CODEWITHSUNDEEP
javaxmljackson2jackson-dataformat-xml
ulquiorra
ulquiorra

Reputation: 945

how to allow duplicate keys with jackson while deserializing xml

I have a simple xml file structured like this with multiples B keys

<?xml version="1.0" encoding="UTF-8"?>
<Example>
   <ExampleId>0000</ExampleId>
   <Name>test</Name>
   <Important1>
      <Id>
         <Number>1111111111</Number>
      </Id>
   </Important1>
   <B>
      <Name>test2</Name>
      <Important2>
         <Id>
            <Number>2222222222</Number>
         </Id>
      </Important2>
   </B>
   <B>
      <Name>test3</Name>
      <Important2>
         <Id>
            <Number>3333333333</Number>
         </Id>
      </Important2>
   </B>
</Example>

I need to retrieve all values from Number keys to a List
I tried multiple ways

JsonNode numberParser = xmlMapper.readTree(inputStream);
JsonNode node1 = numberParser.get("Example");
            
//first try
List<String> list =node1.findValuesAsText("Number");
            
//second try
List<String> list =new ArrayList<String>();
node1.fields().forEachRemaining(firstEntry -> {
    if(firstEntry.getKey().equals("Important1")){
            JsonNode fieldsNode = firstEntry.getValue();
            String numberFromImportant1 = fieldsNode.findValue("Number").asText();
            list.add(numberFromImportant1);
    }
    if(firstEntry.getKey().equals("B")){
            JsonNode fieldsNode = firstEntry.getValue();
            String numberFromImportant2 = fieldsNode.findValue("Number").asText();
            list.add(numberFromImportant2);
    }
});

However , it seems that jackson takes only the last value from Number if it encounters duplicate B nodes while deserializing Any way to prevent that and allow Jackson to take duplicate into account as well ?

Upvotes: 1

Views: 450

Answers (1)

Greg
Greg

Reputation: 4518

If all you want to do is retrieve the value for the node number, then consider using xpath. Appending double forward space to the XML node will search for the node (in the entire xml file). E.g: //Number

For example:

Document xmlDocument = builder.parse(inputStream);
this.clean(xmlDocument);
XPath xPath = XPathFactory.newInstance().newXPath();
String expression = "//Number";
nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);

Upvotes: 1

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