How to compile multiple proto files in single command?

Question

I have two proto files inside single directory and I am looking for a way to generate classes from those files in a single command. It seems that this can be accomplished with the --proto_path argument. As per the documentation:

You must provide one or more .proto files as input. Multiple .proto files can be specified at once. Although the files are named relative to the current directory, each file must reside in one of the IMPORT_PATHs [specified by the --proto_path argument] so that the compiler can determine its canonical name.

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

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

but I am getting following error

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 unable to figure out when two paths (e.g. absolute and relative) are equivalent (it's harder than you think).

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

See developers.google.com/protocol-buffers/docs/proto#generating about what the --proto_path is used for. The flag that instructs protoc which files to compile is actually just the @filename argument. --proto_path is for where to look in order to resolve imports. — Yehuda Makarov
– Yehuda Makarov, Commented Sep 4, 2020 at 2:43

Cosmin Lehene · Accepted Answer · 2019-02-16 03:36:59Z

39

Here's an option using find

protoc --js_out=js \
        -Iproto/ \
        $(find proto/google -iname "*.proto")

answered Feb 16, 2019 at 3:36

Cosmin Lehene

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1 Comment

Dave Over a year ago

This should be the correct answer because all the related .proto files need to run with the same command in order for the directories to be correct, if there are sub-directories of .proto files.

Kenton Varda · Accepted Answer · 2014-07-25 22:35:23Z

33

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

answered Jul 25, 2014 at 22:35

Kenton Varda

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7 Comments

Samoht Over a year ago

It doesn't work: /main/resources/proto/*.proto: Invalid argument The only solution I found is to add every files one by one on the same line.

Kenton Varda Over a year ago

@Samoht It is the shell's responsibility to expand globs (*). If you are running the command through a context that doesn't invoke a shell, you may need to expand globs manually.

William Over a year ago

It's the shell's responsibility to expand globs on Unix. On Windows, it's up to the program receiving the glob. Also, fwiw, I'm running the latest Windows protoc as of 2018-06-19, and protoc does not seem to understand (or make any attempt to expand) globs.

Maxim Zhukov Over a year ago

Does not work for me on Windows. protoc does not recognize *.proto wildcard at cdm.

Kenton Varda Over a year ago

@Saeed Yes, as described earlier in this thread, this appears to be a bug in the Windows version of protoc.

|

PHPirate · Accepted Answer · 2018-10-05 17:10:26Z

6

Commands for Protobuf >= 3.5

It seems to me that the normal command on Windows only worked for Protobuf <= 3.4, and in the newer versions you cannot use the wildcard * but you have to put all the filenames separately. Fortunately it's still easy using a for loop (from here), using relative directories:

for /f %i in ('dir /b proto_trial\*.proto') do protoc proto_trial\%i --java_out=proto_trial\java_op

Alternatively, from here, you could also try to use Git Bash if you have it installed as it could expand the wildcard properly, and then use the command as you have before:

protoc proto_trial\*.proto --java_out=proto_trial\java_op

answered Oct 5, 2018 at 17:10

PHPirate

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VIPLIKE · Accepted Answer · 2023-03-20 10:23:47Z

3

for mac:

protoc --dart_out=grpc:../lib/client **/*.proto

answered Mar 20, 2023 at 10:23

VIPLIKE

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2 Comments

bramh Over a year ago

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

Moses Over a year ago

To add some clarity, run this command in the root folder where your proto files are protoc --dart_out=[YOUR DESTINATION FOLDER] **/*.proto

hetal · Accepted Answer · 2019-10-22 07:08:43Z

1

find . -name "*.proto" | xargs -I {} protoc "{}"

answered Oct 22, 2019 at 7:08

hetal

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sa_penguin · Accepted Answer · 2017-06-22 14:48:25Z

0

Download the windows version of the compiler from this link https://github.com/google/protobuf/releases/tag/v3.3.0

Start command prompt from bin folder of the downloaded application. Use the potoc command to generate the classes.

    protoc --proto_path=<Directory name where your proto file is residing> 
           --java_out=<Directory name where you want your output classes to get generated> 
           <absolute path of your protofile with extention>

Use wild character * for multiple protofiles

answered Jun 22, 2017 at 14:48

sa_penguin

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Syscall · Accepted Answer · 2018-04-27 06:03:21Z

0

Simplest way:

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

edited Apr 27, 2018 at 6:03

Syscall

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answered Apr 27, 2018 at 5:53

Ashok Kumar Pant

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Philippe · Accepted Answer · 2023-11-28 15:44:43Z

0

In my case adding multiple times the --proto_path each with the individual file did the trick.

protoc \
  --ng_out=../your/output/gen \
  --proto_path=../files/first.proto \
  --proto_path=../files/second.proto

answered Nov 28, 2023 at 15:44

Philippe

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