> of lists list="1 2 3" of integers. You have a couple options to parse this XML and extract a List<int> list for each <LIST> element.

Firstly, You can use XmlSerializer to deserialize your XML. First define the followng data model:

[XmlRoot("LISTS")]
public class ListRoot
{
    [XmlElement("LIST")]
    public List<ListEntry> Lists { get; set; } = new List<ListEntry>();
}

public class ListEntry
{
    [XmlAttribute("list")]
    public List<int> ListItems { get; set; } = new List<int>();
}

Then you can deserialize from a file located at fileName as follows:

ListRoot listRoot;
var serializer = new XmlSerializer(typeof(ListRoot));
using (var stream = File.OpenRead(fileName))
    listRoot = (ListRoot)serializer.Deserialize(stream);

By marking the List<int> ListItems with [XmlAttribute("list")] , XmlSerializer will automatically deserialize the space-delimited list="2 4..." attributes into the inner ListItems list.

Now you can access the list items as follows:

foreach (var entry in listRoot.Lists)
{
    List<int> list = entry.ListItems; // The contents of each list="1 2 ..." /> attribute materialized as a List<int>
    // Process the List<int> list however you want:
    Console.WriteLine(string.Join(",", list));
}

Which outputs:

213,82,14,2,3,4,18,4,1,5,5,3,6,5,2
2,4,5,1,4,0,0,0,2,3,5,532,7,10,0
10,511,4,8,2,6,4,721,2,5,7,10,10,4

Demo fiddle #1 here .

Alternatively, you could use LINQ to XML to load your XML into an XDocument and query it using LINQ :

var listRoot = XDocument.Load(fileName);

var listItems = listRoot.Root
    .Elements("LIST") // Select the <LIST> child nodes
    .Select(e => e.Attribute("list")) // Select the list=".." attribute
    .Select(a => a.Value.Split(' ').Select(s => XmlConvert.ToInt32(s)).ToList()) // Convert the space-delimited integer string to a list of integers
    .ToList(); // And materialize the query

foreach (List<int> list in listItems)
{
    // Process the List<int> list however you want:
    Console.WriteLine(string.Join(",", list));
}

Demo fiddle #2 here .

I don't really recommend using the older XmlDocument for new code as LINQ to XML is easier to use and more performant.

Answer 2

It is better to use LINQ to XML API. It is available in the.Net Framework since 2007.

Here is your starting point.

c#

void Main()
{
    const string fileName = @"e:\temp\input.xml";
    // Load XML file directly from the file system
    //XDocument xdoc = XDocument.Load(fileName);
    
    XDocument xdoc = XDocument.Parse(@"<LISTS>
            <LIST list='213 82 14 2 3 4 18 4 1 5 5 3 6 5 2'/>
            <LIST list='2 4 5 1 4 0 0 0 2 3 5 532 7 10 0'/>
            <LIST list='10 511 4 8 2 6 4 721 2 5 7 10 10 4'/>
        </LISTS>
        ");
            
    foreach (XElement xelem in xdoc.Descendants("LIST"))
    {
        Console.WriteLine("list: '{0}'", xelem.Attribute("list").Value);
    }
}

Output

list: '213 82 14 2 3 4 18 4 1 5 5 3 6 5 2'
list: '2 4 5 1 4 0 0 0 2 3 5 532 7 10 0'
list: '10 511 4 8 2 6 4 721 2 5 7 10 10 4'