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opcUaClient.WriteValues() success=true ,Unable to receive notification of point

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07 Nov 2023 15:22 #12339 by support

Replied by support on topic opcUaClient.WriteValues() success=true ,Unable to receive notification of point

Hello.

Yes, 0x800A0000 is for BadTimeout. Unless you have an indication to the contrary, it means what it says in the log message. I.e. the server has not responded soon enough to the Publish or Republish service call.

It would only be possible to tell more if we had a network capture (Wireshark) with the actual communication during that session. However, most likely outcome of such analysis is simply that the server has not responded soon enough to the service call. Less likely, but of course still possible, is that the server has responded in time, but somehow the client did not interpret the response correctly.

Best regards

The following user(s) said Thank You: xpycy001

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07 Nov 2023 15:05 #12338 by xpycy001

Replied by xpycy001 on topic opcUaClient.WriteValues() success=true ,Unable to receive notification of point

"trace sequence number 5699 "
The disconnect log is different from other logs in that it shows connection timeout. There is a detailed tracking process in the attachment log. Can you help analyze why TimeOut occurred

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07 Nov 2023 10:45 #12336 by support

Replied by support on topic opcUaClient.WriteValues() success=true ,Unable to receive notification of point

Hello,
I do not quite understand the nature of your issue. So I will start with the part that seems clear. You wrote

"30 attempts were made to write values to the point, and each write operation returned success, but the point was not actually successful".

If the value was not actually written, but the Write call succeeded, you have a faulty server. You need to fix the server. There is no way this can be fixed on the client side.

Regards

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07 Nov 2023 10:25 #12335 by xpycy001

Replied by xpycy001 on topic opcUaClient.WriteValues() success=true ,Unable to receive notification of point

Supplementary description

1. There are no write failures at other points on the target machine

2. For machines of the same brand at the same process section location, there will be daily Write fail

3. The target point will become true every 15 seconds, opcUaClient. writeValue() false

4. The total number of machines is about 90, among which 30 machines have problems, and write failures occur approximately 5-10 times a day

5. I am not sure how to solve this thorny problem, here is my attempt

6. Adjust HoldPeriod==SessionTimeOut=120 seconds

7. retryDelay=6s, SessionConnectionTimeOut=6s

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07 Nov 2023 09:56 - 07 Nov 2023 10:01 #12334 by xpycy001

opcUaClient.WriteValues() success=true ,Unable to receive notification of point was created by xpycy001

Dear support :
The problem is as follows,
30 attempts were made to write values to the point, and each write operation returned success, but the point was not actually successful
1. 2023/11/7 13:40:15 call opcUaClient.WriteValues() success=true
2. Subscribe Side not received valueChange
3. retry more times …
4. trace tips: timeout
5. trace sequence number 5699
6 opcUaclient used default parameters . (session too much and create again and again)
7 at last I used HoldPeriod =120s && SessionTimeOut=120s ,but write fail again
【opc.tcp://10.xx.xxx.xx:4840】,【ns=4;s=|var|Inovance-PLC.Application.PR109_OPC_MES.vs_OnSlotFinish_B】
trace：

13:40:30.407,EvId:6002,Information The OPC-UA subscription on client session with connect sequence number 5699 to endpoint URL "opc.tcp://10.31.199.205:4840" has been successfully created and checked, and given subscription ID 2. Revised publishing interval is 7500, keep-alive count is 10 and lifetime count is 1000.

13:42:00.691,EvId:4051,Warning The OPC-UA client session with connect sequence number 5699 on endpoint URL "opc.tcp://10.31.199.205:4840" has encountered a publish error with code 0x800A0000 for subscription ID 0 and sequence number 0. The error might be from the Publish or the Republish service. Further publish errors with this code on this session will not be logged.