influxd TypeError：<lambda>（）得到了意外的关键字参数'key_key_password'

Question

我正在使用指南并从文档中生成的代码中复制行和令牌，但是当我运行时，我遇到此错误，该错误引用了此行 write_api.write(bucket=bucket, org=org, record=point)。 influxdb 2.7.5

client = influxdb_client.InfluxDBClient(url=url, token=token, org=org)
write_api = client.write_api(write_options=SYNCHRONOUS)
query_api = client.query_api()
delete_api = client.delete_api()
buckets = client.buckets_api()


print("buckets", buckets.find_buckets())

for value in range(5):
  point = (
    Point("measurement1")
    .tag("id", "1")
    .field("field1", value)
  )
  write_api.write(bucket=bucket, org=org, record=point)
  time.sleep(1) # separate points by 1 second

write_api.close()

上面的代码是在 python 的用户设置中给出的，如果你复制它并运行它，它将生成以下错误。

Traceback (most recent call last):
  File "/influxManager.py", line 21, in <module>
    print("buckets", buckets.find_buckets())
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/client/bucket_api.py", line 119, in find_buckets
    return self._buckets_service.get_buckets(**kwargs)
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/service/buckets_service.py", line 558, in get_buckets
    (data) = self.get_buckets_with_http_info(**kwargs)  # noqa: E501
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/service/buckets_service.py", line 586, in get_buckets_with_http_info
    return self.api_client.call_api(
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/_sync/api_client.py", line 343, in call_api
    return self.__call_api(resource_path, method,
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/_sync/api_client.py", line 173, in __call_api
    response_data = self.request(
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/_sync/api_client.py", line 365, in request
    return self.rest_client.GET(url,
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/_sync/rest.py", line 268, in GET
    return self.request("GET", url,
  File "/opt/anaconda3/lib/python3.9/site-packages/influxdb_client/_sync/rest.py", line 235, in request
    r = self.pool_manager.request(method, url,
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/request.py", line 66, in request
    return self.request_encode_url(method, url, fields=fields,
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/request.py", line 89, in request_encode_url
    return self.urlopen(method, url, **extra_kw)
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/poolmanager.py", line 313, in urlopen
    conn = self.connection_from_host(u.host, port=u.port, scheme=u.scheme)
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/poolmanager.py", line 229, in connection_from_host
    return self.connection_from_context(request_context)
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/poolmanager.py", line 240, in connection_from_context
    pool_key = pool_key_constructor(request_context)
  File "/opt/anaconda3/lib/python3.9/site-packages/urllib3/poolmanager.py", line 105, in _default_key_normalizer
    return key_class(**context)
TypeError: <lambda>() got an unexpected keyword argument 'key_key_password'

Answer 1

我正在使用命令 python filename.py 但是当我使用 python3 filename.py 时没有问题！