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I tried the following command:
processing-java --sketch=mysketch --output=zzout
The program reports:
mysketch does not exist
but I am in the directory containing the directory mysketch, and it is definitely there (and the output directory does not exist as it claims it should not).
Answers
CTRL+E
"I am in the directory containing the directory mysketch"
The --sketch option specifies the sketch folder. So if you are already in the directory, it would search mysketch/mysketch...
Type processing-java alone to get a concise help.
The paths must be absolute:
C:\Java\processing-2.2.1\processing-java.exe --sketch=H:\PhiLhoSoft\Processing_QuickExperiments_Java\ArrayCopy --output=H:\Temp\PE --force --build
@Philho you are right about the absolute paths, though implying that the executable has to be an absolute path is wrong. I can, of course, run processing-java if it's in my path, but on windows it defaults to its own directory for the output.
As I stated, I was in the parent directory containing the folder, but the relative path wasn't doing it.
Thanks!
"implying that the executable has to be an absolute path is wrong"
It wasn't my intent, I just copied the command line I used for experiment. I don't put this exe in my path, even less as I have several versions of Processing side by side...
And yes, even though I quoted it (via drag'n'drop, my usual way...), I misread your information, sorry for that.