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Infralution Support
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MrDon
Guest
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 Posted: Tue May 16, 2006 6:36 am Post subject: GetLicense questions (2) |
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Sorry for posting twice. I think this can be a common interest question.
I successfully used the thecnique described in the topic "Can I license all the controls in a library with one key?" of the FAQ.
What I need now is simply retrieve licensing informations about:
- a given component of the library
- all the components of the library
To achieve this, I'm going to write a generic procedure that can be used:
- from an external application that uses the library
- from a procedure inside the library project, ut outside a component code
This is the code I tried:
| Code: |
Dim sProvider As ControlLicenseProvider 'Inherits from EncryptedLicenseProvider
Dim sType As System.Type
Dim sAssembly As System.Reflection.Assembly
Dim sLicense As License
Dim sEncryptedLicense As EncryptedLicense
sType = GetType(Components.MyControl1)
sAssembly = System.Reflection.Assembly.GetAssembly(sType)
sProvider = New ControlLicenseProvider
sProvider.SetParameters(LICENSE_PARAMETERS)
sLicense = sProvider.GetLicense(LicenseManager.CurrentContext, sType, sAssembly, True)
sEncryptedLicense = CType(sLicense, EncryptedLicense)
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...and this is another attempt:
| Code: |
Dim sProvider As New ControlLicenseProvider
Dim sContext As System.ComponentModel.LicenseContext
Dim sAssembly As System.Reflection.Assembly = System.Reflection.Assembly.GetExecutingAssembly
sProvider.GetLicense(sContext, sAssembly, GetType(Components.MyControl1))
Dim sEncryptedLicense As EncryptedLicense = sProvider.GetLicense(LICENSE_PARAMETERS, "ControlCollection.lic")
ControlLicenseProvider.SetParameters(LICENSE_PARAMETERS)
Dim sComponent As Components.MyControl1 = New Components.MyControl1
mIsValid = LicenseManager.IsValid(GetType(Components.MyControl1), sComponent, mLicense)
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They both does not works.
If possible, check the code, and say me what I missing, because I'm becoming crazy!
Thank you |
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Infralution
Joined: 28 Feb 2005
Posts: 5027
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| Posted: Tue May 16, 2006 9:19 am Post subject: |
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See reply on previous post
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