Java @JacksonXmlProperty(isAttribute=true)使用Jackson XmlMapper附加唯一id
您好,我正在尝试使用jackson dataformat xml 2.7.3 XmlMapper将POJO转换为xml。我在POJO类中使用jackson注释,如下代码所示,但我在列表的每个标记中都添加了一些唯一的ID。如何删除这些唯一ID //下面是元素标签类Java @JacksonXmlProperty(isAttribute=true)使用Jackson XmlMapper附加唯一id,java,jackson,jackson-dataformat-xml,xmlmapper,Java,Jackson,Jackson Dataformat Xml,Xmlmapper,您好,我正在尝试使用jackson dataformat xml 2.7.3 XmlMapper将POJO转换为xml。我在POJO类中使用jackson注释,如下代码所示,但我在列表的每个标记中都添加了一些唯一的ID。如何删除这些唯一ID //下面是元素标签类 import java.util.List; import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper; import com.f
import java.util.List;
import
com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlProperty;
/**
*Element class
*/
public class ElementTag {
@JacksonXmlProperty(localName = "FL")
@JacksonXmlElementWrapper(useWrapping = false)
private List<ProfessionalLeadDetails> pf;
/**
* @return the pf
*/
public List<ProfessionalLeadDetails> getPf() {
return pf;
}
/**
* @param pf the pf to set
*/
public void setPf(List<ProfessionalLeadDetails> pf) {
this.pf = pf;
}
}
import java.io.Serializable;
import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlProperty;
public class ProfessionalLeadDetails implements Serializable {
/** The Constant serialVersionUID. */
private static final long serialVersionUID = 1L;
@JacksonXmlProperty(isAttribute = true)
private String val;
private String value;
/**
* @return the val
*/
public String getVal() {
return val;
}
/**
* @param val the val to set
*/
public void setVal(String val) {
this.val = val;
}
/**
* @return the value
*/
public String getValue() {
return value;
}
/**
* @param value the value to set
*/
public void setValue(String value) {
this.value = value;
}
}
//在主方法内部使用XmlMapper转换为xml
XmlMapper xmlMapper = new XmlMapper();
ElementTag et = new ElementTag();
List<ProfessionalLeadDetails> pfList = new
ArrayList<ProfessionalLeadDetails>();
ProfessionalLeadDetails pf = new ProfessionalLeadDetails();
pf.setVal("First Name");
pf.setValue("Sandeep");
pfList.add(pf);
pf = new ProfessionalLeadDetails();
pf.setVal("Email");
pf.setValue("Sandeep@gmail.com");
pfList.add(pf);
pfList.add(pf2);
et.setPf(pfList);
try {
System.out.println(xmlMapper.writer()
.with(SerializationFeature.WRAP_ROOT_VALUE)
.withRootName("Leads").writeValueAsString(et));
} catch (JsonProcessingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
XmlMapper XmlMapper=newxmlmapper();
ElementTag et=新的ElementTag();
列表pfList=新建
ArrayList();
ProfessionalLeadDetails pf=新的ProfessionalLeadDetails();
pf.setVal(“名字”);
设定值(“Sandeep”);
pfList.add(pf);
pf=新的专业领导详情();
pf.setVal(“电子邮件”);
pf.setValue(“Sandeep@gmail.com");
pfList.add(pf);
pfList.add(pf2);
et.setPf(pfList);
试一试{
System.out.println(xmlMapper.writer()
.with(SerializationFeature.WRAP\u ROOT\u值)
.withRootName(“Leads”).writeValueAsString(et));
}捕获(JsonProcessingException e){
//TODO自动生成的捕捉块
e、 printStackTrace();
}
但我得到了一些附加在val之前的唯一ID,如zdef1999262822:如下所示:
输出
<Leads xmlns=""><FL zdef2041716767:val="First Name"><value>Sandeep</value></FL><FL zdef1999262822:val="Email"><value>Sandeep@gmail.com</value></FL></Leads>
SandeepSandeep@gmail.com
所需输出:
<Leads xmlns=""><FL val="First Name"><value>Sandeep</value></FL><FL val="Email"><value>Sandeep@gmail.com</value></FL></Leads>
SandeepSandeep@gmail.com
提前谢谢 确保使用Woodstox Stax实现,而不是将Stax实现与JDK结合使用。这通常是通过添加Maven依赖项来显式包含woodstox jar来实现的。这在XML模块中进行了解释
<dependency>
<groupId>org.codehaus.woodstox</groupId>
<artifactId>woodstox-core-asl</artifactId>
<version>4.4.1</version>
</dependency>
org.codehaus.woodstox
woodstox core asl
4.4.1
确保使用Woodstox Stax实现,而不是将Oracle的Stax实现捆绑到JDK中。这通常是通过添加Maven依赖项来显式包含woodstox jar来实现的。这在XML模块中进行了解释
<dependency>
<groupId>org.codehaus.woodstox</groupId>
<artifactId>woodstox-core-asl</artifactId>
<version>4.4.1</version>
</dependency>
org.codehaus.woodstox
woodstox core asl
4.4.1