这是我的 Xml 输出
<worklist>
<work>
<relation-list target-type="artist">
<relation type="composer">
<name>Пётр Ильич Чайковский</name>
<sort-name>Пётр Ильич Чайковский</sort-name>
</relation>
<relation type="writer">
<name>Frank Turner/name>
<sort-name>Turner Frank</sort-name>
</relation>
</relation-list>
</work>
</worklist>
我正在尝试使用 EclipseLInk Moxy 获取此 json 输出
{
"work" : [ {
"relations" : [ {
"type" : "composer",
"name" : "Пётр Ильич Чайковский",
"sort-name" : "Пётр Ильич Чайковский"
}, {
"type" : "writer",
"name" : "frank turner",
"sort-name" : "turner, frank"
} ]
} ]
}
但我能做的最好的就是
{
"work" : [ {
"relationList" : [ {
"relations" : [ {
"type" : "composer",
"name" : "Пётр Ильич Чайковский",
"sort-name" : "Пётр Ильич Чайковский"
}, {
"type" : "writer",
"name" : "frank turner",
"sort-name" : "turner, frank"
} ]
} ],
} ]
}
注意我已经删除了relationList的目标类型属性,并且我已经将关系重命名为关系,但是如果我尝试使用将relationList与其父级合并
<java-type name="Work">
<java-attributes>
<xml-element java-attribute="relationList" xml-path="."/>
</java-attributes>
</java-type>
我收到以下异常
Caused by: java.lang.NullPointerException
at java.io.Writer.write(Writer.java:140)
at org.eclipse.persistence.oxm.record.JSONWriterRecord.writeKey(JSONWriterRecord.java:580)
at org.eclipse.persistence.oxm.record.JSONFormattedWriterRecord.openStartElement(JSONFormattedWriterRecord.java:147)
at org.eclipse.persistence.internal.oxm.XPathNode.startElement(XPathNode.java:359)
at org.eclipse.persistence.internal.oxm.XMLCompositeCollectionMappingNodeValue.marshalSingleValue(XMLCompositeCollectionMappingNodeValue.java:292)
at org.eclipse.persistence.internal.oxm.XMLCompositeCollectionMappingNodeValue.marshal(XMLCompositeCollectionMappingNodeValue.java:91)
at org.eclipse.persistence.internal.oxm.NodeValue.marshal(NodeValue.java:151)
at org.eclipse.persistence.internal.oxm.NodeValue.marshal(NodeValue.java:104)
at org.eclipse.persistence.internal.oxm.record.ObjectMarshalContext.marshal(ObjectMarshalContext.java:60)
at org.eclipse.persistence.internal.oxm.XPathNode.marshal(XPathNode.java:350)
at org.eclipse.persistence.internal.oxm.TreeObjectBuilder.buildRow(TreeObjectBuilder.java:467)
at org.eclipse.persistence.internal.oxm.XMLCompositeCollectionMappingNodeValue.marshalSingleValue(XMLCompositeCollectionMappingNodeValue.java:299)
at org.eclipse.persistence.internal.oxm.XMLCompositeCollectionMappingNodeValue.marshal(XMLCompositeCollectionMappingNodeValue.java:91)
at org.eclipse.persistence.internal.oxm.NodeValue.marshal(NodeValue.java:151)
at org.eclipse.persistence.internal.oxm.NodeValue.marshal(NodeValue.java:104)
at org.eclipse.persistence.internal.oxm.record.ObjectMarshalContext.marshal(ObjectMarshalContext.java:60)
at org.eclipse.persistence.internal.oxm.XPathNode.marshal(XPathNode.java:350)
at org.eclipse.persistence.internal.oxm.TreeObjectBuilder.buildRow(TreeObjectBuilder.java:467)
at org.eclipse.persistence.internal.oxm.XMLCompositeObjectMappingNodeValue.marshalSingleValue(XMLCompositeObjectMappingNodeValue.java:224)
at org.eclipse.persistence.internal.oxm.XMLCompositeObjectMappingNodeValue.marshal(XMLCompositeObjectMappingNodeValue.java:144)
at org.eclipse.persistence.internal.oxm.NodeValue.marshal(NodeValue.java:104)
at org.eclipse.persistence.internal.oxm.record.ObjectMarshalContext.marshal(ObjectMarshalContext.java:60)
at org.eclipse.persistence.internal.oxm.XPathNode.marshal(XPathNode.java:350)
at org.eclipse.persistence.internal.oxm.TreeObjectBuilder.buildRow(TreeObjectBuilder.java:467)
at org.eclipse.persistence.oxm.XMLMarshaller.marshal(XMLMarshaller.java:1074)
at org.eclipse.persistence.oxm.XMLMarshaller.marshal(XMLMarshaller.java:689)
at org.eclipse.persistence.oxm.XMLMarshaller.marshal(XMLMarshaller.java:602)
at org.eclipse.persistence.jaxb.JAXBMarshaller.marshal(JAXBMarshaller.java:584)
我想我明白这个问题了。因为 Relation-List 元素包含关系元素和目标类型属性,所以直接转换为 json 需要关系列表元素封装这两个元素,即我认为如果关系列表元素没有目标类型属性在模型中它工作没有问题。
但是我无论如何都不想输出目标类型,所以它应该可以工作,有办法解决这个问题吗?
编辑: 感谢您的回答,但它似乎与我正在做的事情相同,除了我已经尝试过的示例代码工作正常。我尝试向示例代码中添加更多元素来尝试破坏它,但没有成功。
我最初的问题显示了一个比现实更简单的情况,所以下面我实际上输出了完整的输出,以防有人可以看到可能导致问题的原因
{
"count" : 1,
"offset" : 0,
"work" : [ {
"id" : "4ff89cf0-86af-11de-90ed-001fc6f176ff",
"type" : "Opera",
"score" : "100",
"title" : "Symphony No. 5",
"language" : "eng",
"iswcs" : [ "T-101779304-1", "B-101779304-1" ],
"disambiguation" : "demo",
"aliases" : [ "Symp5" ],
"relation-list" : [ {
"relations" : [ {
"type" : "composer",
"direction" : "backward",
"attributes" : [ "additional" ],
"artist" : {
"id" : "1f9df192-a621-4f54-8850-2c5373b7eac9",
"name" : "Пётр Ильич Чайковский",
"sort-name" : "Пётр Ильич Чайковский"
}
}, {
"type" : "writer",
"direction" : "backward",
"artist" : {
"id" : "abcdefgh-a621-4f54-8850-2c5373b7eac9",
"name" : "frank",
"sort-name" : "turner"
}
} ]
} ],
"tags" : [ {
"count" : 10,
"name" : "classical"
} ]
} ]
}
工作正常,但添加
<java-type name="Work">
<java-attributes>
<xml-element java-attribute="relationList" xml-path="."/>
</java-attributes>
</java-type>
导致堆栈跟踪。
也许堆栈跟踪包含解决方案
更新
完整项目可在线获取
数据模型位于http://svn.musicbrainz.org/mmd-schema/trunk/brainz-mmd2-jaxb/ 下载并运行 mvn install
那么使用这个的项目在http://svn.musicbrainz.org/search_server/trunk/ 下载并运行mvn包,
这包含三个子项目,问题之一是 servlet 项目,复制该问题:
cd servlet
编辑 src/main/resources/oxml.xml 更改
<xml-element java-attribute="relationList" name="relationList"/>
到
<xml-element java-attribute="relationList" xml-path="."/>
mvn包
src/test/java/org/musicbrainz/search/servlet/FindWorkTest 现在将失败 testOutputAsJsonNew() 和 testOutputAsJsonNewPretty()
进一步更新 Blaise 解决了示例没有显示(而且我不知道)的问题,即您实际上可以有多个关系列表元素,即 Xml 可以是
<worklist>
<work>
<relation-list target-type="artist">
<relation type="composer">
<name>Jane Doe</name>
<sort-name>Doe Jane</sort-name>
</relation>
<relation type="writer">
<name>Frank Turner</name>
<sort-name>Turner Frank</sort-name>
</relation>
</relation-list>
<relation-list target-type="release">
<relation type="cover">
<name>Hey Jude</name>
<sort-name>Hey Jude</sort-name>
</relation>
</relation-list>
</work>
</worklist>
其中我想要输出的是
{
"work" : [ {
"relations" : [ {
"type" : "composer",
"name" : "Jane Doe",
"sort-name" : "Doe Jane"
}, {
"type" : "writer",
"name" : "Frank Turner",
"sort-name" : "Turner Frank"
}, {
"type" : "cover",
"name" : "Hey Jude",
"sort-name" : "Hey Jude"
} ]
} ]
}
而我能得到的最好的
{
"work" : [ {
"relationLists" : [ {
"relations" : [ {
"type" : "composer",
"name" : "Jane Doe",
"sort-name" : "Doe Jane"
}, {
"type" : "writer",
"name" : "Frank Turner",
"sort-name" : "Turner Frank"
} ]
}, {
"relations" : [ {
"type" : "cover",
"name" : "Hey Jude",
"sort-name" : "Hey Jude"
} ]
} ]
} ]
}
所以本质上我想合并所有关系,以便它们位于一个列表中并丢失关系列表标签。我们合并了不同的关系列表并丢弃了标识每个关系列表的目标类型属性,但我们不需要能够识别这些子列表,这可能看起来很奇怪。
可以这样做吗?
最佳答案
更新#2
根据调查结果,我更新了示例代码以使用 XmlAdapter
而不是 .
XPath。这种方法将更适合您的用例。
更新#1
从堆栈跟踪看来,您可能在集合属性上设置了 self(“.”)XPath,这是不允许的(我们将更改代码以便抛出更好的异常)。
在 Work
类中,您有一个名为 relationList
的集合属性
@XmlElement(name = "relation-list")
protected List<RelationList> relationList;
出现此问题是因为您在元数据文件中尝试将 xml-path
设置为 '.'
。
<xml-element java-attribute="relationList" xml-path="."/>
相反,您需要在引用保存集合的对象的属性上使用 '.
' 路径。请参阅下面的示例:
XML 适配器
您的用例将由 XmlAdapter
更好地处理,而不是使用 .
XPath。 XmlAdapter
允许我们将一种对象结构转换为另一种更好地映射到所需输入/输出的结构。对于此用例,XmlAdapter
看起来像这样。
package forum12338288;
import java.util.*;
import javax.xml.bind.annotation.adapters.XmlAdapter;
public class WorkAdapter extends XmlAdapter<WorkAdapter.AdaptedWork, Work> {
public static class AdaptedWork {
public List<Relation> relations = new ArrayList<Relation>();
}
@Override
public AdaptedWork marshal(Work work) throws Exception {
AdaptedWork adaptedWork = new AdaptedWork();
for(RelationList relationList : work.relationList) {
for(Relation relation : relationList.relation) {
adaptedWork.relations.add(relation);
}
}
return adaptedWork;
}
@Override
public Work unmarshal(AdaptedWork adaptedWork) throws Exception {
Work work = new Work();
RelationList relationList = new RelationList();
for(Relation relation : adaptedWork.relations) {
relationList.relation.add(relation);
}
work.relationList.add(relationList);
return work;
}
}
映射文档
您的映射文档应如下所示:
oxm.xml
<?xml version="1.0"?>
<xml-bindings
xmlns="http://www.eclipse.org/eclipselink/xsds/persistence/oxm"
package-name="forum12338288">
<java-types>
<java-type name="WorkList">
<java-attributes>
<xml-element java-attribute="work">
<xml-java-type-adapter value="forum12338288.WorkAdapter"/>
</xml-element>
</java-attributes>
</java-type>
<java-type name="RelationList">
<java-attributes>
<xml-transient java-attribute="targetType"/>
<xml-element java-attribute="relation" name="relations"/>
</java-attributes>
</java-type>
</java-types>
</xml-bindings>
演示代码
下面的演示代码演示了如何利用映射文档读取 XML 并输出所需的 JSON。
演示
package forum12338288;
import java.io.File;
import java.util.*;
import javax.xml.bind.*;
import org.eclipse.persistence.jaxb.JAXBContextProperties;
public class Demo {
public static void main(String[] args) throws Exception {
// XML
JAXBContext jc = JAXBContext.newInstance(WorkList.class);
Unmarshaller unmarshaller = jc.createUnmarshaller();
File xml = new File("src/forum12338288/input.xml");
WorkList workList = (WorkList) unmarshaller.unmarshal(xml);
// JSON
Map<String, Object> properties = new HashMap<String, Object>();
properties.put(JAXBContextProperties.OXM_METADATA_SOURCE, "forum12338288/oxm.xml");
properties.put(JAXBContextProperties.MEDIA_TYPE, "application/json");
properties.put(JAXBContextProperties.JSON_INCLUDE_ROOT, false);
JAXBContext jsonJC = JAXBContext.newInstance(new Class[] {WorkList.class}, properties);
Marshaller marshaller = jsonJC.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(workList, System.out);
}
}
input.xml
<worklist>
<work>
<relation-list target-type="artist">
<relation type="composer">
<name>Jane Doe</name>
<sort-name>Doe Jane</sort-name>
</relation>
<relation type="writer">
<name>Frank Turner</name>
<sort-name>Turner Frank</sort-name>
</relation>
</relation-list>
<relation-list target-type="release">
<relation type="cover">
<name>Hey Jude</name>
<sort-name>Hey Jude</sort-name>
</relation>
</relation-list>
</work>
</worklist>
输出
{
"work" : [ {
"relations" : [ {
"type" : "composer",
"name" : "Jane Doe",
"sort-name" : "Doe Jane"
}, {
"type" : "writer",
"name" : "Frank Turner",
"sort-name" : "Turner Frank"
}, {
"type" : "cover",
"name" : "Hey Jude",
"sort-name" : "Hey Jude"
} ]
} ]
}
JAVA模型
下面是我用来确保一切正常的 Java 模型。
工作列表
package forum12338288;
import java.util.List;
import javax.xml.bind.annotation.*;
@XmlRootElement(name="worklist")
@XmlAccessorType(XmlAccessType.FIELD)
public class WorkList {
List<Work> work;
}
工作
package forum12338288;
import java.util.*;
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.FIELD)
public class Work {
@XmlElement(name="relation-list")
List<RelationList> relationList = new ArrayList<RelationList>();
}
关系列表
package forum12338288;
import java.util.List;
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.FIELD)
public class RelationList {
@XmlAttribute(name="target-type")
String targetType;
List<Relation> relation;
}
关系
package forum12338288;
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.FIELD)
public class Relation {
@XmlAttribute
String type;
String name;
@XmlElement(name="sort-name")
String sortName;
}
jaxb.properties
要将 MOXy 指定为您的 JAXB 提供程序,您需要在与您的域模型相同的包中包含一个名为 jaxb.properties
的文件,其中包含以下条目(请参阅:http://blog.bdoughan.com/2011/05/specifying-eclipselink-moxy-as-your.html)。
javax.xml.bind.context.factory=org.eclipse.persistence.jaxb.JAXBContextFactory
关于java - 在 MOXy 中,如果元素包含子属性,则无法合并元素,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12338288/