Closed hs-CN closed 1 year ago
Describe the bug
def send(self, method, arguments, on_invocation=None, invocation_id=str(uuid.uuid4())) -> InvocationResult:
invocation_id=str(uuid.uuid4()) this code will generate a uuid str just only once.
To Reproduce print( invocation_id), and call send method without specify ‘invocation_id’.
def on_open(): print("signalr open") client.send("ClearWatch", []) def call_back(result: CompletionMessage): # print("callback") # print(result.type) # print(result.invocation_id) # print(result.error) # print(result.result) pass client.send("WatchOnPLC", [[]], call_back)
PS D:\Github\signalrcore> python .\test.py signalr open 7ebd47dd-1401-4e55-beb0-9de07fbfd539 7ebd47dd-1401-4e55-beb0-9de07fbfd539
Desktop (please complete the following information):
Fix in PR #95
@mandrewcito can this be released pls? The problem is fixed in main branch but still persists in the current latest version.
Describe the bug
invocation_id=str(uuid.uuid4()) this code will generate a uuid str just only once.
To Reproduce print( invocation_id), and call send method without specify ‘invocation_id’.
Desktop (please complete the following information):