protoc-gen-swagger doesn't output to specified --swagger_out directory · grpc-ecosystem/grpc-gateway#383

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bughelp wantedopenapi

説明

Invoking protoc-gen-swagger with:

protoc -I/usr/local/include -I. -I$GOPATH/src \
  -I$GOPATH/src/github.com/grpc-ecosystem/grpc-gateway/third_party/googleapis \
  --swagger_out=logtostderr=true:. \
  mycorp/api/user/v1/user_service.proto

The outputted file ends up in mycorp/api/user/v1/, not the current working directory.

コントリビューターガイド

技術スタック: gogrpc
領域: toolingdocumentation
Issue 種別: bug
難度: 2
推定時間: 1-3 hours
活動状況: stale
明確さ: clear
前提条件: Go programmingprotoc usagegRPC gateway project structure
初心者向け度: 55
調査方針: The issue is that protoc gen swagger outputs files to the same directory as the proto file instead of the directory specified via swagger out. To fix, investigate the output path logic in the generator code, likely in 'protoc gen swagger/main.go' or 'generator.go'. Look for how the output file path is constructed from the proto file path and the swagger out directory. The fix should ensure that the output directory is used as the base for the generated file.