Question

如何使用JAXB，

从此xml文件创建java对象

<?xml version="1.0" encoding="ISO-8859-1" ?>
<Parameters>
<Varremise>
     <Name>Amazigh</Name>
     <Var>Notprovided</Var>
     <Var>Notprovided</Var>     
</Varremise>
<Varheader>
     <Name>CdAppOr</Name>
     <Var>P</Var>
     <Var>A</Var>
     <Var>O</Var>
</Varheader>
<Vardetail>
     <Name>TopUC</Name>
     <Var>U</Var>
     <Var>C</Var>
</Vardetail>
</Parameters>

我试过这个，但它不起作用，

参数类：

import java.util.Set;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlElementWrapper;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "parametersContainer")
@XmlRootElement(name = "Parameters")
public class Parameters {

@XmlElement(name = "Varremise")
@XmlElementWrapper(name = "Varremise")
private Set<VarRemise> varremise;

@XmlElement(name = "Varheader")
@XmlElementWrapper(name = "Varheader")
private Set<VarHeader> varheader;

@XmlElement(name = "Vardetail")
@XmlElementWrapper(name = "Vardetail")
private Set<VarDetail> vardetail;


public Set<VarHeader> getVarheader() {
    return varheader;
}

public void setVarheader(final Set<VarHeader> varheader) {
    this.varheader = varheader;
}

public Set<VarDetail> getVardetail() {
    return vardetail;
}

public void setVardetail(final Set<VarDetail> vardetail) {
    this.vardetail = vardetail;
}

public Set<VarRemise> getVarremise() {
    return varremise;
}

public void setVarremise(final Set<VarRemise> varremise) {
    this.varremise = varremise;
}

}

AllVarsDefinition类：

import java.util.Set;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public abstract class AllVarsDefinition {

@XmlElement(name = "Name", required = true)
private String name;

@XmlElement(name = "Var", required = true)
private Set<String> var;

public String getName() {
    return name;
}

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

public Set<String> getVar() {
    return var;
}

public void setVar(final Set<String> var) {
    this.var = var;
}

}

VarHeader类：

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;

@XmlAccessorType(XmlAccessType.FIELD)
public class VarHeader extends AllVarsDefinition {

}

与Vardetail和varRemise类相同的代码，

当我执行此代码时，它会创建一个空对象

谢谢，

Answer 1

问题是您的XML不是预期的。您已指定了包装器和具有相同名称的元素。

@XmlElement(name = "Varheader")
@XmlElementWrapper(name = "Varheader")

预期的xml实际上是以下形式：

<Varheader>
    <Varheader></Varheader>
</Varheader>

修改

基于一些评论，解决方案应该是：

删除WrapperElements

将Set设置为List，因为JAXB将自动以这种方式解析列表。

Answer 2

参数类

import java.util.List;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

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


@XmlElement(name = "Varremise")
private List<VarRemise> varremise;


public List<VarRemise> getVarremise() {
    return varremise;
}

public void setVarremise(final List<VarRemise> varremise) {
    this.varremise = varremise;
}

@XmlElement(name = "Varheader")
private List<VarHeader> varheader;

@XmlElement(name = "Vardetail")
private List<VarDetail> vardetail;

public List<VarHeader> getVarheader() {
    return varheader;
}

public void setVarheader(final List<VarHeader> varheader) {
    this.varheader = varheader;
}

public List<VarDetail> getVardetail() {
    return vardetail;
}

public void setVardetail(final List<VarDetail> vardetail) {
    this.vardetail = vardetail;
}

}

VarRemise类

import java.util.Set;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;


@XmlAccessorType(XmlAccessType.FIELD)
public class VarRemise {


@XmlElement(name = "Name", required = true)
private String name;

@XmlElement(name = "Var", required = true)
private Set<String> vars;

public String getName() {
    return name;
}

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

public Set<String> getVars() {
    return vars;
}

public void setVars(final Set<String> vars) {
    this.vars = vars;
}

}

与vardetail和varhead类相同的代码，它可以工作

谢谢cvesters

使用Jaxb将xml文件转换为Java对象

2 个答案: