Question

我必须发出一个SOAP POST请求，如下所示

POST /sample/demo.asmx HTTP/1.1
Host: www.website.org
Content-Type: text/xml; charset=utf-8
Content-Length: length
SOAPAction: "https://www.website.org/Method"

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
  <soap:Body>
    <Method xmlns="https://www.ourvmc.org/">
      <item1>string</item1>
      <item2>string</item2>
      <item3>string</item3>
      <item4>string</item4>
    </Method>
  </soap:Body>
</soap:Envelope>

我已经实现了

POST /sample/demo.asmx HTTP/1.1
Host: www.website.org
Content-Type: text/xml; charset=utf-8
Content-Length: length
SOAPAction: "https://www.website.org/Method"

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
<soap:Body>
<Method xmlns="http://tempuri.org/"/>
</soap:Body>
</soap:Envelope>

我有一个名为Method和getter setter的POJO类，用于所有item1，item2，item3，item4。我需要将该POJO转换为xml格式

<item1>string</item1>
  <item2>string</item2>
  <item3>string</item3>
  <item4>string</item4>

然后发布它。任何人都可以建议怎么做？我已经研究过，但没有找到任何有用的解决方案。

Answer 1

使用以下XMLUtil将pojo转换为xml

用法示例：

1）假设Pojo是学生

@XmlRootElement(name = "Student")
@XmlAccessorType(XmlAccessType.FIELD)
public class Student {

    @XmlAttribute
    private String type;

    @XmlElement(name="Name")
    private String name;

    public void setType(String type) {
        this.type = type;
    }

    public void setName(String name) {
        this.name = name;
    }
 }

2）XMLUtil

import java.io.StringWriter;    
import java.util.logging.Logger;    
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;


public class XMLUtil {


    public static String toXML(Object data) {
        String xml = "";
        try {
            LOGGER.info("Generating xml for: " + data.getClass());
            JAXBContext jaxbContext = JAXBContext.newInstance(data.getClass());
            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            //jaxbMarshaller.marshal(data, System.out);
            StringWriter sw = new StringWriter();
            jaxbMarshaller.marshal(data, sw);
            xml = sw.toString();
        } catch (JAXBException e) {
            //handle your exception here
        }
        return xml;
    }

}

3）将数据设置为student对象并传递给util

   Student st = new Student();
   st.setType("schoolStudent");
   st.setName("Devendra");

   String studentXml = XMLUtil.toXML(st);

Answer 2

考虑使用JAXB：https://jaxb.java.net/2.2.11/docs/ch01.html#documentation

对于实际发布，请考虑Apache CXF：http://cxf.apache.org/docs/index.html

有关使用JAXB + CXF从POJO生成XML的更多信息：http://cxf.apache.org/docs/jaxb.html

将POJO转换为XML

2 个答案: