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Refine error handling in find_provider function to suggest a single similar provider name
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+3
-3
@@ -13,9 +13,9 @@ def find_provider(provider_name: Union[str, None]):
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return provider_class()
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providers_name = [provider.NAME.lower() for provider in providers]
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providers_founds = difflib.get_close_matches(provider_name.lower(), providers_name)
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provider_found = difflib.get_close_matches(provider_name.lower(), providers_name, n=1) # Show only one suggestion
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if providers_founds:
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raise ValueError(f"Provider {provider_name} not found. Maybe you try to use is '{providers_founds[0]}'?")
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if provider_found:
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raise ValueError(f"Provider {provider_name} not found. Maybe you try to use is '{provider_found[0]}'?")
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else:
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raise ValueError(f"Provider {provider_name} not found.")
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