I can't access my c# application in another pc

Ouma

New member
Joined
Nov 29, 2021
Messages
3
Programming Experience
1-3
I need your help PLEASE .
I developed a desktop application ( c# ; .NETFramework 4.7.2 ) with Visual studio2019 & SqlServer2017 ; and I tried to deploy it but it doesn't work in another laptop .
I get the error message :

Capture.PNG

************* Débogage JIT **************
Pour activer le débogage juste-à-temps (JIT), le fichier de configuration pour cette
application ou cet ordinateur (machine.config) doit avoir la valeur
jitDebugging définie dans la section system.windows.forms.
L'application doit également être compilée avec le débogage
activé.

Par exemple :

<configuration>
<system.windows.forms jitDebugging="true" />
</configuration>

Lorsque le débogage juste-à-temps est activé, les exceptions non gérées
seront envoyées au débogueur JIT inscrit sur l'ordinateur
plutôt que d'être gérées par cette boîte de dialogue.
 
Last edited:
Unfortunately, this is an English language forum. Can you post the error message translation in English?
 
You didn't even post the actual error message as text, so that we could translate it for ourselves. Given what I've seen in the past and the fact that there appears to be a SQL Server instance name specified in the error message, I'm guessing that you have hard-coded a connection string for your development machine and not considered that it would need to be different once the app is deployed. We shouldn't have to guess though.
 
Unfortunately, this is an English language forum. Can you post the error message translation in English?
Hello , here the message :

************* JIT Debugging **************
To enable just-in-time (JIT) debugging, the configuration file for this
application or this computer (machine.config) must have the value
jitDebugging defined in the system.windows.forms section.
The application must also be compiled with debugging
activated.

For example :

<configuration>
<system.windows.forms jitDebugging = "true" />
</configuration>

When just-in-time debugging is enabled, unhandled exceptions
will be sent to the JIT debugger registered on the computer
rather than being managed by this dialog box.

My DataBase's name : Article . I already detached it and put it in the DEBUG folder with the project .
I tried all those strings connection , but without vain .


1 : @"Data Source=LocalDB\MSSQLLocalDB;AttachDbFilename=GestEleve.mdf;Integrated Security=True"

2 : @" Data Source=.\SQLEXPRESS; AttachDbFilename=|DataDirectory|\Article.mdf; Initial catalog = Article ;Integrated Security=True"

3: @ "Data Source=.\SQLEXPRESS;Initial catalog =Article;Integrated Security=True");

4 :" Data Source=(LocalDB)\\MSSQLLocalDB;AttachDbFilename=|DataDirectory|Article.mdf;Integrated Security=True");

5 : "Data Source=.\\SQLSERVER;AttachDbFilename=|DataDirectory|Article.mdf;Integrated Security=True"

May you help me please .
 
Last edited:
The more interesting thing to get a translation of was the error in the screenshot regarding "Article" and "DESKTOP-NV2DOE2\Urez".

Also in your original post you said you were using SQL Server 2017, but some of the connection strings you were using above were for LocalDB and SQL Express which are different from the full blown SQL Server 2017.

What is the connection string that you are currently using that works on your computer?
 
The more interesting thing to get a translation of was the error in the screenshot regarding "Article" and "DESKTOP-NV2DOE2\Urez".

Also in your original post you said you were using SQL Server 2017, but some of the connection strings you were using above were for LocalDB and SQL Express which are different from the full blown SQL Server 2017.

What is the connection string that you are currently using that works on your computer?
Hello, "Article" is my data base's name, "DESKTOP-NV2DOE2 \ Urez" the computer in which I installed the application.
my connection string that works on my computer : SqlConnection con = new SqlConnection(@"Data Source=.\sqlexpress;Initial catalog =Article;Integrated Security=True");
 

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