Remove GraphQL debug output
Let's get this ready to go since we don't want to accidentally leak private information.
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@ -334,7 +334,6 @@ module GitHub
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def self.open_graphql(query, variables: nil, scopes: [].freeze, raise_errors: true)
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data = { query:, variables: }
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result = open_rest("#{API_URL}/graphql", scopes:, data:, request_method: "POST")
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odebug "GraphQL Query Response", result
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if raise_errors
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if result["errors"].present?
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