小编Sla*_*yev的帖子

我正在尝试使用xstream 1.4.2将xml转换为对象.它确实对我来说非常好,直到我将对象的类文件放在一个单独的包中,而不是主代码运行的位置.然后我得到一个CannotResolveClassException.我尝试过使用其他人推荐的setClassLoader方法,但这没有用.

Exception in thread "main" com.thoughtworks.xstream.mapper.CannotResolveClassException: result
at com.thoughtworks.xstream.mapper.DefaultMapper.realClass(DefaultMapper.java:56)
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 com.thoughtworks.xstream.core.TreeUnmarshaller.start(TreeUnmarshaller.java:133)
at com.thoughtworks.xstream.core.AbstractTreeMarshallingStrategy.unmarshal(AbstractTreeMarshallingStrategy.java:32)
at com.thoughtworks.xstream.XStream.unmarshal(XStream.java:1052)
at com.thoughtworks.xstream.XStream.unmarshal(XStream.java:1036)
at com.thoughtworks.xstream.XStream.fromXML(XStream.java:912)
at com.thoughtworks.xstream.XStream.fromXML(XStream.java:903)
at main.readClass(main.java:48)
at main.main(main.java:28)

答案: xstream期望xml结构相对于它(对象)所源自的包.因此必须使用xstream.alias才能为xml结构添加别名.

xstream.alias("something", Something.class);

否则xstream将期望"Something"在默认包中,而不是它所属的包.

我试图通过Linux上的exec调用运行ffmpeg.但是我必须在命令中使用引号(ffmpeg需要它).我一直在浏览java doc for processbuilder和exec以及stackoverflow上的问题,但我似乎无法找到解决方案.

我需要跑

ffmpeg -i "rtmp://127.0.0.1/vod/sample start=1500 stop=24000" -re -vcodec copy -acodec copy -f flv rtmp://127.0.0.1/live/qltv

我需要在下面的参数字符串中插入引号.注意,由于processbuilder解析和运行命令的性质,简单地添加单引号或双引号前面加一个反斜杠不起作用.

String argument = "ffmpeg -i rtmp://127.0.0.1/vod/"
                    + nextVideo.getFilename()
                    + " start=" + nextVideo.getStart()
                    + " stop=" + nextVideo.getStop()
                    + " -re -vcodec copy -acodec copy -f flv rtmp://127.0.0.1/live/qltv";

任何帮助将不胜感激.

当我对NOW()执行查询时,mysql返回的时间比服务器上当前时间大约-30秒.有任何想法吗？我试着查看配置文件但没有发现任何内容.我正在运行5.1.37版

SELECT NOW()