go php protobuf,go 了解protobuf导入和输出相对路径 - 糯米PHP

Bertha 。 2023-01-18 06:38 127阅读 0赞

我可以肯定这是操作员错误，并且我当时的想法还不清楚。

这是设置：

$GOPATH/src/github.com///proto/a.proto

$GOPATH/src/github.com///proto/b.proto

等等

现在，在原始文件中，我正在使用类似go的导入(也许是问题)，它a.proto具有：import "github.com///b.proto"

我可能有两个不同的问题。

我无法使用导入正确编译导入go:generate protoc

我无法将输出a.pb.go文件放置在$GOPATH/src/github.com///proto/路径中。

我尝试了多种配置，可能没有正确的组合。Using option go\_package = "github.com///proto"在每个.proto文件中

go生成多种变体；

go:generate protoc --proto\_path=.:$GOPATH/src --go\_out=$GOPATH/src a.proto

go:generate protoc --proto\_path=.:$GOPATH/src --go\_out=. a.proto

go:generate protoc --go\_out=import\_prefix=github.com//:. api.proto

I clearly have a poor understanding on how protoc looks at import paths and file outputs. Anyone point me in the direction of what I am doing wrong?

Thanks!

Update \#1

In a.proto;

option go\_package = "github.com///proto";

import "github.com///proto/b.proto";

and the go generate;

//go:generate protoc --proto\_path=$GOPATH/src --go\_out=$GOPATH/src/github.com///proto a.proto

Which is called from a go file in the proto directory with the a.proto.

I received the error;

a.proto: File does not reside within any path specified using --proto\_path (or -I). You must specify a --proto\_path ch encompasses this file. Note that the proto\_path must be an exact prefix of the .proto file names -- protoc is too dumb to figure out when two paths (e.g. absolute and relative) are equivalent (it's harder than you think).

I have confirmed $GOPATH is to the expected location.

Solution

感谢Shivam Jindal向我指出正确的方向。导入与他的解决方案完全相同。输出是我滥用--go\_out和选项go\_package的问题。我使用go\_package来指定输出位置，并使用--go\_out来指定类似于--proto\_path的根目录。现在一切正常。

option go\_package = "github.com///proto";

和

//go:generate protoc --proto\_path=$GOPATH/src/ --go\_out=$GOPATH/src/ $GOPATH/src/github.com///proto/a.proto

谢谢！