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Programming-Idioms

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Idiom #103 Load XML file into object

Read from the file data.xml and write its contents into the object x.
Assume the XML data is suitable for the type of x. 

import "encoding/xml"
import "os"
buffer, err := os.ReadFile("data.xml")
if err != nil {
	return err
}
err = xml.Unmarshal(buffer, &x)
if err != nil {
	return err
}

buffer is a []byte.
&x is the address of x.
You must check errors after each step.
using System.Xml.Linq;
XDocument x = XDocument.Load("data.xml");
New implementation...
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programming-idioms.org 
buffer, err := os.ReadFile("data.xml")
if err != nil {
	return err
}
err = xml.Unmarshal(buffer, &x)
if err != nil {
	return err
}
XDocument x = XDocument.Load("data.xml");


XmlArchive archive = new XmlArchive!(char);
archive.doc.parse(read("data.xml"));
Serializer ser = new Serializer(archive);
ser.deserialize(x);
def x = new XmlSlurper().parse(new File('data.xml'))
(let ((x (plump:parse #p"data.xml")))
  ...)
my $x = XML::LibXML->load_xml(
location => 'data.xml');
x = lxml.etree.parse('data.xml')
val xml = XML.loadFile("data.xml")