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java - 使用 Jaxb 库 Java 从 xml 文件返回 map

转载 作者:行者123 更新时间:2023-12-04 08:25:20 25 4
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我正在尝试直接从 XML 文件返回 map 。我已经通过创建自定义适配器进行了尝试,但问题是该适配器没有从 XML 接收带有值的输入参数。
对于引用,这里是问题的屏幕截图。
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这是我试图解析的 XML

<?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 的。
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");
}
}

最佳答案

我不知道是否可以在 xs:sequence 上使用 XmlAdapter。我确实设法在 Entry 上将 XmlAdapter 设置了一级元素:

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;
}
}
使用的测试 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>

关于java - 使用 Jaxb 库 Java 从 xml 文件返回 map ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/65288722/

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