如何在 Java 中将 XML 转换为 JSON 并避免解析器尝试将字符串解析为数字

Question

I'm using org.json.XML library to parse XML to JSON. http://www.json.org/javadoc/org/json/XML.html

In my XML doc there is an ID field that is randomly generated with [0-9][az]. It is intended to be a String. Everything works fine until there's this unlucky ID 123456789e1234 that happens to be a scientific notation of a number. Here's a piece of test code:

public class XmlToJsonTest {
    public static String testXML = "<MyXML><ID>123456789e1234</ID></MyXML>";
    @Test
    public void testXMLtoJSON() throws JSONException {
        JSONObject testJsonObject = XML.toJSONObject(testXML);
    }
}

Here's the exception:

org.json.JSONException: JSON does not allow non-finite numbers.

The XML lib toJson() method first attempts to converts the String to Integer, Long or Double and if none works, it gives up parsing it as number and treats it like a String. In this case the String 123456789e1234 is parsable as a Double. And later when the lib checks if the double is infinite by Double.isInfinite(), it throws the JSONException since obviously 123456789e1234 is greater than Double's standard of finite.

How do I force it not to parse the value as number at all? Is there any java library out there that correctly converts XML to JSON in this case?

Answer 1

Try this ...

import org.json.me.JSONObject;

import cjm.component.cb.json.ToJSON;

public class DummyClass
{
public static void main(String[] args)
{
    try
    {
        String xml = "<MyXML><ID>123456789e1234</ID></MyXML>";

        JSONObject obj = (new ToJSON().convertToJSON(xml));

        System.out.println(obj);
    }
    catch (Exception e)
    {
        e.printStackTrace();
    }
}
}

Output

 -------- XML Detected -------- 
 -------- JSON created Successfully -------- 
{"MyXML":{"ID":"123456789e1234"}}

Get the Conversion Box Jar for that ....

Answer 2

Not sure, but this works for me:

String testXML = "<MyXML><ID>123456789e1234</ID></MyXML>";
JSONObject testJsonObject = org.json.XML.toJSONObject(testXML);
System.out.println(testJsonObject.getJSONObject("MyXML").getString("ID"));

Answer 3

There is an underscore-java library with static method U.xmlToJson(xml) . I am the maintainer of the project.

public class XmlToJsonTest {
    public static String testXML = "<MyXML><ID>123456789e1234</ID></MyXML>";

    public void main(String[] args) {
        String testJson = U.xmlToJson(testXML);
        System.out.println(testJson);
    }
}

Output:

{
  "MyXML": {
    "ID": "123456789e1234"
  },
  "#omit-xml-declaration": "yes"
}

Answer 4

If you need the XML to be parsed as String, you can pass the second parameter to XML.toJSONObject as XMLParserConfiguration.KEEP_STRINGS

public static void main(String[] args) {
    System.out.println(XML.toJSONObject("<MyXML><ID>123456789e1234</ID></MyXML>", XMLParserConfiguration.KEEP_STRINGS));
}

Output

{"MyXML":{"ID":"123456789e1234"}}