[python] always respect typespec namesapce even if package-name is set #5649
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| { | ||
| "package-name": "headasbooleantrue", | ||
| "head-as-boolean": "true", | ||
| "enable-typespec-namespace": "false", |
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instead of adding in enable-typespec-namespace: false here, can we instead switch package-name: headasbooleantrue to namespace: headasbooleantrue?
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I am afraid not. Since for unbranded, we always respect typespec namespace if enable-typespec-namespace is undefined, you can see the namespace of the 3 packages are all type.model.visibility which will conflict with each other if we don't set enable-typespec-namespace as false.
| if ( | ||
| self.code_model.namespace.count(".") >= self._namespace_from_package_name.count(".") | ||
| or self.code_model.options["enable_typespec_namespace"] | ||
| ): |
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the idea here is to no longer get namespace from package name, the preference should be
namespacein tspconfig- @clientNamespace in tsp (i know there are issues differentiating this rn from the original namspace in the spec)
namespacein spec
Now by default, we get package-name from whatever namespace out of the top 3 wins. Only if you want to override that namesapce, then you would use the package-name flag. We do have to see how many specs this will influence and add namespace to tspconfigs so we don't break
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It is similar with my proposal shared in teams channel. Let us discuss about it in tomorrow's scrum meeting.
We have discussed that we shall always respect typespace namespace even if the package-name is different with it.