Please help - total newbie to php

[xenok]

Hi All,

I'm just getting to grips with the basics of php, and thus setting myself the task of creating a dynamic menu in Flash, php, and MySQL.

Question is, what am I doing wrong here?

<?

$dbHost = "localhost";
$dbUser = "";
$dbPass = "";
$dbName = "***";

function dbConnect() {

$link = @mysql_connect($dbHost, $dbUser, $dbPass) or die ("Unable to connect!";
mysql_select_db("$dbName" or die ("Unable to select database!";

link . "|";
{

// print output as form-encoded data
echo "&items=" . urlencode($labelStr) . "&urls=" .
urlencode($linkStr);
}

// close connection
mysql_close($connection);
?>

Any help greatly appreciated!

Cheers,

Will

[pjleonhardt]

r u getting any errors.. what happens when u run the page

[kobmat]

I think these lines of your code is messed up

$link = @mysql_connect($dbHost, $dbUser, $dbPass) or die ("Unable to connect!"; 
mysql_select_db("$dbName" or die ("Unable to select database!"; 

link . "|"; 

Change this to:

$link = @mysql_connect($dbHost, $dbUser, $dbPass) or die("Unable to connect!");

mysql_select_db($dbName) or die("Unable to select database!link");

Let me know if that fixes the error

[xenok]

<?

$dbHost = "localhost";
$dbUser = "";
$dbPass = "";
$dbName = "***";

function dbConnect() {

$link = @mysql_connect($dbHost, $dbUser, $dbPass) or die ("Unable to connect!");
mysql_select_db($dbName) or die ("Unable to select database!");

$labelStr = mysql_query("SELECT label FROM menu")
or die("Invalid query: " . mysql_error());

$linkStr = mysql_query("SELECT link FROM menu")
or die("Invalid query: " . mysql_error());

link . "|";
{

// print output as form-encoded data
echo "&items=" . urlencode($labelStr) . "&urls=" .
urlencode($linkStr);
}

// close connection
mysql_close($connection);
?>

So I've put in my version of a (non-working!) query, where table = menu and link_identifiers are fields "label" and "link". That's as far as I've got!

Thanks for the help so far, any more advice greatly appreciated.

[kobmat]

Try this..

<? 

$dbHost = "localhost"; 
$dbUser = "***"; 
$dbPass = "***"; 
$dbName = "***"; 

$link = @mysql_connect($dbHost, $dbUser, $dbPass) or die ("Unable to connect!"); 
mysql_select_db($dbName) or die ("Unable to select database!"); 

$SQLstr = "SELECT label, link FROM menu";

$res1 = mysql_query("SELECT label FROM menu") ;

$xml_str = "<?xml version\"1.0\" ?>";
$xml_str .= "<menu>";
while($rows = mysq_fetch_assoc($res1))
{
$xml_str .= '<item name="'.$rows['label'].'" link="'.$rows['link'].'">';
}
$xml_str .= "</menu>";

mysql_close($connection); 
header("Content-Type: text/plain");
echo $xml_str;
?> 

on the flash side use the XML object, for example;

var i = new XML();
i.ignoreWhite = true;
i.load("thescript.php");

i.onload = function()
{
_root.menuitems = new Array();

toplevel = this.firstchild;
for(count = 0; count < toplevel.childNodes.length; count++)
{
temp = toplevel.childNodes[count];

root.menuitems[count] = new Array();
root.menuitems[count]["label"] = temp.attributes["label"];
_root.menuitems[count]["link"] = temp.attributes["link"];
}

//Continue with the movie
}

Basically the php script return an XML document and then the actionscript side puts it into an array _root.menuitems;

so access the label for example, if you have a dynamic text box;

menutext.text = _root.menuitems[0]["label"];

Hope that helps

[silent]

Just noticed that you have variables
(ie dbname etc..)
and then you try to call them in a function.

You should use 'global' first :

$varname = 'hello there';

function say_hello() {
global $varname;

echo $varname;
}

Hope it helps.

[kobmat]

I noticed i lack another parameter to mysql_select DB, change

mysql_select_db($dbName) or die ("Unable to select database!");

to

mysql_select_db($dbName, $link) or die ("Unable to select database!");