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satisfactory: implement the 'resources' query

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Danny Robson 2019-04-28 15:48:48 +10:00
parent 65eb55e409
commit 8cb5d5a76c
2 changed files with 7 additions and 14 deletions
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3
describe.py
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@ -9,5 +9,6 @@ if __name__ == '__main__':
recipe_root = os.path.join(root, 'data', 'recipes') recipe_root = os.path.join(root, 'data', 'recipes')
cookbook = satisfactory.Cookbook(recipe_root) cookbook = satisfactory.Cookbook(recipe_root)
print(i for i in cookbook.resources()) for i in cookbook.resources():
print(i)

18
satisfactory/__init__.py
View File

@ -26,19 +26,11 @@ class Cookbook(object):
def is_component(self, name): def is_component(self, name):
return 'component' in self.__recipes[name]['type'] return 'component' in self.__recipes[name]['type']
def is_resource(self, name):
return ''
def components(self): def components(self):
found = set() return (i for i in self.all() if self.is_component(i))
for _, descriptor in self.__recipes.items(): def is_resource(self, name):
for variation in descriptor['recipes']: return 'resource' in self.__recipes[name]['type']
for need, _ in variation['input'].items():
if need in found:
continue
found.add(need)
if not self.is_component(need):
continue
yield need
def resources(self):
return (i for i in self.all() if self.is_resource(i))