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I'm trying to add a caching feature for http requests (for one of my projects) and I thought of using Etag as the hash value. But if the Etag is not there i thought of using the payload to generate a unique hash value. As we all know same xml pay loads might have different structures. For example Sample A and Sample B are same. But their string structures are not the same. What I need is a way to generate the same hash key from both xml samples.
Sample A
<note>
<to>Tove</to>
<from>Jani</from>
<heading>Reminder</heading>
<body>Don't forget me this weekend!</body>
</note>
Sample B
<note>
<to>Tove</to>
<heading>Reminder</heading>
<from>Jani</from>
<body>Don't forget me this weekend!</body>
</note>
571
asked Apr 14 '26 20:04
org.w3c.dom.document.normalizeDocument() document does not alter the order of child elements.
You could do this by a recursive parse of the document. However, consider whether this is more expensive than the operation you're trying to cache in the first place...
Method
java.util.List implementation, i.e. ArrayList. This is required because org.w3c.dom.NodeList does not allow modificationCollections.sort()Note this does not deal with multiple elements of same name with different contents, but does solve your example
For example:
public static void main(String[] args) throws Exception {
Document doc = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(new File("test.xml"));
sort(doc);
TransformerFactory tf = TransformerFactory.newInstance();
Transformer transformer = tf.newTransformer();
transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
StringWriter writer = new StringWriter();
transformer.transform(new DOMSource(doc), new StreamResult(writer));
System.out.println(writer);
}
private static void sort(Node doc) {
List<Node> children = new ArrayList<>();
for (int i = 0; i < doc.getChildNodes().getLength(); i++) {
children.add(doc.getChildNodes().item(i));
}
for (Node child : children) {
doc.removeChild(child);
}
Collections.sort(children, (a, b) -> {
return a.getNodeName().compareTo(b.getNodeName());
});
for (Node child : children) {
doc.appendChild(child);
}
for (Node child : children) {
sort(child);
}
}
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