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Cannot read resource as stream when running jar


Cannot read resource as stream when running jar

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Cannot read resource as stream when running jar
Tag : java , By : Nandor Devai
Date : December 01 2020, 04:50 PM

will be helpful for those in need I'm using NanoHTTPD as my app server and everything is working fine from IDE. When I'm running the same code form jar (which was generated with maven-assembly-plugin), the resources are not loaded. , Figured out the solution was:
InputStream in = this.getClass().getClassLoader().getResourceAsStream("web" + session.getUri());

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How to read local xml file is resource folder as a input stream in android?


Tag : android , By : dexteryy
Date : March 29 2020, 07:55 AM
seems to work fine Put the xml file into /res/raw folder. It looks like openRawResource opens resources from that folder only. You can also try getResources().getXml(com.MYCLass.R.xml.programs); which will return you an instance of XML parser.
A piece of code taken from @Krupa
InputStream object = this.getResources()
                     .openRawResource(R.raw.fileName);

Given an x264 stream and an ogg vorbis stream, how do I make a muxed stream that mplayer/VLC can read?


Tag : cpp , By : user109285
Date : March 29 2020, 07:55 AM
this one helps. x264 is not a stream format. It is a piece of software. This software encodes video to the H.264 video format. AFAIK, it does not mux video+audio into MP4 or AVI container files. Look into ffmpeg/libav for a full suite. There are other programs to mux video and audio streams.
Here's an experiment I performed:
youtube-dl "http://www.youtube.com/watch?v=0Bmhjf0rKe8"
avconv -i 0Bmhjf0rKe8.flv -vn -c:a libvorbis -b:a 64k 0Bmhjf0rKe8.ogg
avconv -i 0Bmhjf0rKe8.flv -c:v copy -bsf:v h264_mp4toannexb -an 0Bmhjf0rKe8.h264
avconv -i 0Bmhjf0rKe8.h264 -i 0Bmhjf0rKe8.ogg -c copy 0Bmhjf0rKe8.mkv
mplayer 0Bmhjf0rKe8.mkv
avconv -i 0Bmhjf0rKe8.flv -i 0Bmhjf0rKe8.ogg -c copy -map 0:0 -map 1:0 0Bmhjf0rKe8.mp4
mplayer 0Bmhjf0rKe8.mp4

Elixir: can I use Stream.resource to progressively read a large data file?


Tag : elixir , By : Tim Benninghoff
Date : March 29 2020, 07:55 AM
should help you out As a side note, it is a common task to create a stream from a file. This has been taken care of, so you can simply use File.stream!/3 to create the stream, no need to use Stream.resource/3 directly.
Regarding your original question: yes you are correct, Stream.chunk_every/2 is the way to go here. It will lazily split the stream into chunks of the supplied size:
File.stream!("./data/fidap011.mtx") |> Stream.chunk_every(5) |> Enum.each(fn chunk ->
  # do something with chunk
end)

how to read Stream[String] after running scala process


Tag : scala , By : Justin Bowers
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I followed this page http://www.scala-lang.org/api/current/index.html#scala.sys.process.package
res4: Stream[String] = Stream()
res4.foreach(println(_))

Prevent Spring from attempting to read old resource stream


Tag : java , By : hammer_1968
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Turns out there is a configuration for this, in an unexpected place.
In my WebMvcConfigurerAdapter-extending config class, in the addResourceHandlers method:
registry.addResourceHandler("**/pluginresource/**")
.setCacheControl(CacheControl.noStore() )
.resourceChain(false)
.addResolver(pluginResourceResolver);
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