How can check whether a specific user is deactivated(released) on server?

Follow

Dammika Prasad Wijethunga

February 11, 2019 06:06

- Created a Activation Key and activated on 2 computers.

- Now both computers are working (License keys are activated)

- Now i want to release(deactivate) one computer's key from server. After the release, the computer is disappeared from server.

- But still "ValidateLicenseOnServer()" returns TRUE. (even the app connected to server successfully)

How can detect whether the specific computer is released from server?

 

I tried following method, but no luck..

IsIllegalComputer(), IsLicenseKeyActivated().

 

Thank you.

 

 

0

• Official comment

  

  Soraco Technologies February 11, 2019 12:23

  If the license is not activated and there are licenses still available on the server,  ValidateLicenseOnServer will automatically consume a license and activate the computer.

  ValidateLicenseOnServerEx has an extra argument that allows you to control this behavior and not automatically reactivate the license.

  Comment actions Permalink
• 

  Dammika Prasad Wijethunga February 11, 2019 12:36

  Thanks for the reply..

  I can't see "ValidateLicenseOnServerEx". Only i can see following..

   

  [SecuritySafeCritical]
  public string ValidateLicense(string licenseKey);
  [SecuritySafeCritical]
  public string ValidateLicenseEx(string licenseKey, string computerID);
  [SecuritySafeCritical]
  public string ValidateLicenseEx2(string licenseKey, string computerID, bool skipWrites, bool skipExpiryValidation);
  [SecuritySafeCritical]
  public string ValidateLicenseEx3(string licenseKey, ELicenseBinding licenseBinding, bool skipWrites, bool skipExpiryValidation);
  public bool ValidateLicenseOnServer(string webServiceUrl, string activationKey, string computerKey, string computerID, out EServerErrorCode errorCode, out string returnMsg);

  0

  Comment actions Permalink
• 

  Soraco Technologies February 11, 2019 17:15

  This function was recently added. Please upgrade to the latest version.

  0

  Comment actions Permalink
• 

  Dammika Prasad Wijethunga February 13, 2019 02:02

  Perfect.. I used 11.1.19032.2 version and it worked.. Thank you.

  0

  Comment actions Permalink

Please sign in to leave a comment.