PHP - Define Data-type In Parameter?

I have a payment form with the following code...

Code: [Select]
<!-- Expiration Month -->
<span class="group">
<label class="innerlabel" for="expMonth">Month</label><!-- -->
<select id="expMonth" name="expMonth">
<option></option>
<option>01</option>
<option>02</option>
<option>03</option>
<option>04</option>
<option>05</option>
<option>06</option>
<option>07</option>
<option>08</option>
<option>09</option>
<option>10</option>
<option>11</option>
<option>12</option>
</select>
</span>

How is that data represented as far as Data-Type?? 

Is it treated as a String or an Integer??

How can I ensure that when I sent it to the Payment Gateway that it is treated as a Two-Digit value??

Thanks,



Debbie

I am using a query to select everything out of a table in to an array. I have 4 fields as follows name, email, comment and datatime. What i want to do is use one of the stated function to grab the datetime elements and split them at the break point, but i am displaying more than one record with in the table, i think this is were i am fucking up.

for example i tryied this:
Code: [Select]
                $commentsTable = "comments";
            $comments = "SELECT * FROM  $commentsTable";
$commentResults = mysql_query($comments);
$commentRows = mysql_fetch_array($commentResults);

$dateTimeSplit = explode(" ", $commentRows['datetime']);

               
               <?php while($commentRows = mysql_fetch_array($commentResults)){?>
    <div id="comments">
    <div id="CommentWrapper">
            <div id="comment">
                <div id="UserComment"><? echo $commentRows['comment'];?></div>
                 // here is were am having the problem!!!
                <div id="PostDateTime"><i>Posted: </i><? echo $dateTimeSplit['0'];?><i> Time: </i><? echo $dateTimeSplit['1'];?></div>
            </div>
        </div>
    </div>
<?php } mysql_close(); ?>

any sudgestions?

In many cases, variable have to be explicitly defined for whatever reason.

However, often it is not required for a simple web page script.

Yet, even in those cases, if a variable is NOT explicitly defined, the web server error log (example: my_domain.com-error_log) will rack up a lot of entries such as this: Quote

[Fri Dec 31 01:31:43 2010] [error] [client 123.123.123.123] PHP Notice:  Undefined variable: some_variable_name in /home/my_domain/public_html/some_page.php on line 6


A busy site could result in large error logs...

What is considered "best practice"? Should all variables always be explicitly defined with the var key word?

Hi all,
 
I was reading a bit in the manual about the defined function. Since i have seen it being used quite a lot to disallow direct access. But one of the comments made me think if there are any alternatives to get the same.
it's this comment: http://www.php.net/manual/en/function.defined.php#89886

If i read it correct it tells that defined() seems to be pretty slow. I was thinking, if that is the case what are the alternatives...

Does anyone know how to interpret this comment and if there are alternatives? I am not yet a guru tofully understand the results.

cheers!

Hello all,

I just created this function that I want to use in CLi mode. It works perfectly in regular browser mode, but gives me this error on Cli mode if I do convertToCamelCaps('TEST STRING');

PHP Catchable fatal error:  Argument 1 passed to convertToCamelCaps() must be an instance of string, string given in file...


if (!function_exists('convertToCamelCaps')) {
function convertToCamelCaps(string $string, $initialCaps=FALSE) {
if ($words = explode(' ',$string)) {
foreach ($words as $index => &$word) {
$word = strtolower($word);
if($index==0 && $initialCaps===FALSE) { continue; }
$word = ucwords($word);
}
return implode('',$words);
}
return FALSE; 
}
}


If I remove the string datatype requirement in the function before the function argument list, it works fine in CLi mode.

Hello,   I recently posted here about an issue I was having with my database orientated products page.   I have now run into another problem where say if, /db.php was typed or /db.php?p=IDoNotExist was typed, it returns blank.   I have in my code the desired content to be displayed, but it just doesn't seem to want to make a show.   I was also wondering if it is possible to show different content for whatever the URL is, so for no parameter, the content about the products, and a non existent one, maybe "Product not found"?   Here is my code:

<?php
	$db=mysql_connect  ("localhost", "webwibco_charlie",  "Hello123") or die ('I cannot connect to the database  because: ' . mysql_error()); 
	$mydb=mysql_select_db("webwibco_products"); 

	include("header.php");

	$status = htmlspecialchars( @$_GET ['p'] );

	if ($status == "floorpuzzles") {

		     echo "<h1>Our Floor Puzzles</h1>";
		$sql="SELECT  ID, Name, Tags, Description, Category FROM products WHERE Category LIKE '%" . FloorPuzzles .  "%'"; 
		$result=mysql_query($sql); 
		while($row=mysql_fetch_array($result)){ 
			$Name  =$row['Name'];
			$ID  =$row['ID']; 
			$Description  =$row['Description'];  
			echo "<div class=\"box\">";
			echo "<h1>$Name</h1>";
			echo "<div class=\"floorbox\"><a href=\"?p=$ID\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"small\"></a></div>";
			echo "<h2>$Description</h2>";
			echo "</div>";
		}
?>
<?
}else{



	if ($status == $_GET["p"]) {
		$sql="SELECT  ID, Name, Tags, Description, Pieces, Size, Barcode, Category FROM products WHERE ID = '" . $_GET['p'] .  "'";   
		$result=mysql_query($sql);   
		while($row=mysql_fetch_array($result)){  
			$Name  =$row['Name'];
			$ID  =$row['ID']; 
			$Description  =$row['Description']; 
			$Pieces  =$row['Pieces']; 
			$Size  =$row['Size']; 
			$Barcode  =$row['Barcode'];   
			echo "<div class=\"1\">";
			echo "<h1>$Name</h1>";
			echo "<div class=\"bigbox\">";
			echo "<div class=\"floorbox\"><img src=\"images/products/catalogue/big/floorpuzzles/$ID.jpg\" class=\"big\"></div>";
			echo "</div>";
			echo "</div>";
			echo "<div class=\"2\">";
			echo "<p>Puzzle Pieces: $Pieces</p>
				  <p>Puzzle Size: $Size</p>
				  <p>Barcode: $Barcode</p>";
			echo "</div>"; 
		} 
		}else{
?>
<?
			echo"<h1>Our Products</h1>
				<p>Our jigsaw puzzles are hand cut by skilled craftsmen and therefore each one is unique with self-correcting pieces. There is a strict quality control process at all stages by our highly experienced staff. The puzzles are durable and provide fun and excitement, enhancing learning and a child’s development.<p>
				<p>All of our jigsaws are made using materials from sustainable resources grown in managed forests. Where possible we support companies in the UK and source our components locally, most of our suppliers are in the East Midlands, many in Derbyshire and Nottinghamshire. We keep packaging to a minimum and take our environmental and ethical responsibilities very seriously.</p>
				<p>Reducing waste and recycling was a way of life for us before it became fashionable. We are constantly searching for new ideas and consult teachers when developing our jigsaws, which are often used within the national curriculum.</p> 
				<p>As well as making our own range, we manufacture for leading suppliers to the education market. Check for \"Made in Britain\" and it is probably made by us.</p>
				<p>We have a wide variety of products available for viewing, from classic floor puzzles to innovative inset trays. You can take a look at all our products on this page, simply use the navigation buttons to your left.</p>";

				
				}}

include("footer.php");
?>

The final echo is what I wish to be displayed on the URL without or with an invalid parameter.   Here is my site URL: http://www.webwib.co...saws/search.php (note that only the "Floor Puzzles" category has content within it).   Thank you in advance for assistance.

I have the following code just to insert a username and hashed password into the database but somehow I am getting this error and I couldn't find out where I am doing it wrong...can someone please give me a hand?   I tried it in two ways and both errors... the first few lines are just connecting database which worked fine and a password.php so I can use password_hash() with my php version

$hash = password_hash('xx', PASSWORD_BCRYPT, array('cost' => 10));
$username = 'xx';


$insertQuery = $db->prepare("
		INSERT INTO et_todo (username, password)
		VALUES (:username, :hash)
	");

$insertQuery->execute(array(
		'username' => $username,
		'password' => $hash
	));

also tried

$hash = password_hash('xx', PASSWORD_BCRYPT, array('cost' => 10));

$insertQuery = $db->prepare("
		INSERT INTO et_todo (username, password)
		VALUES ('xx', :hash)
	");

$insertQuery->execute(array(
		'username' => 'xx',
		'password' => $hash
	));

hi there,

i am fairly new to OOPs in php, i get an error when i declare the argument type (as object) in a function and pass the same type (object).



class eBlast {
       public static function getEmail(object $result) {
           return $result->email;
       }
}

$r = mysql_fetch_object($query);
eBlast::getEmail($r);

echo gettype($r); // outputs: object



error is :

Code: [Select]
Catchable fatal error: Argument 1 passed to eBlast::getEmail() must be an instance of object, instance of stdClass given, called in C:\wamp\www\integra\client\pl_eblast\admin\send_emails.php on line 145 and defined in C:\wamp\www\integra\client\pl_eblast\app\app.eBlast.php on line 8
if i remove the type declaration in the function it works, but just would like to know why it shows error when pass the same type, also isnt mysql_fetch_object is the instance of stdclass?

thanks in advance!

I have at the very top of my web page the following. Not inside any function in case that would have a scope issue.

Define ('GROUP',  '1');
Define ('ROUNDOF16',  '2');
Define ('QUARTERFINAL',  '3');
Define ('SEMIFINAL',  '4');
Define ('FINAL',  '5');

Lower down I have
if($Round == 5)

This works but this
if($Round == FINAL) has error

Parse error: syntax error, unexpected T_FINAL in  .......

I thought this was correct !

im doing it wrong can some one help me ? , thanks
{
Code: [Select]
define('userlevel'( 9="admin" , 1="guest"));}

Hi, I have the define(); function  where i need to  put the variable for directory name which is relative ... How to do that ...

Code: [Select]
<?php
if (TARGET_PLATFORM == FILE_SYSTEM_PLATFORM)
{
/**
  * if FILE_SYSTEM_PLATFORM is selected as TARGET_PLATFORM,
  * then assign root folder adress to FILE_ROOT macro
  */
 
define("FILE_ROOT", "$diro");

}
public function __construct($fileRoot) {
if (is_dir($fileRoot)) 
{
$this->fileRoot = $fileRoot;
//if ($this->fileRoot[strlen($this->fileRoot) - 1] != '/' ) {
// $this->fileRoot .= '/';
//}

}
else {
die($fileRoot .' is not a directory ');
}

}
?>

When i go with this, it says this: " 'dir name ' is not a directory".  But I need the variable for this to work as I want to. Does anyone have an idea?

To make it clear, this is only the small part of code, which is a part of the directory root for the directory tree.

Ive got a simple function that's counting percentages of the results, and what I want is when the first line does $variable / 100 - to go on 2 decimals(ex. 0.72142141 what I want is to write 0.72).

Code: [Select]
function postotak(){
$p = $bodovi / 100;
$postotak = $p * 100;

Imagine a site programmed using full OOP.

Where would you store information like this:

$companyName = "ABC Ltd";
$companyPhone = "02476 999 999";
$companyAddress etc...

Would you just define them in the public scope? (and use global or pass them in when needed??)

Is there a better way?

They will be needed in more places than just a navigation bar and would be needed by more than one class.

im having a problem getting this code to work...i know it may not make sence why i would do this but it is for a larger script...is there any reason why it wont echo please?

define ('TEST_VAR', 'hello');
$var = "TEST";
echo {$var}_VAR;