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java - JAXB @XmlAdapter : Map -> List adapter?(仅限编码)

转载 作者:塔克拉玛干 更新时间:2023-11-03 03:45:09 31 4
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我有一个 Map<String, String> .
每个人的第一个想法是将其转换为 List<Pair<String,String>> (Pair 是自定义类)。

我试过 @XmlAdapter像这样:

public class MapPropertiesAdapter extends XmlAdapter<List<Property>, Map<String,String>> { ... }

但是 Eclipse MOXy,我使用的 JAXB impl,以 ClassCastException 结束。 - “无法将 HashMap 转换为 Collection”。

JAXB 支持这种转换吗?还是我忽略了一些解释为什么不是这样的文档部分?

附言:我想得到这样的 XML:

<properties>
<property name="protocol"/>
<property name="marshaller"/>
<property name="unmarshaller"/>
<property name="timeout"/>
...
</properties>

我明白了,只需要使用中级类(class)。也描述于 Handle NPE in XMLCompositeObjectMappingNodeValue.marshalSingleValue( XMLCompositeObjectMappingNodeValue.java:161)

最佳答案

与其将 Map 适配为 List,不如将其适配为具有 List 属性的对象。

XmlAdapter (MapPropertiesAdapter)

import java.util.*;
import java.util.Map.Entry;
import javax.xml.bind.annotation.*;
import javax.xml.bind.annotation.adapters.XmlAdapter;

public class MapPropertiesAdapter extends XmlAdapter<MapPropertiesAdapter.AdaptedProperties, Map<String, String>>{

public static class AdaptedProperties {
public List<Property> property = new ArrayList<Property>();
}

public static class Property {
@XmlAttribute
public String name;

@XmlValue
public String value;
}

@Override
public Map<String, String> unmarshal(AdaptedProperties adaptedProperties) throws Exception {
if(null == adaptedProperties) {
return null;
}
Map<String, String> map = new HashMap<String, String>(adaptedProperties.property.size());
for(Property property : adaptedProperties.property) {
map.put(property.name, property.value);
}
return map;
}

@Override
public AdaptedProperties marshal(Map<String, String> map) throws Exception {
if(null == map) {
return null;
}
AdaptedProperties adaptedProperties = new AdaptedProperties();
for(Entry<String,String> entry : map.entrySet()) {
Property property = new Property();
property.name = entry.getKey();
property.value = entry.getValue();
adaptedProperties.property.add(property);
}
return adaptedProperties;
}

}

域模型(根)

下面是一个带有 Map 属性的模型对象。 @XmlJavaTypeAdapter 注释用于指定 XmlAdapter

import java.util.Map;
import javax.xml.bind.annotation.*;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Root {

@XmlJavaTypeAdapter(MapPropertiesAdapter.class)
private Map<String, String> properties;

}

演示

import java.io.File;
import javax.xml.bind.*;

public class Demo {

public static void main(String[] args) throws Exception {
JAXBContext jc = JAXBContext.newInstance(Root.class);

Unmarshaller unmarshaller = jc.createUnmarshaller();
File xml = new File("src/forum17024050/input.xml");
Root root = (Root) unmarshaller.unmarshal(xml);

Marshaller marshaller = jc.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(root, System.out);
}

}

input.xml/输出

<?xml version="1.0" encoding="UTF-8"?>
<root>
<properties>
<property name="A">a</property>
<property name="B">b</property>
</properties>
</root>

关于java - JAXB @XmlAdapter : Map -> List adapter?(仅限编码),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17024050/

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