如何在单个命令中编译多个proto文件？ [英] How to compile multiple proto files in single command?

问题描述

我在单个目录中有两个proto文件，我正在寻找一种在单个命令中从这些文件生成类的方法。 Protobuf文档说我们需要使用 - proto_path 参数。

I have two proto files inside single directory and I am looking for a way to generate classes from those files in single command. Protobuf documentation says that we need to use --proto_path argument for this.

C:\shekhar\proto_trial>dir
 Volume in drive C is C

 Directory of C:\shekhar\proto_trial

07/25/2014  12:16 PM    <DIR>          .
07/25/2014  12:16 PM    <DIR>          ..
07/25/2014  12:16 PM    <DIR>          java_op
07/25/2014  12:16 PM               230 map.proto
07/23/2014  04:24 PM               161 message.proto
07/25/2014  12:17 PM             1,228 response.proto
               3 File(s)          1,619 bytes
               3 Dir(s)  50,259,398,656 bytes free

我使用了 - proto_path 参数，如下所示

I used --proto_path argument as shown below

C:\shekhar\proto_trial>protoc 
                       --proto_path=C:\shekhar\proto_trial 
                       --java_out=C:\shekhar\proto_trial\java_op 
                       *.proto

但我收到以下错误

message.proto: File does not reside within any path specified using --proto_path (or -I). 
You must specify a --proto_path which 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).

请建议一些方法将所有原型文件一次性编译。

Please suggest some way to compile all the proto files together in single shot.

推荐答案

问题是你指定 - proto_path 作为绝对路径，但你的原型文件作为相对路径。你可以删除 - proto_path 参数（无论如何都默认为当前目录），或者你可以这样做：

The problem is that you are specifying --proto_path as an absolute path but your proto files as relative paths. You can either drop the --proto_path argument (it defaults to the current directory anyway), or you can do:

protoc --proto_path=C:\shekhar\proto_trial
       --java_out=C:\shekhar\proto_trial\java_op
       C:\shekhar\proto_trial\*.proto

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