使用simplexml 2.1.0将xml映射到Java

推子

我正在尝试使用SimpleXML 2.1.0将Web服务给出的xml响应映射到java对象，但是我被卡住了。

这是xml：

<QueryINVAP-WS-ASSETResponse baseLanguage="ES" creationDateTime="2016-11-08T17:10:09-03:00">
<INVAP-WS-ASSETSet>
    <ASSET rowstamp="265381811">
        <ASSETID>1529</ASSETID>
        <ASSETNUM>2503</ASSETNUM>
        <DESCRIPTION>POWER CHASSIS()</DESCRIPTION>
        <ITEMNUM>A000232</ITEMNUM>
        <LOCATION>LOCATIONTEST</LOCATION>
        <SERIALNUM>123456789</SERIALNUM>
        <SITEID>TVD</SITEID>
    </ASSET>
</INVAP-WS-ASSETSet>

可以想象，我要转换为Java对象的唯一数据是assetid，assetnum，描述，itemnum，位置，SerialNum和siteid。

到现在为止，这就是我所拥有的：

@Root(name = "QueryINVAP-WS-ASSETResponse", strict = false)
public class Activos {

    @ElementList(inline = true, entry = "ASSET", type = Activo.class)
    private List<Activo> activos;

    @Root
    class Activo {

        @Attribute(name = "rowstamp")
        public String rowstamp;

        @Element(name = "ASSETID")
        public Integer assetid;

        @Element(name = "ASSETNUM")
        public String assetnum;

        @Element(name = "DESCRIPTION")
        public String description;

        @Element(name = "LOCATION")
        public String location;

        @Element(name = "SERIALNUM")
        public String serialnum;

        @Element(name = "SITEID")
        public String siteid;
    }
}

我得到的错误是：

org.simpleframework.xml.core.ValueRequiredException: Unable to satisfy @org.simpleframework.xml.ElementList(data=false, empty=true, entry=ASSET, inline=true, name=, required=true, type=class invap.invapgestionmovil.modelos.Activos$Activo) on field 'activos' private java.util.List invap.invapgestionmovil.modelos.Activos.activos for class invap.invapgestionmovil.modelos.Activos at line 1

我究竟做错了什么？

推子

好吧，终于我解决了这个问题。我创建了三个类：

@Root(name = "QueryINVAP-WS-ASSETResponse", strict=false)
public class QueryINVAPWSASSETResponse {

    @Element(name = "INVAP-WS-ASSETSet")
    private INVAPWSASSETSet set;
}

@Root(name = "INVAP-WS-ASSETSet", strict=false)
public class INVAPWSASSETSet {

    @ElementList(inline=true, name = "ASSET")
    private List<Activo> activos;
}

@Root(name = "ASSET", strict = false)
public class Activo {

    @Attribute(name = "rowstamp")
    public String rowstamp;

    @Element(name = "ASSETID")
    public Integer assetid;

    @Element(name = "ASSETNUM")
    public String assetnum;

    @Element(name = "DESCRIPTION")
    public String description;

    @Element(name = "LOCATION")
    public String location;

    @Element(name = "SERIALNUM")
    public String serialnum;

    @Element(name = "SITEID")
    public String siteid;

}

现在，当我进行呼叫时，我将资产与其他两个类“包装在一起”。

希望这对其他人有帮助:)问候

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