Below attached config file
[default]
PersistMessages=Y
ConnectionType=initiator
UseDataDictionary=Y
[SESSION]
ConnectionType=initiator
FileStorePath=store
FileLogPath=fixlog
StartTime=00:00:00
EndTime=00:00:00
BeginString=FIXT.1.1
AppDataDictionary=FIX50SP2.xml
TransportDataDictionary=FIXT.1.1.xml
DefaultApplVerID=FIX.5.0SP2
SenderCompID=xxxxx
TargetCompID=yyyyy
DeliverToCompID=zzzzz
Username=xxxxxx
Password=yyyyyy
SocketConnectHost=aaaa
SocketConnectPort=xxxxx
HeartBtInt=20
#ReconnectInterval=30
ResetOnLogon=Y
#ResetOnLogout=Y
#ResetOnDisconnect=Y
[SESSION]
ConnectionType=initiator
FileStorePath=store
FileLogPath=fixlog
StartTime=00:00:00
EndTime=00:00:00
BeginString=FIXT.1.1
AppDataDictionary=FIX50SP2.xml
TransportDataDictionary=FIXT.1.1.xml
DefaultApplVerID=FIX.5.0SP2
SenderCompID=aaaaa
TargetCompID=bbbb
Username=xxxxx
Password=cccccc
DeliverToCompID=yyyyy
SocketConnectHost=xxxxx
SocketConnectPort=dddddd
HeartBtInt=20
#ReconnectInterval=30
ResetOnLogon=Y
TO logout one session i'm sending
QuickFix.Session.LookupSession(priceSessionID).Logout();
i received logout for the particular session. Here my question is, how to logon to the same session without logging out another session?? and with out stopping the initiator.
答案 0 :(得分:0)
QF并非真正用于手动登录或注销的用途。
预期的工作流程是: