PHP - Convert Variable Into Utf-8

I have just re-installed Xampp and suddenly my sites are now displaying lots of:

Notice: Use of undefined constant name - assumed 'name' in ...
Notice: Use of undefined constant price - assumed 'price' in ...

this is an example of the line its refering too:

$defineProducts[1001] = array(name=>'This is a product', price=>123);

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

 

Olá,   I have this codE:

        $cb = 'str_replace("' . $this->LE . '", "", "<$1>");';
        $encoded_value[$key] = preg_replace("/<([^>]+)>/e", $cb, $encoded_value[$key]);

error:   "PHP message: PHP Deprecated:  preg_replace(): The /e modifier is deprecated, use preg_replace_callback instead   How can i transform this code to preg_replace_callback ?   thanks

I have the following code in javascript, I would like to rewrite it in PHP, if someone can help thank you very much.

my code: http://pastebin.com/Jnt8RiYj

How can I echo this as seconds?

$date_at_timezone = new DateTimeZone('Asia/Shanghai');
$date = new DateTime(); 
$date->setTimezone($date_at_timezone);
echo $date->format('l dS F Y h:i:s A') . "Asia/Shanghai<br>";

Hi Coders,     any body can convert the following code into codeigniter. 

if($_GET['type'] == 'country'){
	$result = mysql_query("SELECT name FROM country where name LIKE '".strtoupper($_GET['name_startsWith'])."%'");	
	$data = array();
	while ($row = mysql_fetch_array($result)) {
		array_push($data, $row['name']);	
	}	
	echo json_encode($data);
}

Thanks

I have come across a little feature I need to create to be able to upload .doc files and convert them to .pdf using PHP.

I was wondering if anyone had any idea how I would go about achieving this, I am not asking for the code, I'm just asking on how it could be accomplished...I've searched Google and other PHP related forums but none have the answer to which I am looking for.

Any ideas or suggestions are greatly appreciated.

Regards, PaulRyan.

I have a form that creates rows of data input textboxes depending on a user input number of things.  I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row.  All this is working fine.

My problem is I need to get the data inserted into this table of textboxes into an array.

Here's my code where I attempt to to this (it does not work):

Code: [Select]
$temp = $_SESSION['Num_Part'];
$count = 1;
while ($count <= $temp){
  $temp2[$count] = "'Participant_P".$count."'";
  //echo $temp2[$count]."<br/>";
  $temp3[$count]=$_POST[$temp2[$count]];  //here's the problem
  $temp4[$count] = "'Result_P".$count."'";
  $temp5[$count]=$_POST[$temp4[$count]];  //here's the problem
  //echo $temp4[$count]."<br/>";
  $count++;
}

The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes.

Can a variable be used in a POST argument and if so what is the correct syntax?

If not, is there some other simple solution to harvest the data into an array.  I understand I can harvest by explicitly accessing each key in the post assoc array.  But this could be dozens of rows of input fields.

Thanks in advance for your help here.  I couldn't find anything online re this topic.

I have a script that adds points together based upon the placing. This is the actual script:

Code: [Select]
<? $points = 0;

if($place === '1st') {$points = $points + 50;}
elseif($place === '2nd') {$points = $points + 45;}
elseif($place === '3rd') {$points = $points + 40;}
elseif($place === '4th') {$points = $points + 35;}
elseif($place === '5th') {$points = $points + 30;}
elseif($place === '6th') {$points = $points + 25;}
elseif($place === '7th') {$points = $points + 20;}
elseif($place === '8th') {$points = $points + 10;}
elseif($place === '9th') {$points = $points + 10;}
elseif($place === '10th') {$points = $points + 10;}
elseif($place === 'CH') {$points = $points + 50;}
elseif($place === 'RCH') {$points = $points + 40;}
elseif($place === 'TT') {$points = $points + 30;}
elseif($place === 'T5') {$points = $points + 30;}
elseif($place === 'Champion') {$points = $points + 50;}
elseif($place === 'Reserve Champion') {$points = $points + 40;}

echo "Total HF Points: $points";
?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end.

It is included into a file, in this area:

Code: [Select]
<div class="tabbertab">
<h2>Records</h2>
<? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE";
$result92 = mysql_query($query92) or die (mysql_error());

echo "<table class='record'>
<tr><th>Show</th> <th>Class</th> <th>Place</th></tr>
";

while($row92 = mysql_fetch_array($result92)) {
$class = $row92['class'];
$place = $row92['place'];
$entries = $row92['entries'];
$race = $row92['show'];
$purse = number_format($row92['purse'],2);

echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>";
}
?>
<tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr>
</table>
</div>

This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong?

Hello everyone,

I can get Test 2 to successfully operate the if statement using a variable variable.

But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed?

Code: [Select]

<?php
# TEST 1
$_SESSION[test_variable] = "abcd";
$session_variable_name = "_SESSION[test_variable]";
if ($$session_variable_name == "abcd")
{
echo "<br>line 373, abcd<br>";
}

# TEST 2
$test_variable = "efgh";
$test_variable_name = "test_variable";
if ($$test_variable_name == "efgh")
{
echo "<br>line 379, efgh<br>";
}

?>
Many thanks,

Stu

 

Hello guys.

Im not familiar with PHP but I was asked to convert our Homepage which uses a Database to store tickets from 5.x to 7.3.9 .

Now I get this error:

 

I'd appreaciate any help from you guys, tia. :)

I am trying to convert colors from Hex to HSL (Not HSL to Hex). I am using a PHP function for this purpose. But It's not working properly for some colors. For example for #e04c4c the HSL should be (0, 70%, 59%) which it isn't the case with the function.

function hexToHsl($hex) {
    $red = hexdec(substr($hex, 0, 2)) / 255;
    $green = hexdec(substr($hex, 2, 2)) / 255;
    $blue = hexdec(substr($hex, 4, 2)) / 255;

    $cmin = min($red, $green, $blue);
    $cmax = max($red, $green, $blue);
    $delta = $cmax - $cmin;

    if ($delta === 0) {
        $hue = 0;
    } elseif ($cmax === $red) {
        $hue = (($green - $blue) / $delta) % 6;
    } elseif ($cmax === $green) {
        $hue = ($blue - $red) / $delta + 2;
    } else {
        $hue = ($red - $green) / $delta + 4;
    }

    $hue = round($hue * 60);
    if ($hue < 0) {
        $hue += 360;
    }

    $lightness = (($cmax + $cmin) / 2) * 100;
    $saturation = $delta === 0 ? 0 : ($delta / (1 - abs(2 * $lightness - 1))) * 100;
    if ($saturation < 0) {
        $saturation += 100;
    }

    $lightness = round($lightness);
    $saturation = round($saturation);

    //return "hsl(${hue}, ${saturation}%, ${lightness}%)";
    return array($hue, $saturation, $lightness);
}

This is how I used it:
 

$templatePrimaryColor = '#e04c4c';

$templatePrimaryColor = str_replace("#", "",$templatePrimaryColor);

list($h,$s,$l) = hexToHsl($templatePrimaryColor);

$primaryColor = "hsl(${h}, ${s}%, ${l}%)";

This is the output:

echo '<pre>',print_r(hexToHsl($templatePrimaryColor)).'</pre>';

Array
(
    [0] => 0
    [1] => 99
    [2] => 59
)
1

Does anyone know what is the problem there?

Thank you.

Hi,
  I want to place the following HTML statement inside an echo statement, i.e. in PHP I want to output an image that is also a link.
        <a href="#"><img src="Somefolder/animage.jpg" width="125" height="156" alt="some image" /></a>

  All ideas will be greatly appreciated.

How do I update a table and convert say, 66.249.75.215, from one column to another column as 0x249f94bd7. Both columns are varbinary. I'm using PDOConnect and this isn't working:

    $stmt = $pdo->prepare("UPDATE `download` SET `ip_address` = inet_aton(`ADDRESS`) WHERE ip_address is null;");
    $stmt->execute() ;

I'm getting the following error:

Fatal error: Uncaught PDOException: SQLSTATE[HY000]: General error: 1411 Incorrect string value: '`test`.`download`.`ADDRESS`' for function inet_aton in /home/larry/web/test/public_html/index.php:31 Stack trace: #0 /home/larry/web/test/public_html/index.php(31): PDOStatement->execute() #1 {main} thrown in /home/larry/web/test/public_html/index.php on line 31