[PHP-DB] MySQL query result with some errors!
Hi,
Im trying to get a query result from mysql database to
show it on a flash animation, so far i have this code
that its supposed to connect to mysql server , make
the query and get the results, store them on variables
so that flash can read them. Basically i have these
main fileds like : lastname, firstname and so on... a
guy here told me correctly that i should in case to
have more than one entry on the database, use the
output like lastname1, lastname2, etc. I gues this
code is doing that, but when i want to show that info
on flash i get eg:
lastname: smithn=0
lastname: jacksonn=1
when it should be
lastname: smith
lastname: jackson
...
...
So i guess im doing something wrong, can someone point
me in the right direcction??
Here is the php code:
?php
$conn = @mysql_connect(localhost, mysql,
ullneverguessit);
if (!$conn) {
echo( PNo se pudo conectar .
al servidor MySQL./P );
exit();
}
if (! @mysql_select_db(llamadas) ) {
echo( PNo se puede encontrar .
la base de datos clientes!/P );
exit();
}
$qr = mysql_query(select * from datos);
$nrows =
mysql_num_rows($qr);
$rString = n=.$nrows;
for ($i=0; $i $nrows; $i++) {
$row = mysql_fetch_array($qr);
$rstring .= Id.$i.=.$row['id'];
$rString .=
Dia.$i.=.$row['dia'].Mes.$i.=.$row['mes'].Ano.$i.=.$row['ano'];
$rString .=
Hora.$i.=.$row['hora'].Minutos.$i.=.$row['minutos'];
$rString .= Receptor.$i.=.$row['receptor'].
$rString .=
Destinatario.$i.=.$row['destinatario'].
$rString .= Apellido.$i.=.$row['apellido'].
$rstring .= Nombre.$i.=.$row['nombre'];
$rstring .= Telefono.$i.=.$row['telefono'];
$rstring .= Asunto.$i.=.$row['asunto'];
}
echo $rString;
mysql_free_result($qr);
?
Thanks.
Juan
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[PHP-DB] MySQL query result with some errors!
Hi,
Im trying to get a query result from mysql database to
show it on a flash animation, so far i have this code
that its supposed to connect to mysql server , make
the query and get the results, store them on variables
so that flash can read them. Basically i have these
main fileds like : lastname, firstname and so on... a
guy here told me correctly that i should in case to
have more than one entry on the database, use the
output like lastname1, lastname2, etc. I gues this
code is doing that, but when i want to show that info
on flash i get eg:
lastname: smithn=0
lastname: jacksonn=1
when it should be
lastname: smith
lastname: jackson
...
...
So i guess im doing something wrong, can someone point
me in the right direcction??
Here is the php code:
?php
$conn = @mysql_connect(localhost, mysql,
ullneverguessit);
if (!$conn) {
echo( PNo se pudo conectar .
al servidor MySQL./P );
exit();
}
if (! @mysql_select_db(llamadas) ) {
echo( PNo se puede encontrar .
la base de datos clientes!/P );
exit();
}
$qr = mysql_query(select * from datos);
$nrows =
mysql_num_rows($qr);
$rString = n=.$nrows;
for ($i=0; $i $nrows; $i++) {
$row = mysql_fetch_array($qr);
$rstring .= Id.$i.=.$row['id'];
$rString .=
Dia.$i.=.$row['dia'].Mes.$i.=.$row['mes'].Ano.$i.=.$row['ano'];
$rString .=
Hora.$i.=.$row['hora'].Minutos.$i.=.$row['minutos'];
$rString .= Receptor.$i.=.$row['receptor'].
$rString .=
Destinatario.$i.=.$row['destinatario'].
$rString .= Apellido.$i.=.$row['apellido'].
$rstring .= Nombre.$i.=.$row['nombre'];
$rstring .= Telefono.$i.=.$row['telefono'];
$rstring .= Asunto.$i.=.$row['asunto'];
}
echo $rString;
mysql_free_result($qr);
?
Thanks.
Juan
___
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[PHP-DB] MySQL query result with some errors!
Hi,
Im trying to get a query result from mysql database to
show it on a flash animation, so far i have this code
that its supposed to connect to mysql server , make
the query and get the results, store them on variables
so that flash can read them. Basically i have these
main fileds like : lastname, firstname and so on... a
guy here told me correctly that i should in case to
have more than one entry on the database, use the
output like lastname1, lastname2, etc. I gues this
code is doing that, but when i want to show that info
on flash i get eg:
lastname: smithn=0
lastname: jacksonn=1
when it should be
lastname: smith
lastname: jackson
...
...
So i guess im doing something wrong, can someone point
me in the right direcction??
Here is the php code:
?php
$conn = @mysql_connect(localhost, mysql,
ullneverguessit);
if (!$conn) {
echo( PNo se pudo conectar .
al servidor MySQL./P );
exit();
}
if (! @mysql_select_db(llamadas) ) {
echo( PNo se puede encontrar .
la base de datos clientes!/P );
exit();
}
$qr = mysql_query(select * from datos);
$nrows =
mysql_num_rows($qr);
$rString = n=.$nrows;
for ($i=0; $i $nrows; $i++) {
$row = mysql_fetch_array($qr);
$rstring .= Id.$i.=.$row['id'];
$rString .=
Dia.$i.=.$row['dia'].Mes.$i.=.$row['mes'].Ano.$i.=.$row['ano'];
$rString .=
Hora.$i.=.$row['hora'].Minutos.$i.=.$row['minutos'];
$rString .= Receptor.$i.=.$row['receptor'].
$rString .=
Destinatario.$i.=.$row['destinatario'].
$rString .= Apellido.$i.=.$row['apellido'].
$rstring .= Nombre.$i.=.$row['nombre'];
$rstring .= Telefono.$i.=.$row['telefono'];
$rstring .= Asunto.$i.=.$row['asunto'];
}
echo $rString;
mysql_free_result($qr);
?
Thanks.
Juan
___
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http://correo.yahoo.com.ar
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RE: [PHP-DB] MySQL query result with some errors!
The code looks correct so my guess is that you are parsing it inorrectly in
actionscript...
Bastien
From: Juan Stiller [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] MySQL query result with some errors!
Date: Tue, 11 Oct 2005 18:14:42 -0300 (ART)
Hi,
Im trying to get a query result from mysql database to
show it on a flash animation, so far i have this code
that its supposed to connect to mysql server , make
the query and get the results, store them on variables
so that flash can read them. Basically i have these
main fileds like : lastname, firstname and so on... a
guy here told me correctly that i should in case to
have more than one entry on the database, use the
output like lastname1, lastname2, etc. I gues this
code is doing that, but when i want to show that info
on flash i get eg:
lastname: smithn=0
lastname: jacksonn=1
when it should be
lastname: smith
lastname: jackson
...
...
So i guess im doing something wrong, can someone point
me in the right direcction??
Here is the php code:
?php
$conn = @mysql_connect(localhost, mysql,
ullneverguessit);
if (!$conn) {
echo( PNo se pudo conectar .
al servidor MySQL./P );
exit();
}
if (! @mysql_select_db(llamadas) ) {
echo( PNo se puede encontrar .
la base de datos clientes!/P );
exit();
}
$qr = mysql_query(select * from datos);
$nrows =
mysql_num_rows($qr);
$rString = n=.$nrows;
for ($i=0; $i $nrows; $i++) {
$row = mysql_fetch_array($qr);
$rstring .= Id.$i.=.$row['id'];
$rString .=
Dia.$i.=.$row['dia'].Mes.$i.=.$row['mes'].Ano.$i.=.$row['ano'];
$rString .=
Hora.$i.=.$row['hora'].Minutos.$i.=.$row['minutos'];
$rString .= Receptor.$i.=.$row['receptor'].
$rString .=
Destinatario.$i.=.$row['destinatario'].
$rString .= Apellido.$i.=.$row['apellido'].
$rstring .= Nombre.$i.=.$row['nombre'];
$rstring .= Telefono.$i.=.$row['telefono'];
$rstring .= Asunto.$i.=.$row['asunto'];
}
echo $rString;
mysql_free_result($qr);
?
Thanks.
Juan
___
1GB gratis, Antivirus y Antispam
Correo Yahoo!, el mejor correo web del mundo
http://correo.yahoo.com.ar
--
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