本文介绍了使用Jaxb库Java从xml文件返回映射的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试直接从XML文件返回地图.我已经通过创建自定义适配器进行了尝试,但是问题是此适配器未从XML接收带有值的输入参数.
I am trying to return a map directly from an XML file. I've tried it by creating a custom Adapter but the problem is this adapter is not receiving Input Parameter with values from XML.
对于引用,这是问题的屏幕截图.
For Referencing, here are the screenshots of the problem.
这是我要解析的XML
Here is the XML which I am trying to parse
<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<GroupRoot>
<Root>
<Group>
<UUID>ahsdlfkjadslkfjalkdsjflakjdslf</UUID>
<Name>Root Group 1</Name>
<Entry>
<UUID>1212135454==</UUID>
<String>
<Key>Notes</Key>
<Value>Notes from Manager</Value>
</String>
<String>
<Key>Item1</Key>
<Value>Item1Value</Value>
</String>
<String>
<Key>Item2</Key>
<Value>Item2Value</Value>
</String>
<String>
<Key>Item3</Key>
<Value>Item3Value</Value>
</String>
<String>
<Key>Item4</Key>
<Value>Item4Value</Value>
</String>
</Entry>
<Entry>
<UUID>45645466546546464==</UUID>
<String>
<Key>Notes</Key>
<Value>Notes from Manager</Value>
</String>
<String>
<Key>Item1</Key>
<Value>Item1Value1</Value>
</String>
<String>
<Key>Item2</Key>
<Value>Item2Value1</Value>
</String>
<String>
<Key>Item3</Key>
<Value>Item3Value1</Value>
</String>
<String>
<Key>Item4</Key>
<Value>Item4Value1</Value>
</String>
</Entry>
</Group>
</Root>
</GroupRoot>
&这是我的Pojo.
& Here is my Pojo's.
package com.parser.xml.model;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "GroupRoot")
public class KeePassFile {
@XmlElement(name = "Root")
public Root root;
}
package com.parser.xml.model;
import java.util.List;
import javax.xml.bind.annotation.XmlElement;
public class Root {
@XmlElement(name = "Group")
public List<Group> groups;
}
package com.parser.xml.model;
import java.util.List;
import javax.xml.bind.annotation.XmlElement;
public class Group {
@XmlElement(name = "Name")
public String name;
@XmlElement(name = "Group", nillable = false )
public List<Group> subGroups;
@XmlElement(name = "Entry",nillable = false)
public List<Entry> entries;
}
package com.parser.xml.model;
import java.util.HashMap;
import java.util.Map;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.adapters.XmlAdapter;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
public class Entry {
@XmlElement(name = "UUID")
public String UUID;
@XmlElement(name = "String")
@XmlJavaTypeAdapter(StringMapperAdapter.class)
public Map<String, String> MappedItems;
}
class MapperString {
@XmlElement(name = "Key")
public String key;
@XmlElement(name = "Value")
public String value;
public MapperString() {
}
public MapperString(String key, String value) {
this.key = key;
this.value = value;
}
}
class StringMapperAdapter extends XmlAdapter<MapperString[], Map<String, String>> {
@Override
public Map<String, String> unmarshal(MapperString[] v) throws Exception {
HashMap<String, String> map = new HashMap<String, String>();
for (MapperString element : v) {
map.put(element.key, element.value);
}
return map;
}
@Override
public MapperString[] marshal(Map<String, String> v) throws Exception {
// TODO Auto-generated method stub
return null;
}
}
主要方法
package com.parser.xml;
import java.io.File;
import java.io.IOException;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Unmarshaller;
import org.springframework.core.io.ClassPathResource;
import com.parser.xml.model.KeePassFile;
public class RunAppMain {
private static final String FILE_PATH = "TestXML.xml";
public static void main(String[] Args) throws IOException, JAXBException
{
File resource = new ClassPathResource(FILE_PATH).getFile();
JAXBContext context = JAXBContext.newInstance(KeePassFile.class);
Unmarshaller un = context.createUnmarshaller();
KeePassFile entry = (KeePassFile) un.unmarshal(resource);
System.out.println("Testing");
}
}
推荐答案
我不知道是否可以在xs:sequence上使用XmlAdapter.我确实设法在Entry
元素上将XmlAdapter设置为一个级别:
I don't know if it's possible to use an XmlAdapter on an xs:sequence. I did manage to set the XmlAdapter one level up on the Entry
element:
import java.io.File;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.adapters.XmlAdapter;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
public class Test {
public static void main(String[] args) throws Exception {
Root root = (Root) JAXBContext.newInstance(Root.class).createUnmarshaller().unmarshal(new File("test.xml"));
System.out.println("" + root.entryList.get(0).map);
System.out.println("" + root.entryList.get(1).map);
}
}
@XmlRootElement(name = "Root")
class Root {
@XmlElement(name = "Entry")
public List<Entry> entryList = new ArrayList<>();
}
@XmlJavaTypeAdapter(EntryAdapter.class)
class Entry {
public String uuid;
public Map<String, String> map = new TreeMap<>();
}
class EntryMappableByJaxb {
@XmlElement
public String UUID;
@XmlElement(name = "String")
public List<KeyValue> stringList = new ArrayList<>();
}
class KeyValue {
@XmlElement
public String Key;
@XmlElement
public String Value;
}
class EntryAdapter extends XmlAdapter<EntryMappableByJaxb, Entry> {
@Override
public Entry unmarshal(EntryMappableByJaxb entryMappableByJaxb) throws Exception {
Entry entry = new Entry();
entry.uuid = entryMappableByJaxb.UUID;
entryMappableByJaxb.stringList.forEach(i -> entry.map.put(i.Key, i.Value));
return entry;
}
@Override
public EntryMappableByJaxb marshal(Entry v) throws Exception {
return null;
}
}
测试使用的XML:
<Root>
<Entry>
<UUID>uuid 1</UUID>
<String>
<Key>1.1</Key>
<Value>value 1.1</Value>
</String>
<String>
<Key>1.2</Key>
<Value>value 1.2</Value>
</String>
</Entry>
<Entry>
<UUID>uuid 2</UUID>
<String>
<Key>2.1</Key>
<Value>value 2.1</Value>
</String>
<String>
<Key>2.2</Key>
<Value>value 2.2</Value>
</String>
</Entry>
</Root>
这篇关于使用Jaxb库Java从xml文件返回映射的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!