Question

我有一个以下格式的xml文件，作为服务的响应。它不是传统的xml格式，其中值包含在相应的标记内。它只是一个示例，而实际文件将有数百个元素。如何以最有效的方式到达我需要的节点（让我们说“TDE-2＆＃39;”）并将其值放在{map(TenderID, TDE-2), map(ContactID, null)}

这样的地图中

<xml version="1.0" encoding="UTF-8"?>
<report>
<report_header>
<c1>TenderID</c1>
<c2>ContactID</c2>
<c3>Address</c3>
<c4>Description</c4>
<c5>Date</c5>
</report_header>
<report_row>
<c1>TDE-1</c1>
<c2></c2>
<c3></c3>
<c4>Tender 1</c4>
<c5>09/30/2016</c5>
</report_row>
<report_row>
<c1>TDE-2</c1>
<c2></c2>
<c3></c3>
<c4>Tender 2</c4>
<c5>10/02/2016</c5>
</report_row>
</report>

Answer 1

JAXB允许您将XML反序列化为Java对象。如果您创建Java POJO以匹配XML文档模型，则可以使用JAXB在POJO中解组XML。

例如：

的POJO：

<强> Report.java ：

import java.util.List;

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

@XmlRootElement
public class Report {

    private List<ReportRow> reportRows;

    public List<ReportRow> getReportRows() {
        return reportRows;
    }

    @XmlElement(name = "report_row")
    public void setReportRows(List<ReportRow> reportRows) {
        this.reportRows = reportRows;
    }
}

<强> ReportRow.java

import javax.xml.bind.annotation.XmlElement;

public class ReportRow {

private String c1;
private String c2;
private String c3;
private String c4;

public String getC1() {
    return c1;
}

@XmlElement
public void setC1(String c1) {
    this.c1 = c1;
}

public String getC2() {
    return c2;
}

@XmlElement
public void setC2(String c2) {
    this.c2 = c2;
}

public String getC3() {
    return c3;
}

@XmlElement
public void setC3(String c3) {
    this.c3 = c3;
}

public String getC4() {
    return c4;
}

@XmlElement
public void setC4(String c4) {
    this.c4 = c4;
}

}

读取XML并将其绑定到java对象的代码：

import java.io.File;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Unmarshaller;

import org.junit.Test;

public class JaxbTest {

    @Test
    public void testFoo() throws JAXBException {

        File xmlFile = new File("src/test/resources/reports.xml");
        JAXBContext context = JAXBContext.newInstance(Report.class, ReportRow.class);
        Unmarshaller jaxbUnmarshaller = context.createUnmarshaller();
        Report report = (Report) jaxbUnmarshaller.unmarshal(xmlFile);
        ReportRow reportYouWant = report.getReportRows().stream().filter(reportRow -> reportRow.getC1().equals("TDE-1"))
                .findFirst().get();

    }
}

您还需要将以下依赖项添加到构建脚本中：

compile group: 'javax.xml', name: 'jaxb-impl', version: '2.1'
compile group: 'javax.xml', name: 'jaxb-api', version: '2.1'

在Java

1 个答案: