How to compile a sketch to Java

edited September 2014 in Using Processing

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).

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Answers

  • edited September 2014 Answer ✓

    "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

  • edited September 2014

    @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!

  • edited September 2014

    "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.

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