How to serialize List<T> in c# without the ArrayOf wrapper?

Question

I'm attempting to serialize a List<T> of some custom classes with:

List<T> l = new List<T>();
...
XmlSerializer serializer = new XmlSerializer(l.GetType());
serializer.Serialize(writer, list);

And it outputs:

<ArrayOfT>
    <T>
        ...
    </T>
    <T>
        ...
    </T>
</ArrayOfT>

How do I get the same thing, but without the ArrayOfT elements, so it would look like:

<T>
    ...
</T>
<T>
    ...
</T>

I'm attempting to do this because it's happening in a wrapper class so my actual output is coming in the format of:

<PropertyName>
    <ArrayOfT>
        <T>
            ...
        </T>
        <T>
            ...
        </T>
    </ArrayOfT>
</PropertyName>

And the ArrayOfT element is redundant (logically it would have the same name as PropertyName.)

Use the [XmlElement("T")] attribute if l is a property in a class. — TheLethalCoder
– TheLethalCoder, Commented Jan 11, 2018 at 17:26
@TheLethalCoder I'm trying to completely remove the ArrayOfT element, not rename it, also l is not a property, it is a variable within an implementation of IXmlSerializable.WriteXml(XmlWriter writer) which is in the class which makes up the property. — CoryG
– CoryG, Commented Jan 11, 2018 at 17:35
@juharr Apparently so, I'm chalking this up to my unfamiliarity with the XmlSerializer, if you want to post this as an answer I'll accept it. — CoryG
– CoryG, Commented Jan 11, 2018 at 17:46
Possible duplicate of XML Serialization - Disable rendering root element of array — ASpirin
– ASpirin, Commented Jan 11, 2018 at 18:53

Ctznkane525 · Accepted Answer · 2018-01-11 18:08:58Z

2

You could do this approach perhaps:

XmlSerializer serializer = new XmlSerializer(l.GetType());
foreach (T item in list)
{
    serializer.Serialize(writer, item);
}

This way you are serializing the items but not the outer object.

answered Jan 11, 2018 at 18:08

Ctznkane525

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1 Comment

CoryG Over a year ago

This is correct but while I voted it up I'm not marking it as accepted yet just because @juharr mentioned it first in a comment and I want to give him time to submit it as an answer, I'll check back tomorrow and if he hasn't posted it I'll accept this.

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