验证 List(Jaxb) 中未编组的条目
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【中文标题】验证 List(Jaxb) 中未编组的条目【英文标题】：Verifying unmarshalled entries from List( Jaxb) 【发布时间】：2021-11-05 06:28:57 【问题描述】：我有 4 个 XML 文件，其结构如下：
<?xml version="1.0"?>
<Registry>
<Insitutions>
<Insitiution>
<name>A</name>
<location>B</location>
</Insitiution>
<Insitiution>
<name>C</name>
<location>D</location>
</Insitiution>
<Insitiution>
<name>E</name>
<location>F</location>
</Insitiution>
</Insitutions>
</Registry>
我已经为各个 XML 标记编写了类，并使用 jaxb 将它们解组。 我解组所有文件并将所有条目存储在一个列表中，如下所示：
private static List<String> loadBaseNodesets() throws JAXBException
log.info("Loading nodesets");
Registry xmlentries = null;
JAXBContext jaxbContext = null;
List<Registry> entries = new ArrayList<>();
List<String> privateBaseNodeSets= new ArrayList<>();
File dir = new File("XMLFiles");
if (dir.exists() && dir.isDirectory())
FileFilter filter = new FileFilter()
public boolean accept(File file)
return file.isFile() && file.getName().endsWith(".xml");
;
log.info("iterating through all files");
File [] files = dir.listFiles(filter);
if (files != null)
for (int i =0;i <1;i++)
jaxbContext = JAXBContext.newInstance(Registry.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
xmlentries = (Registry) jaxbUnmarshaller.unmarshal(files[i]);
privateBaseNodeSets.add(xmlentries.toString());
log.info(files[i].getName()+" NodeSet loaded");
entries.add(xmlentries);
log.info("Nodeset loading complete");
return privateBaseNodeSets;
我需要将所有 xmlentries 放到一个字符串列表中。
现在从主程序中，我想检查是否可以获得相同格式的 XML 条目。
public static void main(String[] args) throws JAXBException
log.info("Started local test main");
List<String> baseNodesets = loadBaseNodesets();
ListIterator<String> litr = baseNodesets.listIterator();
while (litr.hasNext())
JAXBContext jaxbContext=JAXBContext.newInstance(Registry.class);
Marshaller marshaller = jaxbContext.createMarshaller();#
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(, System.out);// DONT KNOW THE FIRST PARAMETER
我无法在此处获取第一个参数。有人可以帮忙吗？
【问题讨论】：
答案对您有帮助吗？能否解决问题？ 当然做到了！抱歉有点耽误了。再次感谢:) 【参考方案1】：您的方法marshal 需要一个java 对象作为第一个参数转换为XML。
这会将您的条目列表转换为 xml 列表。
final JAXBContext jaxbContext=JAXBContext.newInstance(Registry.class);
final Marshaller marshaller = jaxbContext.createMarshaller();
List<String> xmlEntries = entries.stream().map(entrie ->
OutputStream os = new ByteArrayOutputStream();
marshaller.marshal(entrie, os); <-- your object to transform to XML
return os.toString();
.collect(Collectors.toList());
【讨论】：
【参考方案2】：我不确定我是否正确理解了你的问题，但我猜你正在尝试做这样的事情：
-
逐个读取每个文件并解组并存储在
entries列表中。
然后循环遍历entries List 并一一编组。
将转换后的 XML 存储到 List 中
我在一个文件夹中存储了 2 个相同的 XML 文件并尝试读取它
XML 文件：
<?xml version="1.0"?>
<Registry>
<Insitutions>
<Insitiution>
<name>A</name>
<location>B</location>
</Insitiution>
<Insitiution>
<name>C</name>
<location>D</location>
</Insitiution>
<Insitiution>
<name>E</name>
<location>F</location>
</Insitiution>
</Insitutions>
</Registry>
根：
@Data
@XmlRootElement(name = "Registry")
@XmlAccessorType(XmlAccessType.FIELD)
public class Root
@XmlElementWrapper(name = "Insitutions")
@XmlElement(name = "Insitiution")
private List<Insitiution> Insitiution;
机构：
@Data
@XmlAccessorType(XmlAccessType.FIELD)
public class Insitiution
private String name;
private String location;
编组（主要）：
public class Marshalling
public static void main(String[] args) throws JAXBException, XMLStreamException, FactoryConfigurationError, IOException, SAXException
final File dir = new File("/Users/Downloads/test");
final List<Root> entries = new ArrayList<>();
final Unmarshaller unmarshaller = JAXBContext.newInstance(com.newJAXB.Root.class).createUnmarshaller();
final StringWriter singleXmlEvent = new StringWriter();
final List<String> xmlOutput = new ArrayList<>();
final Marshaller marshaller = JAXBContext.newInstance(Root.class).createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
for (File file : dir.listFiles())
if (!file.isDirectory() && file.getName().endsWith(".xml"))
//Storing all the files into entries List
final DataInputStream inputStream = new DataInputStream(new FileInputStream(file.getAbsolutePath()));
final XMLStreamReader xmlStreamReader = XMLInputFactory.newInstance().createXMLStreamReader(inputStream);
final com.newJAXB.Root root = unmarshaller.unmarshal(xmlStreamReader, com.newJAXB.Root.class).getValue();
entries.add(root);
System.out.println(root.toString());
//Loop through each entry in entries and marshal it
for (Root item : entries)
marshaller.marshal(item, singleXmlEvent);
xmlOutput.add(singleXmlEvent.toString());
// Clear the StringWriter for next event
singleXmlEvent.getBuffer().setLength(0);
//Final all the XML
System.out.println(xmlOutput);
输出：
Root(Insitiution=[Insitiution(name=A, location=B), Insitiution(name=C, location=D), Insitiution(name=E, location=F)])
Root(Insitiution=[Insitiution(name=A, location=B), Insitiution(name=C, location=D), Insitiution(name=E, location=F)])
[<?xml version="1.0" encoding="UTF-8"?>
<Registry>
<Insitutions>
<Insitiution>
<name>A</name>
<location>B</location>
</Insitiution>
<Insitiution>
<name>C</name>
<location>D</location>
</Insitiution>
<Insitiution>
<name>E</name>
<location>F</location>
</Insitiution>
</Insitutions>
</Registry>
, <?xml version="1.0" encoding="UTF-8"?>
<Registry>
<Insitutions>
<Insitiution>
<name>A</name>
<location>B</location>
</Insitiution>
<Insitiution>
<name>C</name>
<location>D</location>
</Insitiution>
<Insitiution>
<name>E</name>
<location>F</location>
</Insitiution>
</Insitutions>
</Registry>
]
【讨论】：
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