PHP - How To Add Class To A Link?

If a class has a constructor but also has a static method, if I call the static method does the constructor run so that I can use an output from the constructor in my static method?

--Kenoli

Ok. I know you can pass the object of a class as an argument. Example:


class A {
    function test() {
        echo "This is TEST from class A";
    }
}


class B {
    function __construct( $obj ) {
        $this->a = $obj;
    }

    function test() {
       $this->a->test();
    }

}


Then you could do:

$a = new A();
$b = new B($a);


Ok so that's one way i know of.

I also thought that you could make a method static, and do this: (assuming class A's test is 'static')

class B {
    function test() {
       A::test(); 
   }
}


But that is not working.


I'd like to know all possible ways of accomplishing this. Any hints are appreciated. thanks

Hi

Can you call Class A's methods or properties from Class B's methods?

Thanks.

I have an existing instance of my class Database, now I want to call that instance in my Session class, how would I go about doing this?

Hi,

I need to be able to call a class based on variables.  E.G.

I would normally do:
Code: [Select]
$action = new pattern1()
but i would like to be able to do it dynamicaly:
Code: [Select]
$patNum = 1;
$action = new pattern.$patNum.()
Im wondering if that's possible?  If so what would the correct syntax be?

Many Thanks.

Hi people!

class FirstOne{

public function FunctionOne($FirstInput){
//do stuff and output value
return $value1;
}

}


Then:-


class SecondOne{

public function FunctionTwo($AnotherInput){
//do stuff and output value
return $value2;
}

}


What I want to know is this, if I want to use FunctionOne() in Class SecondOne do I do it like this:-

(Assume as I have instantiated the first class using $Test = new FirstOne(); )


class SecondOne{

function SecondedFunction(){
global $Test;
return $Test->FunctionOne();
}

public function FunctionTwo($AnotherInput){
//do stuff and output value
return $value2;
}

public function FunctionThree(){

//some code here
$this->Test->SecondedFunction();<--I think as I can omit the $this-> reference
}
}


My point is: Do I have to do it this way or is there way of having this done through __construct() that would negate the need for a third party function?

I have a version working, I just think that it is a little convoluted in the way as I have done it, so I thought I would ask you guys.

Any help/advice is appreciated.

Cheers
Rw

I have two classes:

## Admin.php

<?php
class Admin
{
public function __construct()
{
include("Config.php");
}

/**
* deletes a client
* @returns true or false
*/
function deleteClient($id)
{
return mysql_query("DELETE FROM usernames WHERE id = '$id'");
}
}
?>


## Projects.php

<?php
class Projects
{
public function __construct()
{
include("Config.php");
$this->admin = $admin;
$this->dataFolder = $dataFolder;
}

/**
* Deletes a project
* @returns true or false
*/
function deleteProject($id)
{
$root  = $_SERVER['DOCUMENT_ROOT'];
$theDir  = $root . $this->dataFolder;
$sql  = mysql_query("SELECT * FROM projectData WHERE proj_id = '$id'");
while ($row = mysql_fetch_array($sql))
{
$mainFile = $row['path'];
$thumb = $row['thumbnail'];
if ($thumb != 'null')
{
unlink($theDir . "/" . substr($thumb,13));
}
unlink($theDir . "/" . substr($mainFile,13));
}
$delete = mysql_query("DELETE FROM projectData WHERE proj_id = '$id'");
$getDir = mysql_query("SELECT proj_path FROM projects WHERE id = '$id'");
$res = mysql_fetch_array($getDir);
rmdir($theDir . "/" . $res['proj_path']);
return mysql_query("DELETE FROM projects WHERE id = '$id'");
}
}
?>


How can I call deleteProject() from within Admin.php?

I do know how to do this but I am curious about whether or not there is a "preferred" way to do this. I know there are a couple ways to use a class (I'll call Alpha_Class) within another class (I'll class Beta_Class)

Let's say we have this simple class (Beta_Class):

class beta {
    function foo(){
    
    }
}


If I wanted to use the Alpha Class within the Beta Class, I could any number of things. For example:

class beta {
    function foo(){
        $this->alpha = new alpha;
        //$this->alpha->bar();
    }
}


Or you could simply use the $GLOBALS array to store instantiated objects in:

$GLOBALS['alpha'] = new alpha;
class beta {
    function foo(){
        //GLOBALS['alpha']->bar();
    }
}


You could even declare Alpha_Class as a static class and thus would not need to be instantiated:

static class alpha {
    static function bar(){}
}

class beta {
    function foo(){
        //alpha::bar();
    }
}


Those are the only ways I can think of right now. Are there any other ways to accomplish this? I was wondering which way is the best in terms of readability and maintainability.

Hi all,

I have two classes. Registration and Connection.

Inside a registration.php I include my header.php, which then includes my connection.php...

So all the classes should be declared when the page is loaded.

This is my code:


registration.php:

<?php include ('assets/header.php'); ?> 
        
    <?php
        class registration{                    
            public $fields = array("username", "email", "password");
            public $data = array();
            public $table = "users";
            public $dateTime = "";
            public $datePos = 0;
            public $dateEntryName = "date";
            
            function timeStamp(){
                return($this->dateTime = date("Y-m-d H:i:s"));
            }
            
            function insertRow($data, $table){      
                    
                    foreach($this->fields as $key => $value){
                        mysql_query("INSERT INTO graphs ($this->fields)
                        VALUES ('$data[$key]')");
                    }
                    
                    mysql_close($connection->connect);
            }
            
            function validateFields(){
            
                $connection = new connection(); 
                $connection->connect();
                        
                foreach($this->fields as $key => $value){
                    array_push($this->data, $_POST[$this->fields[$key]]);
                }
                $this->dateTime = $this->timeStamp();
                array_unshift($this->data, $this->dateTime);                
                array_unshift($this->fields, $this->dateEntryName);   
                
                foreach($this->data as $value){
                    echo "$value";
                }
                
                $this->insertRow($this->data, $this->table);
            }
        }
        
        $registration = new registration();
        $registration->validateFields();
    ?>

<?php include ('assets/footer.php'); ?>



At this point I cannot find my connection class defined on another included/included page.



                $connection = new connection(); 
                $connection->connect;





config.php (included within header.php)

<?
    class connection{ 
        public $dbname = '**';   
        public $dbHost = '**';
        public $dbUser = '**';
        public $dbPass = '**';
        public $connect;
        
        function connect(){
            $this->connect = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass) or die ('Error connecting to mysql');
            mysql_select_db($this->dbname, $this->connect);
        }
    }
?>



Any ideas how to call it properly?

How does one go about using one class inside another?

For example, building a class that does some series of functions, and uses a db abstraction layer class in the process?

I have a class in which I have a function called connection. I am now trying to call this function from another class, but it will not work. It works if I put the code in from the other function rather than calling it but that defeats the purpous.

class locationbox

{
   function location()
   {
   
   
   $databaseconnect = new databaseconnect();
         $databaseconnect -> connection();{

$result = mysql_query("SELECT * FROM locations");

while($row = mysql_fetch_array($result)) // line that now gets the error, mysql_fetch_array() expects parameter 1 to be resource, boolean given //in
  {
  echo "<option>" . $row['location'] . "</option>";
    }
   
}
}}

Well the title may seem a bit confusing, but heres an example:

Code: [Select]
<?php
class User{
   public $uid;
   public $username;
   protected $password;
   protected $email;
   public $usergroup;
   public $profile;
   

  public function __construct($id){
      // constructor code inside
  }

  public function getemail(){
      return $this->email;
  } 
   
  public function getusergroup(){
      return $this->usergroup;
  }

  public function getprofile(){
      $this->profile = new UserProfile($this->uid);
  }
}

class UserProfile(){
   protected $avatar;
   protected $bio;
   protected $gender;
   protected $favcolor;
   

  public function __construct($id){
      // constructor code inside
  }

  public function formatavatar(){
      // avatar formatting code inside
  }

  public function formatusername(){
      // format username?
  } 
}
?>

As you can see, the User class(an outer class) has a property called Profile, which can be instantiated as a UserProfile object(an inner class). The two objects have distinct functionalities, but there are times when the UserProfile object needs to access property and methods from the user object. I know its easy for outer class to access methods from inner class by using the single arrow access operator twice, but how about the other way around?

Lets say from the above example the userprofile can format the username displayed to the screen by adding a sun to the left of the username if the usergroup is admin, a moon if the usergroup is mod, and nothing if its just a member. The usergroup property is stored in the outer class, and can be accessed with this $user->getusergroup() method only. I know I can always do the hard way by passing a user object to the method's argument, but is there an easier way for the inner class UserProfile to access properties/methods for outerclass User? If so, how can I achieve that?

Hi

I've got this database I created with fields

ProductId   ProductName  Image

I've managed to get it to list the ID,productname, and Image urls
in a list.

My next step is to have the image field actually display
an image and make it clickable:

heres what I've done so far:

Code: [Select]
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("productfeed", $con);

$result = mysql_query("SELECT * FROM productfeeds");

echo 
"<table border='0'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Image</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ProductID'] . "</td>";
  echo "<td>" . $row['ProductName'] . "</td>";
  echo "<td>" . $row['ImageURL'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>
Heres what I want to do:

Code: [Select]
while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ProductID'] . "</td>";
  echo "<td>" . $row['ProductName'] . "</td>";

 // my changes beneath
  echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>
Can you guys point me in the right direction?

Many thanks

Hello. I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.  One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong. All help is 

Hi Support,

I have a form, where it collects user description input. I can collect the inputs and store it with newline. The issue is - how to collect the http link to actual hyperlink ref during display.  The following is my code:

<textarea name="description" cols="50" rows="10" id="description"><?php echo str_replace("<br>", "\n", $description);?></textarea></td>

For example, User input:

Hi, Check it out - http://www.google.com/

I would like to display google link as href so that Viewers can click the link and go to the page. Right now, it is not href and user need to copy the link to new tabs or pages and then it can come.

Thanks for your help.
Regards,
Ahsan