我在Python3中有一个带有pysimplesoap的soap服务器。我不知道,因为我得到了nexts erros。
代码
from wsgiref.simple_server import make_server
application = WSGISOAPHandler(dispatcher)
wsgid = make_server('', 8008, application)
wsgid.serve_forever()
错误
Traceback (most recent call last):
File "/usr/lib/python3.4/wsgiref/handlers.py", line 138, in run
self.finish_response()
File "/usr/lib/python3.4/wsgiref/handlers.py", line 180, in finish_response
self.write(data)
File "/usr/lib/python3.4/wsgiref/handlers.py", line 266, in write
"write() argument must be a bytes instance"
AssertionError: write() argument must be a bytes instance
答案 0 :(得分:3)
这是因为WSGI是针对python 2编写的,因此在python3中使用它会遇到一些麻烦。 如果您不想像第一个答案那样更改库的行为,解决方法是对所有文本数据进行编码,例如:
def application(environ,start_response):
response_body = 'Hello World'
return [response_body.encode()]
答案 1 :(得分:1)
Wsgi框架是基于Python 2构建的。因此,如果程序中包含不包含Python 3依赖项的内容,请使用Python 2运行该应用程序。
答案 2 :(得分:0)
+++ pysimplesoap/server.py
e['name'] = k
if array:
e[:] = {'minOccurs': "0", 'maxOccurs': "unbounded"}
- if v in TYPE_MAP.keys():
- t = 'xsd:%s' % TYPE_MAP[v]
- elif v is None:
+
+ # check list and dict first to avoid
+ # TypeError: unhashable type: 'list' or
+ # TypeError: unhashable type: 'dict'
+ if v is None:
t = 'xsd:anyType'
elif isinstance(v, list):
n = "ArrayOf%s%s" % (name, k)
n = "%s%s" % (name, k)
parse_element(n, v.items(), complex=True)
t = "tns:%s" % n
+ elif v in TYPE_MAP.keys():
+ t = 'xsd:%s' % TYPE_MAP[v]
else:
raise TypeError("unknonw type v for marshalling" % str(v))
e.add_attribute('type', t)
答案 3 :(得分:0)
在handlers.py第180行
self.write(data.encode())而不是self.write(data)