【发布时间】：2022-08-04 01:57:13
【问题描述】：
无法导入 protobuf 生成的 python 模块。这是代码。
-
error message
这是运行python main.py 后的错误消息。
# python main.py
Traceback (most recent call last):
File \"/usr/lib64/python2.7/multiprocessing/queues.py\", line 266, in _feed
send(obj)
PicklingError: Can\'t pickle <class \'demo_pb2.msg\'>: import of module demo_pb2 failed
^CTraceback (most recent call last):
File \"main.py\", line 31, in <module>
main()
File \"main.py\", line 24, in main
for msg in iter(q.get, None):
File \"/usr/lib64/python2.7/multiprocessing/queues.py\", line 117, in get
res = self._recv()
KeyboardInterrupt
- 代码树
.
├── __init__.py
├── main.py
└── pb
├── demo_pb2_grpc.py
├── demo_pb2.py
├── demo.proto
├── __init__.py
└── run_codegen.py
-
main.py
from multiprocessing import Queue
import os
from threading import Thread
import time
from pb import demo_pb2
q = Queue()
def generate_file_path(path):
for root, dirs, files in os.walk(path):
for dir_ in dirs:
q.put(demo_pb2.msg(path=os.path.join(root, dir_)))
time.sleep(0.1)
for file_ in files:
q.put(demo_pb2.msg(path=os.path.join(root, file_)))
time.sleep(0.1)
q.put(None)
def main():
t = Thread(target=generate_file_path, args=(\'/root/pip\',))
t.start()
for msg in iter(q.get, None):
print(msg)
q.close()
t.join()
if __name__ == \'__main__\':
main()
-
pb/demo.proto
syntax = \"proto3\";
package demo;
message msg {
string path = 1;
}
-
pb/run_codegen.py
from grpc_tools import protoc
protoc.main((
\'-I./\',
\'--python_out=./\',
\'--grpc_python_out=./\',
\'demo.proto\',
))
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你好@iskylite，欢迎来到 StackOverflow！请与我们分享
demo_pb2.msg类的定义。此外，如果您将代码设为minimal reproducible example，我们将不胜感激。 -
您好@David，感谢您的回复。但是我没有找到文件
demo_pb2.msg，文件在哪里？以上是最小的可重现代码。实际上只需要main.py和pb/demo.proto两个文件。 -
请分享文件
pb/demo_pb2.py的内容。里面有应该成为class msg:定义（除非msg是从某处导入的）。
标签： python grpc grpc-python