@JacksonXmlProperty（isAttribute = true）使用Jacson添加唯一标识XmlMapper

嗨，我想使用jackson-dataformat-xml 2.7.3 XmlMapper将我的POJO转换为xml。我在POJO类中使用jackson注释，如下面的代码所示：但是我得到了一些独特的ID，它们被添加到我的列表的每个标签中。我如何删除这些独特的ID。@JacksonXmlProperty（isAttribute = true）使用Jacson添加唯一标识XmlMapper

//下面是ElementTag的类

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; 
} 
}

//下面是ProfessionalLeadDetails类

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; 
} 
}

// 使用主要方法内XmlMapper转换成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("[email protected]"); 
    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(); 
}

但我正在逐渐VAL之前附加像zdef1999262822一些独特的ID：如下图所示：输出

<Leads xmlns=""><FL zdef2041716767:val="First Name"><value>Sandeep</value></FL><FL zdef1999262822:val="Email"><value>[email protected]</value></FL></Leads>

所需的输出：

<Leads xmlns=""><FL val="First Name"><value>Sandeep</value></FL><FL val="Email"><value>[email protected]</value></FL></Leads>

提前感谢！

来源

2017-08-31 Sandeep Soni

确保使用Woodstox Stax实现，而不是使用Stax实现Oracle捆绑JDK。这通常通过添加Maven依赖来明确包含woodstox jar来完成。这在XML模块上进行了解释

<dependency> 
      <groupId>org.codehaus.woodstox</groupId> 
      <artifactId>woodstox-core-asl</artifactId> 
      <version>4.4.1</version> 
     </dependency>

来源

2017-08-31 17:45:53

@JacksonXmlProperty（isAttribute = true）使用Jacson添加唯一标识XmlMapper

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