Question

我使用下面的代码成功地将我的XML转换为Json。

我遇到了这个问题，我在下面的程序中有两个不同的XML字符串。

第一个具有单个用户名值，第二个具有两个值。

第一个xml生成没有Square括号的json，第二个xml生成Square括号。

我希望他们俩都有方括号。有没有办法在输出字符串中强制转换用户名作为数组？

首先，我希望它们看起来像输出json中的数组。

package com.java.json;

import net.sf.json.JSONObject;
import net.sf.json.JsonConfig;
import net.sf.json.xml.XMLSerializer;

public class XmlSerializer {

public static void main(String[] args) throws Exception {
String xmlstring = "<soapenv:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:ser=\"http://services.web.post.list.com\"><soapenv:Header><authInfo xsi:type=\"soap:authentication\" xmlns:soap=\"http://list.com/services/SoapRequestProcessor\"><!--You may enter the following 2 items in any order--><username xsi:type=\"xsd:string\">user@email.com</username><password xsi:type=\"xsd:string\">password</password></authInfo></soapenv:Header></soapenv:Envelope>";
//String xmlstring = "<soapenv:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:ser=\"http://services.web.post.list.com\"><soapenv:Header><authInfo xsi:type=\"soap:authentication\" xmlns:soap=\"http://list.com/services/SoapRequestProcessor\"><!--You may enter the following 2 items in any order--><username xsi:type=\"xsd:string\">user@email.com</username><username xsi:type=\"xsd:string\">user@email.com</username><password xsi:type=\"xsd:string\">password</password></authInfo></soapenv:Header></soapenv:Envelope>";

JsonConfig conf =  new JsonConfig();
XMLSerializer xs = new XMLSerializer();

xs.setForceTopLevelObject(true);   
xs.setSkipWhitespace(true);
xs.setTrimSpaces(true);    
xs.setSkipNamespaces(true);
xs.setRemoveNamespacePrefixFromElements(true);

JSONObject jsonobj = (JSONObject) xs.read(xmlstring);    
String  jsonstring  = jsonobj.toString().replace("\"@", "\"");    

System.out.println(jsonstring);     

}
}

第一个XML应该提供这样的输出。

{"Envelope":{"Header":{"authInfo":{"type":"soap:authentication","username":[null],"password":null}}}}

目前正在这样做。

{"Envelope":{"Header":{"authInfo":{"type":"soap:authentication","username":null,"password":null}}}}

上述程序中的第二个XML很好，它提供了这样的输出。

{"Envelope":{"Header":{"authInfo":{"type":"soap:authentication","username":[null,null],"password":null}}}}

提前致谢。

Answer 1

经过大量研究后，我找到了处理问题的最佳方法。

您可以使用下面的代码段将任何XML转换为JSON，而不会丢失类型信息和数组问题。

我们需要Java POJO像下面这样转换。

在下面的代码中使用AuthInfo.class。我正在使用适当的数据类型获取JSON。

您可以在此方法中将任何复杂的XML转换为JSON。

package com.java.json;

import com.fasterxml.jackson.annotation.JsonInclude.Include;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import com.fasterxml.jackson.dataformat.xml.jaxb.XmlJaxbAnnotationIntrospector;
import com.fasterxml.jackson.module.jaxb.JaxbAnnotationIntrospector;
import com.tda.trident.beb.AuthInfo;

public class XmlDoc {

public static void main(String[] args) throws Exception {
    String xmlstring = "<soapenv:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:ser=\"http://services.web.post.list.com\"><soapenv:Header><authInfo xsi:type=\"soap:authentication\" xmlns:soap=\"http://list.com/services/SoapRequestProcessor\"><!--You may enter the following 2 items in any order--><username xsi:type=\"xsd:string\">user@email.com</username><password xsi:type=\"xsd:string\">password</password></authInfo></soapenv:Header></soapenv:Envelope>";
    //String xmlstring = "<soapenv:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:ser=\"http://services.web.post.list.com\"><soapenv:Header><authInfo xsi:type=\"soap:authentication\" xmlns:soap=\"http://list.com/services/SoapRequestProcessor\"><!--You may enter the following 2 items in any order--><username xsi:type=\"xsd:string\">user@email.com</username><username xsi:type=\"xsd:string\">user@email.com</username><password xsi:type=\"xsd:string\">password</password></authInfo></soapenv:Header></soapenv:Envelope>";
    Object messageObj = null;       

    XmlMapper unmarshaller = new XmlMapper();
    unmarshaller.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    XmlJaxbAnnotationIntrospector xmlIntrospector = new XmlJaxbAnnotationIntrospector();
    unmarshaller.setAnnotationIntrospector(xmlIntrospector);
    JaxbAnnotationIntrospector introspector = new JaxbAnnotationIntrospector();
    ObjectMapper marshaller = new ObjectMapper().enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY);
    marshaller.setSerializationInclusion(Include.NON_NULL);
    marshaller.setAnnotationIntrospector(introspector);

    TransactionTradeMessage rootNode = unmarshaller.readValue(xmlstring, AuthInfo.class);

    messageObj = rootNode.getTransactionTrade();

    System.out.println(marshaller.writeValueAsString(messageObj));

}
}

Xml到JSON单值数组处理

1 个答案: