Hello Juan
If I declare &Telefono=4141414 in one line and in the next line I declare
&Telefono=31453531, then I have overwritten the first variable.
What you need to change is the name of the variable. You need to append your
ID number to the variable name.
So &Telefono01=4141414 and &Telefono0
Thanks for the response,
Now its fixed, and this is the output for the code:
&Id=28&Apellido=sadfsadf&Nombre=asdfsdf&Dni=sadfsdf&Telefono=sadfsadf&Dia=1&Mes=12&Ano=2004&Hora=9&Minutos=50&Abogado=Dra.
Claudia I.
Gorno&Nombre=asdfsdf&Asunto=asdfsd&Donde=Guia Barrial
Palermo&Id=26&Apellido=Sarasa&No
"&Donde={$row['dondeclientesnuevos']}";
}
mysql_free_result($result1);
?>
And this error es shown:
Warning: mysql_num_rows(): supplied argument is not a
valid MySQL result resource
Any hints??
--- Bastien Koert <[EMAIL PROTECTED]> escribió:
>
> remove the @ symb
Fixing this code to show all variables!
Date: Mon, 1 Nov 2004 10:50:12 -0300 (ART)
Hi, im trying to use this code to send a query to a
mysql database, and then display the query results and
put the results on variables, the problem is that i
don´t know how to make it so that if there is more
than one
Hi, im trying to use this code to send a query to a
mysql database, and then display the query results and
put the results on variables, the problem is that i
don´t know how to make it so that if there is more
than one result, it uses xvariables as results are,
using this script just like it is, if
