NameError: name 'Recipe' is not defined
I'm trying to run a script in django using the following command:
python manage.py shell < recipes/launch_bundle.py
The script is as follows:
from recipes.models import Recipe
def create_bundle(recipe_id):
rec = Recipe.objects.get(pk=recipe_id)
print('done')
create_bundle(1)
Notice the first line- the model Recipe should be imported. When run, i get the error:
Traceback (most recent call last):
File "manage.py", line 15, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 381, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 375, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 316, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 353, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/commands/shell.py", line 92, in handle
exec(sys.stdin.read())
File "<string>", line 7, in <module>
File "<string>", line 4, in create_bundle
NameError: name 'Recipe' is not defined
I am confused because I clearly import Recipe, and further because i can launch a shell with python manage.py shell, then copy paste the code and it works fine.
What is the issue here? Why does my code run fine in the shell, but not when called as a script?
Please note i have researched other questions, but they all seem to say: you havent imported the model, or something similar
django
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
add a comment |
I'm trying to run a script in django using the following command:
python manage.py shell < recipes/launch_bundle.py
The script is as follows:
from recipes.models import Recipe
def create_bundle(recipe_id):
rec = Recipe.objects.get(pk=recipe_id)
print('done')
create_bundle(1)
Notice the first line- the model Recipe should be imported. When run, i get the error:
Traceback (most recent call last):
File "manage.py", line 15, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 381, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 375, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 316, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 353, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/commands/shell.py", line 92, in handle
exec(sys.stdin.read())
File "<string>", line 7, in <module>
File "<string>", line 4, in create_bundle
NameError: name 'Recipe' is not defined
I am confused because I clearly import Recipe, and further because i can launch a shell with python manage.py shell, then copy paste the code and it works fine.
What is the issue here? Why does my code run fine in the shell, but not when called as a script?
Please note i have researched other questions, but they all seem to say: you havent imported the model, or something similar
django
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34
add a comment |
I'm trying to run a script in django using the following command:
python manage.py shell < recipes/launch_bundle.py
The script is as follows:
from recipes.models import Recipe
def create_bundle(recipe_id):
rec = Recipe.objects.get(pk=recipe_id)
print('done')
create_bundle(1)
Notice the first line- the model Recipe should be imported. When run, i get the error:
Traceback (most recent call last):
File "manage.py", line 15, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 381, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 375, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 316, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 353, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/commands/shell.py", line 92, in handle
exec(sys.stdin.read())
File "<string>", line 7, in <module>
File "<string>", line 4, in create_bundle
NameError: name 'Recipe' is not defined
I am confused because I clearly import Recipe, and further because i can launch a shell with python manage.py shell, then copy paste the code and it works fine.
What is the issue here? Why does my code run fine in the shell, but not when called as a script?
Please note i have researched other questions, but they all seem to say: you havent imported the model, or something similar
django
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
I'm trying to run a script in django using the following command:
python manage.py shell < recipes/launch_bundle.py
The script is as follows:
from recipes.models import Recipe
def create_bundle(recipe_id):
rec = Recipe.objects.get(pk=recipe_id)
print('done')
create_bundle(1)
Notice the first line- the model Recipe should be imported. When run, i get the error:
Traceback (most recent call last):
File "manage.py", line 15, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 381, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python3.6/site-packages/django/core/management/__init__.py", line 375, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 316, in run_from_argv
self.execute(*args, **cmd_options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/base.py", line 353, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python3.6/site-packages/django/core/management/commands/shell.py", line 92, in handle
exec(sys.stdin.read())
File "<string>", line 7, in <module>
File "<string>", line 4, in create_bundle
NameError: name 'Recipe' is not defined
I am confused because I clearly import Recipe, and further because i can launch a shell with python manage.py shell, then copy paste the code and it works fine.
What is the issue here? Why does my code run fine in the shell, but not when called as a script?
Please note i have researched other questions, but they all seem to say: you havent imported the model, or something similar
django
django
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
asked Dec 28 '18 at 14:47
User632716User632716
2,45211234
2,45211234
possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34
add a comment |
possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34
possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34
possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34
add a comment |
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possible duplicate with this (stackoverflow.com/questions/44728109/…)
– seuling
Dec 28 '18 at 15:34