如何以编程方式将JSON转换为JSON模式以比较两个JSON对象的结构

Question

如何以编程方式将JSON转换为JSON模式以比较Json Structure，而无需 考虑值。例如，我提到了两个json

{ “ name”：“ John”， “年龄”：30岁， “地址”：伦敦 }    和

{ “ name”：“ Robert”， “年龄”：31， “地址”：Southampton }

我已经写了如下代码。方法应该返回true但返回false。因为两个json属性值不同，所以我只想比较结构

public static boolean assertXmlStructureEquals(String expectedXML,String actualXML) {
        try {
            DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
            dbFactory.setNamespaceAware(true);
            DocumentBuilder docBuilder = dbFactory.newDocumentBuilder();
            Node doc = docBuilder.parse(expectedXML);
            Node doc1 = docBuilder.parse(actualXML);
            XpathFormatter expectedXMLFormatter = new XpathFormatter(doc);
            XpathFormatter actualXMLFormatter = new XpathFormatter(doc1);
            if (expectedXMLFormatter.equals(actualXMLFormatter)) {
                return true;
            } else {
                return false;
            }
        } catch (Exception ex) {
            return false;
        }

    }