launcher
01-18-2009, 10:48 AM
Here it is...
The Flash give me this.
error
<root>
<br/>
<b>Warning</b>
: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
<b>C:\wamp\www\Flash\PHP\general.php</b>
on line
<b>26</b>
<br/>
</root>
Here's my code.
ActionScript
var xml:XML;
var xmlLoader:URLLoader = new URLLoader;
xmlLoader.load(new URLRequest("http://localhost/Flash/PHP/general.php"));
xmlLoader.addEventListener(Event.COMPLETE, xmlLoaded);
function xmlLoaded(event:Event):void
{
xml = new XML(event.target.data);
trace(xml);
}
general.php
<?php
$sadarjanie = $_GET['vic'];
$nomer = $_GET['number'];
$iden = $_GET['id'];
$content = $sadarjanie;
$number = $nomer;
$id = $iden;
if($id == 1){$prom = "grafiti";}
else if($id == 2){$prom = "blondinki";}
else if($id == 3){$prom = "bigbrother";}
else if($id == 4){$prom = "znaeteli4e";}
else if($id == 5){$prom = "razni";}
else if($id == 6){$prom = "laforizmi";}
else if($id == 7){$prom = "pc";}
else if($id == 8){$prom = "ivan4o";}
else if($id == 9){$prom = "borci";}
echo $prom;
//the XML
echo "<?xml version=\"1.0\" ?>\n";
echo "<root>\n";
include("C:\wamp\www\Flash\PHP\Includes\conn.php");
$query = mysql_query("select * from $prom");
while($array = mysql_fetch_array($query))
{
echo "<vic id=" . $array['0'] . ">" . $array['1'] . "</vic>\n";
}
echo "</root>\n";
$number = 0;
if($number > 0)
{
if($number == 1){$var = "grafiti";}
else if($number == 2){$var = "blondinki";}
else if($number == 3){$var = "bigbrother";}
else if($number == 4){$var = "znaeteli4e";}
else if($number == 5){$var = "razni";}
else if($number == 6){$var = "laforizmi";}
else if($number == 7){$var = "pc";}
else if($number == 8){$var = "ivan4o";}
else if($number == 9){$var = "borci";}
include("C:/wamp/www/Flash/PHP/Includes/conn.php");
mysql_query("SET NAMES CP1251");
$query = "INSERT INTO `vicove`.`$var` (`id` ,`vic` )VALUES (NULL , '$content')";
$row = mysql_query($query);
mysql_close($conn);
if($row){echo "Вица ви беше оспешно добавен";}
else{echo "Вица ви не беше добавен";}
}
?>
conn.php
<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "vicove";
$conn = mysql_connect($host, $user, $pass);
$select_db = mysql_select_db($db, $conn);
?>
The Flash give me this.
error
<root>
<br/>
<b>Warning</b>
: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
<b>C:\wamp\www\Flash\PHP\general.php</b>
on line
<b>26</b>
<br/>
</root>
Here's my code.
ActionScript
var xml:XML;
var xmlLoader:URLLoader = new URLLoader;
xmlLoader.load(new URLRequest("http://localhost/Flash/PHP/general.php"));
xmlLoader.addEventListener(Event.COMPLETE, xmlLoaded);
function xmlLoaded(event:Event):void
{
xml = new XML(event.target.data);
trace(xml);
}
general.php
<?php
$sadarjanie = $_GET['vic'];
$nomer = $_GET['number'];
$iden = $_GET['id'];
$content = $sadarjanie;
$number = $nomer;
$id = $iden;
if($id == 1){$prom = "grafiti";}
else if($id == 2){$prom = "blondinki";}
else if($id == 3){$prom = "bigbrother";}
else if($id == 4){$prom = "znaeteli4e";}
else if($id == 5){$prom = "razni";}
else if($id == 6){$prom = "laforizmi";}
else if($id == 7){$prom = "pc";}
else if($id == 8){$prom = "ivan4o";}
else if($id == 9){$prom = "borci";}
echo $prom;
//the XML
echo "<?xml version=\"1.0\" ?>\n";
echo "<root>\n";
include("C:\wamp\www\Flash\PHP\Includes\conn.php");
$query = mysql_query("select * from $prom");
while($array = mysql_fetch_array($query))
{
echo "<vic id=" . $array['0'] . ">" . $array['1'] . "</vic>\n";
}
echo "</root>\n";
$number = 0;
if($number > 0)
{
if($number == 1){$var = "grafiti";}
else if($number == 2){$var = "blondinki";}
else if($number == 3){$var = "bigbrother";}
else if($number == 4){$var = "znaeteli4e";}
else if($number == 5){$var = "razni";}
else if($number == 6){$var = "laforizmi";}
else if($number == 7){$var = "pc";}
else if($number == 8){$var = "ivan4o";}
else if($number == 9){$var = "borci";}
include("C:/wamp/www/Flash/PHP/Includes/conn.php");
mysql_query("SET NAMES CP1251");
$query = "INSERT INTO `vicove`.`$var` (`id` ,`vic` )VALUES (NULL , '$content')";
$row = mysql_query($query);
mysql_close($conn);
if($row){echo "Вица ви беше оспешно добавен";}
else{echo "Вица ви не беше добавен";}
}
?>
conn.php
<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "vicove";
$conn = mysql_connect($host, $user, $pass);
$select_db = mysql_select_db($db, $conn);
?>