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use swaggerUiPrefix in template_file #196

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phil2fer opened this issue May 1, 2018 · 2 comments
Open

use swaggerUiPrefix in template_file #196

phil2fer opened this issue May 1, 2018 · 2 comments

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@phil2fer
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@phil2fer phil2fer commented May 1, 2018

I want to use template_file argument with a unique YAML file.
But i'm behind a reverse proxy and i need to use the "swaggerUiPrefix" option in the template.

swaggerUiPrefix=LazyString(lambda : request.environ.get('HTTP_X_SCRIPT_NAME', ''))

But when i use a YAML file, this option doesn't work. flasgger see this option like a string

How can i use the 'swaggerUiPrefix' in YAML file ?

@rochacbruno
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@rochacbruno rochacbruno commented May 2, 2018

Currently there is no way to evaluate lazy values on YAML, we can implement it.

@phil2fer
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@phil2fer phil2fer commented May 2, 2018

thank you for your quick answer.
Why not move this option to the swagger config dictionary ? I think is more a config parameter than a template parameter

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