Question

我有一个以下格式的xml：

<fld-03 m="01" d="01" y="1965" sex-m="" sex-f="1"></fld-03>
<fld-04 last="lastname" first="firstname" middle=""></fld-04>
<fld-05 addr="Richardson" city="Dallas" state="TX" zip="75080" phone=""></fld-05>
<fld-06 self="" spouse="1" child="" other=""></fld-06>
<fld-07 addr="" city="" state="" zip="" phone=""></fld-07>

有33个这样的“fld”，属性差异很大。以下是我尝试继续的示例：

@XmlRootElement(name = "myroot")
class Formxml implements Serializable{

    @XmlElement (name="fld-00")
    private Fld00 fld00 = new Fld00();

    public class Fld00 {
        private String payer;
        @XmlAttribute(name = "payer", required = true)
        String getPayer() {
            return payer
        }
        void setPayer(String payer) {
            this.payer = payer
        }
    }
}

这不起作用，因为类Fld00是非静态内部类。

当我将其更改为静态类并添加XmlAccessType.FIELD时，它会引发"Class has two properties of the same name "payer""异常。下面是带有静态类的代码：

@XmlRootElement(name = "Claims")
class Formxml implements Serializable{

    @XmlElement (name="fld-00")
    private Fld00 fld00 = new Fld00();
    @XmlAccessorType(XmlAccessType.FIELD)
    public static class Fld00 {
        private String payer;
        @XmlAttribute(name = "payer", required = true)
        String getPayer() {
            return payer
        }
        void setPayer(String payer) {
            this.payer = payer
        }
    }
}

如果您找到使用JAXB的方法，请告诉我。

由于

Answer 1

下面是一个能够按照您的描述处理XML的类。如果你将fld-xx的类作为单独的类编写，而不是我称之为RecordType的静态类，那么它将更简单（使用）。

请注意，这些静态类的编写方式之间存在细微差别：注释@XmlAttribute附加到字段而不是方法。

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "RecordType", propOrder = {
    "fld03",
    "fld04"
})
public class RecordType {

@XmlElement(name = "fld-03", required = true)
protected RecordType.Fld03 fld03;
@XmlElement(name = "fld-04", required = true)
protected RecordType.Fld04 fld04;

public RecordType.Fld03 getFld03() { return fld03; }
public void setFld03(RecordType.Fld03 value) {
    this.fld03 = value;
}
public RecordType.Fld04 getFld04() { return fld04; }
public void setFld04(RecordType.Fld04 value) {
    this.fld04 = value;
}

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "")
public static class Fld03 {
    @XmlAttribute(name = "m", required = true)
    protected int m;

    public int getM() { return m; }
    public void setM(int value) { this.m = value; }
}

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "")
public static class Fld04 {
    @XmlAttribute(name = "lastname", required = true)
    protected String lastname;

    public String getLastname() { return lastname; }
    public void setLastname(String value) { this.lastname = value; }

}
}

如何使用JAXB为每个元素解析具有多个属性的XML？

1 个答案: