Tag: xstream

将XML转换为Java Map

我正在尝试将XML转换为Java代码。 此XML位于不同的文件中; 它匹配单词与数字（概率分布），看起来像这样： 1 2 2 3 2 1 5 1 1 22 2 26 ……. 我正在尝试将其转换为Java Map，这是我正在使用的代码： XStream xstream = new XStream(); @SuppressWarnings(“unchecked”) Map englishCorpusProbDist = (Map)xstream.fromXML(new File(“LocationOfFileOnMyComputer/frequencies.xml”)); 目前，每当我尝试运行上述Java代码时，我在控制台中都会遇到以下exception： Exception in thread “main” com.thoughtworks.xstream.mapper.CannotResolveClassException: root at com.thoughtworks.xstream.mapper.DefaultMapper.realClass(DefaultMapper.java:79) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.DynamicProxyMapper.realClass(DynamicProxyMapper.java:55) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.PackageAliasingMapper.realClass(PackageAliasingMapper.java:88) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.ClassAliasingMapper.realClass(ClassAliasingMapper.java:79) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at […]

com.thoughtworks.xstream.converters.ConversionException

[EDITED] 我正在研究的项目是Java J2EE中的一个3文件夹项目，其中包含servlet和Hibernate的持久性。 结构如下： – Admin – >主程序用bean和HTML / CSS – Jar – >用jar，Hibernate工具和类 – War – >用Servlets 在它们之间，我使用Xstream来分享类和重要信息。 我正在使用Eclipse和Tomcat 7。 希望有了这一切，你们都能得到全球的想法。 这就是Xstream调试器所说的： Caused by: com.thoughtworks.xstream.converters.ConversionException: satdata.musicoterapia.hibernate.Terapeuta0 : satdata.musicoterapia.hibernate.Terapeuta0 —- Debugging information —- message : satdata.musicoterapia.hibernate.Terapeuta0 cause-exception : com.thoughtworks.xstream.mapper.CannotResolveClassException cause-message : satdata.musicoterapia.hibernate.Terapeuta0 class : satdata.musicoterapia.hibernate.Usuario required-type : satdata.musicoterapia.hibernate.Usuario converter-type : com.thoughtworks.xstream.converters.reflection.ReflectionConverter path : /list/Usuario[2]/terapeuta class[1] […]

XStream中的单个元素数组错误

如果你有这样的function： List getUsers() {} 如果getUsers返回只包含一个元素的List ，则生成的JSON只是一个JSON对象而不是JSON数组。 是否有一种解决方法使XStream返回JSON数组，无论该函数是否返回单个数组List？

XStream序列化空值

假设我有 class Student { String name; int age; String teacher; } 然后 ： public class App1 { public static void main(String[] args) { Student st = new Student(); st.setName(“toto”); XStream xs = new XStream(); xs.alias(“student”,Student.class); System.out.println(xs.toXML(st)); } } 给我 ： toto 0 有没有办法处理空值？ 我的意思是 ： toto 0 如果我这样做可能 st.setTeacher(“”); 但如果老师是空的，那就没有。 我尝试使用自定义转换器，但似乎空值不会发送到转换器。

将XML String转换为Map并使用Java获取键和值对

我有一个XML字符串。 我正在尝试将该字符串转换为映射，以便我可以获得键和值。 但是它无法转换。 这是我的代码 String xmlString = ” ” def convertStringToDocument = { xmlString -> DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); DocumentBuilder builder; try { builder = factory.newDocumentBuilder(); org.w3c.dom.Document doc = builder.parse(new InputSource(new StringReader(xmlString))); return doc; } catch (Exception e) { e.printStackTrace(); } return null; } def populateDocProofsFromWaiversXML = { xmlString, mandateFlag -> final List documentProofs = […]

从xstream反序列化xml文件

我正在使用Xstream来序列化Job对象。 它看起来很好。 但反序列化，我有一个问题： Exception in thread “main” com.thoughtworks.xstream.io.StreamException: : only whitespace content allowed before start tag and not . (position: START_DOCUMENT seen …. @1:1) at com.thoughtworks.xstream.io.xml.XppReader.pullNextEvent(XppReader.java:78) at com.thoughtworks.xstream.io.xml.AbstractPullReader.readRealEvent(AbstractPullReader.java:137) at com.thoughtworks.xstream.io.xml.AbstractPullReader.readEvent(AbstractPullReader.java:130) at com.thoughtworks.xstream.io.xml.AbstractPullReader.move(AbstractPullReader.java:109) at com.thoughtworks.xstream.io.xml.AbstractPullReader.moveDown(AbstractPullReader.java:94) at com.thoughtworks.xstream.io.xml.XppReader.(XppReader.java:48) at com.thoughtworks.xstream.io.xml.XppDriver.createReader(XppDriver.java:44) at com.thoughtworks.xstream.XStream.fromXML(XStream.java:853) at com.thoughtworks.xstream.XStream.fromXML(XStream.java:845) 你们其中一个人之前遇到过这个问题吗？ 这是我为序列化做的方式： XStream xstream = new XStream(); Writer writer = new FileWriter(new File(“model.xml”)); […]

将xml作为字符串而不是使用xstream的类

我有类似的xml 15 我不需要在父对象中创建消息对象，而是将消息表示为String。 所以，当我做parent.message时，输出是 15 而不是消息对象。

com.thoughtworks.xstream.mapper.CannotResolveClassException

这是我尝试XStream的第一次。 但是当我尝试解析我的xml文件时，我得到了这个exception： Exception in thread “main” com.thoughtworks.xstream.mapper.CannotResolveClassException: root at com.thoughtworks.xstream.mapper.DefaultMapper.realClass(DefaultMapper.java:79) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.DynamicProxyMapper.realClass(DynamicProxyMapper.java:55) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.PackageAliasingMapper.realClass(PackageAliasingMapper.java:88) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.ClassAliasingMapper.realClass(ClassAliasingMapper.java:79) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.ArrayMapper.realClass(ArrayMapper.java:74) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) at com.thoughtworks.xstream.mapper.CachingMapper.realClass(CachingMapper.java:45) at com.thoughtworks.xstream.core.util.HierarchicalStreams.readClassType(HierarchicalStreams.java:29) at […]

Xstream no-args构造函数错误

我试图创建一个’Transacao’实例时出现以下错误 `Error: Cannot construct br.com.cbmp.ecommerce.resposta.Transacao as it does not have a no-args constructor : Cannot construct br.com.cbmp.ecommerce.resposta.Transacao as it does not have a no-args constructor ` —- Debugging information —- message : Cannot construct br.com.cbmp.ecommerce.resposta.Transacao as it does not have a no-args constructor cause-exception : com.thoughtworks.xstream.converters.reflection.ObjectAccessException cause-message : Cannot construct br.com.cbmp.ecommerce.resposta.Transacao as it does not […]

XStream短动态别名

我希望有类的短名称，现在我可以使用别名 XStream x = new XStream(); x.alias(“dic”, Dic.class); 但我必须手动为每个类定义别名，有没有办法配置xstream自动执行？