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Home/ Questions/Q 6169651
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T22:53:35+00:00 2026-05-23T22:53:35+00:00

i am trying … loadRecipe(‘existingpackage’) class NewPackage(PackageRecipe): name = ‘newpackage-test’ p = existingpackage.version print

  • 0

i am trying …

loadRecipe('existingpackage')
class NewPackage(PackageRecipe):
    name = 'newpackage-test'
    p = existingpackage.version
    print p

but getting error, that existingpackage is not defined

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  1. Editorial Team
    Editorial Team
    2026-05-23T22:53:36+00:00Added an answer on May 23, 2026 at 10:53 pm

    You got it right that loadRecipe needs the name of the package. But to access information from the recipe, you should use the class defined there, not the package name or the recipe filename. (That’s also quite natural. Sometimes recipes can define more than one classes.)

    For example, in a firefox plugin, I want the version of firefox, so that the plugin can be installed to the right place.

    loadRecipe('firefox')
    class FirefoxPackageSearch(PackageRecipe):
    
        [snip]
    
        def setup(r):
            [snip]
            r.macros.ff_version = '.'.join(FireFox.version.split('.')[:2])
    

    I load the firefox recipe and use Firefox.version to get what I want.

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