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Might have some problems:
USU.iAcesso_Usuario = Convert.ToInt16(Status_Label_Tipo);
I put in the Github for future reference.
Maybe I should take Status_Label_Tipo.Text
.
Depending on the type of USU.iAcesso_Usuario
maybe the conversion needs to be changed, but you can’t tell with so little information.
I have my doubts if just putting this information is enough to show the user and be able to show the form. Again, you can’t tell with so little information.
click View Detali.... and put the most detailed error.
– Marco Souza
Pay attention to that line:
Convert.ToInt16(Status_Label_Tipo);
. What is theStatus_Label_Tipo
? if it is a control, like a textbox or a label, for example, the right one would be to useStatus_Label_Tipo.Text
. In doing so I believe you will solve– Richard Dias
@Rennanhanna passes her aspx with the kind of her control
Status_Label_Tipo
and as you are carrying it in your code Behind, so will facilitate in answering your question.– Marco Souza
Take a look at [tour]. You can accept an answer if it solved your problem. You can vote on every post on the site as well. Did any help you more? You need something to be improved?
– Maniero