Print a selected record in a tablegrid (Page 1 of 2)

Hi derek, when I press the button I get the error I put in the attachment. It would seem to refer to the "e-mail" field But I put this field in the "Search" button. which displays the search data in the TableGrid.

I think I've discovered the problem: I think they don't like the - in the field names.

Hi, sparrow, I had felt that the - had something to do with it. In fact, I removed it and the error was gone. Thank you for the elucidation and correction.

I sparrow, I didn't understand what you want to say me. Is this not good enough?

Hi Fabio, Hi Sparrow,
What Sparrow is saying is that you have set 'nuovo' to be 'new' but no where do you clear it.
Your code should be

procedure frmInserimentopc_OnShow (Sender: TObject; Action: string);
begin
  if action = 'NewRecord' then nuovo := 'new' else nuovo := '';
end;

(and wherever else in your script you are using 'nuovo').
Derek.

Derek, It's all right, It's better to answer two than no one. )))
In addition, decisions can be different and then there is a choice.

When it works incorrectly, of course it also works, but not as it should.
So far it hasn't been critical, but who knows.
for example this moment:

procedure frmInserimentopc_bSalvaenuovopc_OnAfterClick(Sender: TObject);
begin
   if nuovo = 'new' then
     begin
       start.butInseriscipc.click;
       frmInserimentopc.edNomepc.setfocus;
     end else frmUtenti.close;

Procedure for form "frmInserimentopc" and close another form "frmUtenti". The error is purely technical when copying code from one place to another.

Above I posted the second version with all the fixes. decora2.
We'll help in any way we can.

Èh yes I am here again. The new problem I found is on the print report. It is always found in the last TabSheet "Dipendenti." if I click on the print button for the selected record the report opens without any problems, if, however, I click on the "Stampa tutti" button the report opens full of errors and with all fields saying "File not found." It seems illogical, but this may be my wrong thinking, to do two reports for the same table. However, I don't understand why one report opens well and the other one doesn't. Thank you!

Hi Fabio,
The data that you are passing through to the report, although it comes from the same table, is not called the same thing.
Have a look at the attached screenshot.
For example, in the case of one of the reports your field name in Fast Report is 'nome' but the field that is being sent to Fast Report is called 'dipendenti.nome.
This is because you are using 'report' to generate the report in one instance and 'report sql' in the other (I am not sure why you are using 'report sql' as I would have thought 'report' was easier).
Derek.

Hello everyone again.

It's all about how we send the data to the report.
In a SQL query, there is such a thing as an alias.
Everything is very simple:
Like this - SELECT 'Dipendenti'.'CAP' ... in the response, the column name will be called as the name of the column 'CAP'
And like this - SELECT 'Dipendenti'.'CAP' AS "Dipendenti.CAP" ... the column will get the alias "Dipendenti.CAP"

Therefore, in your request for a single record, the name of the column, and the program generates an alias from the name of the database.column name

SELECT
'Dipendenti'.'CAP' AS "Dipendenti.CAP", 
'Dipendenti'.'Città' AS "Dipendenti.Città",
'Dipendenti'.'Cognome' AS "Dipendenti.Cognome",
'Dipendenti'.'DataNascita' AS "Dipendenti.DataNascita",
'Dipendenti'.'Indirizzo' AS "Dipendenti.Indirizzo",
'Dipendenti'.'InternoTelefono' AS "Dipendenti.InternoTelefono",
'Dipendenti'.'Nome' AS "Dipendenti.Nome",
'Dipendenti'.'NumeroTelefono' AS "Dipendenti.NumeroTelefono",          
'Dipendenti'.'email' AS "Dipendenti.email",
'Dipendenti'.'Settore' AS "Dipendenti.Settore",
'Dipendenti'.'TitoloStudio' AS "Dipendenti.TitoloStudio",
'Dipendenti'.'id'             

FROM 'Dipendenti'

WHERE 'Dipendenti'.'id' = {tbDipendenti}