Hollywood forums

if you have any problem, you should create a small, smallest possible code to demostrate problem here i have modified a bit some samples and yes i see that 8192 seems to be a hardcore limit, or we are doing something wrong the following test files are tested in Windows and as long as i send 819_3_ b...

this confuse me too a bit but i THINK tha logical operatiosn SHOULD be inside brackets

does

If (playing=False) Then

works as expected in your tests ?

i think you should test it wih a faster computer (PC ?) first to verify if there data are correct

additionally you may use a known to work tool, like tcpdump, in the faster PC to verify if this one receives correct data

these additional information will help you understand what is going on

Rename it to lumix_flux.mjpeg and use ffmpeg from aminet latest git version (dated 2012) thats what i ise
In amiga (also in pc works too)

is this part of image (a table ?) what is should display ? https://pasteboard.co/JwyJzzC.jpg , i used ffmpeg to decode images using ffmpeg -i lumix_flux -vcodec copy frame%d.jpg and cot above image15 times all frames are "cut" in 8192 bytes, and image is corrupted, so i guess you are get them with h...

isnt showcase (here) good enough ?

additional as logn as it has amiga version (no matter it may have win version too inside) aminet/hwood is a good place

which function for which part. wher have you stuck ? as said data$ = receiveudpdata(af) return data , the data there have to add them by yourself the following command add data data$ = data$..data$ but how much data consist a single jpeg image ?! uknown. ! According to the link you provide, this guy...

the above link says it simple enough, but you have to work it according to this guy camera send mjpeg, ffmpeg can play mjpeg so avcodec most probably can decode it(untested in hollywood but its the simplest out there) so you have to work out how each frame (a single jpeg image is the one frame in mj...

receiveudp data(af)

data$ = REceveiUDPDAta(id) unload all data (max 8kb) to data$ string, from there you have to know a bit more infromation on what this camera send for you have to know exact protocol camera send's in order to see if there an "easy" way in hollywood for example does your camera send a jpg image ?, doe...