Expose all resources on a service resource. #60
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This change brings back behavior that was removed in Boto 3 - 0.0.7
and allows things like the following to work again:
```python
import boto3
s3 = boto3.resource('s3')
obj = s3.Object('bucket-name', 'key')
```
| allows things like ``s3.Object('bucket-name', 'key')`` to | ||
| work even though the JSON doesn't define it explicitly. | ||
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| @rtype: dict |
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What's with the @rtype: notation?
It is possible to define many subresources on a service that have the same resource type, e.g. `Queue`, `PriorityQueue`, and `Stack` might all be implemented as a single resource type with different default values. We need to make sure all of these are exposed on the service resource.
| if has_def.get('resource', {}).get('type') == name: | ||
| definition[has_name] = has_def | ||
| break | ||
| else: |
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Where is this else coming from? It does not look in line with any if statement.
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It's from the for loop. If it never breaks, then the else is executed. I've since removed it to simplify the code and fix a potential case I hadn't considered so it should now be easier to understand.
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Looks good. 🚢 Thanks for fixing the comments I made earlier |
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Changes Unknown when pulling e2d3706 on service-subresources into * on develop*. |
|
Thanks! 🚢 |
danielgtaylor
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Feb 10, 2015
Expose all resources on a service resource.
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This change brings back behavior that was removed in Boto 3 - 0.0.7
and allows things like the following to work again:
This is required to gain direct access to some resources that are not exposed as subresources nor references (think something like
sqs.Message, which you can now only get fromsqs.Queue('url').receive_messages() => [Message(...)]Without the change in this pull request there is no way to instantiate a message instance yourself).cc @jamesls @kyleknap