RE: [PHP-DB] Changing this php code to show multiple variables!
| From: | Norland, Martin | Date: | Wed, 20 Oct 2004 17:10:42 +0000 |
| Subject: | RE: [PHP-DB] Changing this php code to show multiple variables! | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-36931@lists.php.net to get a copy of this message | ||
I don't pretend to know flashes interface, as I said before - but instead of naming the
variables differently - you should be able to pass them in an array.
&Lastname0=craw&Firstname0=david&Lastname1=Zarlanga&Firstname1=John
becomes
&Lastname[]=craw&Firstname[]=david&Lastname[]=Zarlanga&Firstname[]=John
Assuming flash accepts the parameters in this form. You'll want to make sure you emit every
variable each time, otherwise your data sets will get out of sync (emit them even if they're
blank) - but I still say there surely must be a better way to be populating this data.
http://actionscript-toolbox.com/samplemx_php.php
Seems to be an introductory that may be of some use, found just by googling for 'flash php
mysql'.
- Martin Norland, Database / Web Developer, International Outreach x3257
The opinion(s) contained within this email do not necessarily represent those of St. Jude
Children's Research Hospital.
-----Original Message-----
From: Juan Stiller [mailto:juanstiller@yahoo.com.ar]
Sent: Wednesday, October 20, 2004 11:27 AM
To: php-db@lists.php.net
Subject: [PHP-DB] Changing this php code to show multiple variables!
Hi, i´ve got this php code that sends a query to a
mysql server, and when it gets the query results, it
send it to a flash movieclip, everything works fine
when i´ve got only one entry on the database, but when
there´s more than one enrty, the variable is
overwritten and as reslt, it shows only one entry i
was told that i must create a function to to name
multiple variables, so the result might be:
&Lastname0=craw&Firstname0=david&Lastname1=Zarlanga&Firstname1=John
So here it is the php code:
<?php
$conn = @mysql_connect("***", "***", "***");
if (!$conn) {
echo( "<P>No se pudo conectar " .
"al servidor MySQL.</P>" );
exit();
}
if (! @mysql_select_db("clientes") ) {
echo( "<P>No se puede encontrar " .
"la base de datos clientes!</P>" );
exit();
}
// Request all data
$result1 = mysql_query("select * from
clientesnuevos");
print "Results=";
echo "<h1>Clientes agregados:</h1><br>";
while($row=mysql_fetch_array($result1, MYSQL_ASSOC))
{
echo "&Id={$row['id']}";
echo "&Apellido={$row['apellidoclientesnuevos']}";
echo "&Nombre={$row['nombreclientesnuevos']}";
echo "&Dni={$row['dniclientesnuevos']}";
echo "&Telefono={$row['telefonoclientesnuevos']}";
echo "&Dia={$row['diacitaclientesnuevos']}";
echo "&Mes={$row['mescitaclientesnuevos']}";
echo "&Ano={$row['anocitaclientesnuevos']}";
echo "&Hora={$row['horacitaclientesnuevos']}";
echo "&Minutos={$row['minutoscitaclientesnuevos']}";
echo "&Abogado={$row['abogadoclientesnuevos']}";
echo "&Nombre={$row['nombreclientesnuevos']}";
echo "&Asunto={$row['asuntoclientesnuevos']}";
echo "&Donde={$row['dondeclientesnuevos']}";
}
mysql_free_result($result1);
?>
Can anyone help me with this?
Thanks
Juan
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