PHP - [object Htmlinputelement]

Hey all,

I want to have an object that has a property which is an object containing instances of other objects. I try this:
Code: [Select]
class Blog extends Posts {
public $has_posts;

public function __construct($a,$b,$c){
$has_posts = (object) array_merge((array) $a, (array) $b, (array) $c);
}
}

class Posts {
public $b;

public function __construct($b){
$this->b = $b;
}
}

$post1 = new Posts(1);
$post2 = new Posts(2);
$post3 = new Posts(3);

$blog = new Blog($post1,$post2,$post3);
var_dump($blog->has_posts); //null
foreach($blog->has_posts as $post){ //Invalid argument supplied for foreach()
echo $post->b;
}

But as you see, has_posts is null, not an object containing other objects.

Thanks for response.

I have a Soup object in a Bowl object in a Microwave object.

The Microwave object has the methods:

receiveItem(Bowl $b)
cookItem(???)

I want the Microwave to be able to set the Soup->temperature but I'm not how to do that?

Make sense?



TomTees

I made a small script and it says that my database object isn't an object.

I connect to my pdo database via:

<?php

try {
$db = new PDO("mysql:host=localhost;dbname=database", "root", "root"); 
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
}  
catch(PDOException $e) {  

echo $e->getMessage();
  
}

?>

 

The error I get is

Code: [Select]
Notice: Undefined variable: db in /Users/JPFoster/Sites/jobiji_sandbox/extensions/tasks.php on line 21

Fatal error: Call to a member function query() on a non-object in /Users/JPFoster/Sites/jobiji_sandbox/extensions/tasks.php on line 21


here's where the error is
$grabber = $db->query('SELECT * WHERE user_id =' . $userid);
$grabber->setFetchMode(PDO::FETCH_OBJ);

while ( $row = $grabber->fetch() ){

$display = '<div>Task Name: ' . $row->name .'</div>';
$display .= '<div>Task Details: <p>'. $row->details  .'</p></div>';
//$display .= '<div>'. $this->checkifDone($row->task_status) .'</div>';
$display .= '<div>Task Due: '. $row->task_due . '</div>';
echo $display;
}


could anyone tell me where this issue is originating I followed a fairly simple tutorial that didn't seem to work.


Thanks!

Ok. So, I'm trying to retrieve the id value from an object.  Here is the code I have that works so far:
Code: [Select]
$object = 'object';
$koid = $this->$object;
echo '<pre>'; print_r($koid); echo '</pre>';

and this is what is returns:
Code: [Select]
Order Object
(
    [id:protected] => 5
    [stand_by:protected] => test
    [problem:protected] => test
)

what do I need to do to make $koid equal the id value (5 in this case)?

Hi

I have a login script which goes at the top of several pages in my site.  It works fine but I am trying to save some coding and re-writing out my script at the top of each of the pages.  Instead I have made a page called login.php and put all of the code for the script in there. 

Looks like this....
<?php
//login script 
if(isset($_POST['log'])) {
$emaillog = stripslashes($_POST['emaillog']);
$passwordlog = stripslashes($_POST['passwordlog']);
$emaillog = mysql_real_escape_string($_POST['emaillog']);   
$passwordmd5 = md5($passwordlog);  
//generate random session id which i want to follow the user round the site!
$CheckUser = "SELECT * FROM members WHERE email='".$emaillog."' AND password='".$passwordmd5."'";
$userDetails2 = mysql_query($CheckUser);
$userInfo = mysql_fetch_array($userDetails2);
$count = mysql_num_rows($userDetails2);
 if($count != 0) {
  $_SESSION['memberID'] = $userInfo['memberID'];
$_SESSION['accesslevel']= $userInfo['accesslevel'];
$_SESSION['email']= $userInfo['email'];

//add an id that will be carried throughout the user until the session is destroyed
 function getUniqueCode2($length2 = "")
 { $code2 = md5(uniqid(rand(), true));
if ($length2 != "") return substr($code2, 0, $length2);
else return $code2; }

$randomKey = getUniqueCode2(25);
$_SESSION['key'] = $randomKey;

//the user must have an access level of 2, before they can login
if($_SESSION['accesslevel']=='2')
$url = "loggedin/index.php";        
header("Location: ".$url."");   
$success = '1';
}
else {
$success = '0';
$logged = "incorrect login details" or die(mysql_error());
}

//now we need to update the login table to reflect a login
if($success =='1'){
$time = date("Y-m-d h:m:s");
$qupdate_mem_logins = "INSERT INTO logindetails (`email`,`time`) VALUES ('".$emaillog."','".$time."')";
 $rupdate_mem_logs = mysql_query($qupdate_mem_logins) or die(mysql_error());

        ?>

Now on the index page how do I reffer to this login script when someone fills in the form  At the minute I have the following code but it is not checking or running anything.

<?php
//login script 
if(isset($_POST['log'])) {
include('login.php');
}
}
?>