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,
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 result
...when
you finish the debugging,
then you can add them back in.
bastien
From: Juan Stiller [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] 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
Thanks for the response,
Now its fixed, and this is the output for the code:
Id=28Apellido=sadfsadfNombre=asdfsdfDni=sadfsdfTelefono=sadfsadfDia=1Mes=12Ano=2004Hora=9Minutos=50Abogado=Dra.
Claudia I.
GornoNombre=asdfsdfAsunto=asdfsdDonde=Guia Barrial
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
