| Bug #80921 | Error in connection | ||
|---|---|---|---|
| Submitted: | 31 Mar 2016 10:39 | Modified: | 1 Apr 2016 12:09 |
| Reporter: | Иван Иванов | Email Updates: | |
| Status: | Duplicate | Impact on me: | |
| Category: | MySQL Server | Severity: | S2 (Serious) |
| Version: | 14.14 | OS: | Any |
| Assigned to: | CPU Architecture: | Any | |
[1 Apr 2016 9:32]
Chiranjeevi Battula
Hello Иван Иванов, Thank you for the bug report. I tried to reproduce the issue at my end using Visual Studio 2013 (C#.Net) and Connector/Net 6.9.8 but not seeing any issues in connection. Could you please provide more information like MySQL version, Connector / NET version details to confirm this issue at our end? Please add "port" number in the connection string. Thanks, Chiranjeevi.
[1 Apr 2016 11:24]
Иван Иванов
I am use Visual Studio 2013. Connector .net 6.9.8 Mysql server Ver 14.14 Distrib 5.7.11 O.S. RHEL 5.9 (2.6.18-348.e15)
[1 Apr 2016 11:58]
Chiranjeevi Battula
Hello Иван Иванов, Thank you for your feedback. Try to login into your MySQL database from the command prompt / workbench using your username/password, if it is working no problem in database side. Please make sure your network having sufficient permissions or not. This is most likely duplicate of Bug #75917, please see Bug #75917. Thanks, Chiranjeevi.
[1 Apr 2016 11:59]
Chiranjeevi Battula
Screenshot.
Attachment: 80921.JPG (image/jpeg, text), 148.32 KiB.
[1 Apr 2016 12:09]
Иван Иванов
Connection via MySql Workbench works well . I read all the duplicates , but did not find the solution to your problem . My code errors #0

Description: "Authentication to host '192.168.0.14' for user 'root' using method 'mysql_native_password' failed with message: Reading from the stream has failed." How to repeat: conStrBilder = new MySqlConnectionStringBilder(); conStrBilder.Server = "192.168.0.14"; conStrBilder.Database = "login"; conStrBilder.UserID = "root"; conStrBilder.Password = "root"; DataTable dt = null; var con = new MySqlConnection(conStrBilder.ConnectionString); var command = new MySqlCommand(query, con); try { //здесь вываливается исключение con.Open() var dr = command.ExecuteReader(); If(dr.HasRows) { dt = new DataTable(); dt.Load(dr); } } catch(Exceprion ex) { errorString = ex.Message; } finally { con.Close(); }