我在单个目录中有两个proto文件,我正在寻找一种方法,可以在单个命令中从这些文件生成类. Protobuf文档说我们需要使用–proto_path参数.
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
C:\shekhar\proto_trial>protoc
--proto_path=C:\shekhar\proto_trial
--java_out=C:\shekhar\proto_trial\java_op
*.proto
但我得到以下错误
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).
