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# python
s
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r
The snippet that you’ve shown here should work as written. What about it doesn’t work?
My code is a bit more complicated with some concatenations between Outputs and str
I imagine this is the part that’s actually going wrong, so we’ll need an actual example of what you’re doing and the error you’re getting.
s
wouldn't this fail with sth like "project id not a string, it is an Output object"
r
wouldn’t this fail with sth like “project id not a string, it is an Output object”
Yeah that’s fair, You can’t use an Output as the
resource_name
for another resource. But you can use it for every other argument in a resource. Specifically:
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projects.Service(userProject.project_id,           <-- this will not work                                   
                 disable_dependent_services=True,                                
                 project=userProject.project_id,   <-- this works fine                               
                 service="<http://cloudapis.googleapis.com|cloudapis.googleapis.com>")
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m
So that is exactly the error that I am facing...
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TypeError: Expected resource name to be a string
I am not sure what can I do here... Do you have some tips to share ?
r
I would suggest that you give your
Service
a different name. Is that not an option?
m
It is... This is what I was envisaging 🙂
partypus 8bit 1
Thanks at least for pointing that I cannot use it in the name property
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