Question

全部，我正在尝试生成xml输出，如 -

<Row>
    <City>Boston</City>
    <Runners>7000</Runners>
</Row>

xsd生成的类型是：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "RowType", propOrder = {
    "any"
})
public class RowType {

    @XmlAnyElement(lax = true)
    protected List<Object> any;
    public List<Object> getAny() {
        if (any == null) {
            any = new ArrayList<Object>();
        }
        return this.any;
    }

}

Unmarshalling工作正常，而编组则会产生如下输出。

<Row>
    <any>
        <name>CITY</name>
        <declaredType>java.lang.String</declaredType>
        <scope>javax.xml.bind.JAXBElement$GlobalScope</scope>
        <value>Boston</value>
        <nil>false</nil>
        <globalScope>true</globalScope>
        <typeSubstituted>false</typeSubstituted>
    </any>
    <any>
        <name>RUNNERS</name>
        <declaredType>java.lang.String</declaredType>
        <scope>javax.xml.bind.JAXBElement$GlobalScope</scope>
        <value>7000</value>
        <nil>false</nil>
        <globalScope>true</globalScope>
        <typeSubstituted>false</typeSubstituted>
    </any>
</Row>

知道如何使用Spring启动工作吗？

Answer 1

生成的POJO不对。应该是

<强> ROWTYPE：

public class RowType
{
    private Row Row;

    public Row getRow ()
    {
        return Row;
    }

    public void setRow (Row Row)
    {
        this.Row = Row;
    }

    @Override
    public String toString()
    {
        return "ClassPojo [Row = "+Row+"]";
    }
}

Row.java：

public class Row
{
    private String Runners;

    private String City;

    public String getRunners ()
    {
        return Runners;
    }

    public void setRunners (String Runners)
    {
        this.Runners = Runners;
    }

    public String getCity ()
    {
        return City;
    }

    public void setCity (String City)
    {
        this.City = City;
    }

    @Override
    public String toString()
    {
        return "ClassPojo [Runners = "+Runners+", City = "+City+"]";
    }
}

使用此工具进行交叉检查：http://pojo.sodhanalibrary.com/Convert

jaxb编组xs：任何在spring-boot中

1 个答案: