PHP - Any Reason Why These If Statements Aren't Working?

Hey


I have a script that builds an array but for some reason there is a loose integer in the variable which is confusing me and i think its the cause of my syntax errors in javascript. But i got no idea where it is coming from =/


Here is the function i use:

Code: [Select]
<?php
function img($id) {
      $img = array();
      $img['tiles'] = $this->db->getAll("SELECT DISTINCT t.id, CONCAT('data/tiles/',filename) AS f FROM `map_tiles` AS mt
               INNER JOIN `tiles` AS t ON (mt.tile=t.id)
                WHERE mt.map_id={$this->db->qstr($id)}");

      echo print_r($img); 
      die;            
      //exit(json_encode($img)); // turned off for testing
?>


The echo returns this:
Code: [Select]

Array
(
    [tiles] => Array
        (
            [0] => Array
                (
                    [id] => 10
                    [f] => data/tiles/floor.png
                )


        )

)
1  <----- why does this appear? And could this cause a syntax error if sent back for JS processing?


As you can see there is 1 showing at the end but i don't see why =/ Any ideas if thats suppose to happen ?

Hi all,

Something i forget to ask all the time, but now i don't.
While reading again about image sprites, it tells that it's nice because it decreases the amount of http requests.
Now I thought what about all those requires and includes of php aren't they doing the same thing? Not to mention OOP which would not exist without the 2(4).

Does anyone know if this indeed increases the amount of http requests. And if so, if there is a certain good practise to lower the amount, by just combining functions in 1 file.

I would love to hear from some guru's 

I have this code below in a while loop, that gets looped through about 10 times and is simply resizing a slightly larger image to fit within a 50px max square. There is other code on my page, but I've narrowed the slowness down to this specific bit of code (i.e. if I take this snippet out, the page loads instantly).  With this snippet, the page takes between 5 and 10 seconds to load, which seems absurd for 10 images.  FYI, the original images are no larger than 200 px on either side, so it's not like it's looping through large image files. 

Anyone know why this might be taking so long?  Should I be doing it differently somehow?

Code: [Select]

list($width,$height) = getimagesize($myimage);
if ($width > $height) {
$datasofar .= "width=50 /></span>";
} elseif ($height > $width) {
$datasofar .= "height=50 /></span>";
} else { //height and width must be equal so just set width = 50, but could just as easily set height to 50
$datasofar .= "width=50 /></span>";
}

I normally don't have a problem with concenation, but for some reason I can't get it work in this case.

I have two form variables: firstname_ord and lastname_ord.
I have tried concenating them several ways, and get nothing afterwards.

attempt 1:
$lastname = "firstname_ord" . " " . "lastname_ord";

attempt 2:
$firstname = "firstname_ord";
$lastname = "lastname_ord";
$fullname = $firstname . " " . $lastname;

attempt 3:
$fullname = "firstname_ord";
$fullname = $fullname. " " . "lastname_ord";

attempt 4:
$fullname = "firstname_ord";
$fullname .="lastname_ord";

attempt 5:

$fullname = (string)"firstname_ord";
$fullname .= " ";
$fullname .=(string)"lastname_ord";

Used to be a good option, but don't know anymore as password_hash() is now available.   Agree?   I understand that I shouldn't ever manually salt and disable the functions salting.  That being said, is there any reason to add a bit extra to the user's password (such as an internal ID and some random constant)?

I spent the last hour or so typing this code up, and for some reason I am getting a query error. I have reviewed & revised the code up and down for the past half hour and can't seem to figure out the problem.

Can someone look after this for me and tell me what I could be doing wrong?

Yes, I know my code is a bit sloppy and may use bad practice techniques, but it works for me.

Its a survey that I coded so I could collect data and place it on CPA ad listings. So I need this so work at some point soon.

My code:
<?php
$user = $_POST['user'];
$email = $_POST['email'];
$password = $_POST['pass'];
$paypal = $_POST['paypal'];
$q1 = $_POST['q1[favsite]'];
$q2 = $_POST['q2[isp]'];
$q21 = $_POST['q2.1[bill]'];
$email_services = $_POST['email_services'];
$ebay = $_POST['ebay'];
$amazon = $_POST['amazon'];
$q6 = $_POST['q6[purchase]'];
$q7 = $_POST['q7[social]'];
$q8 = $_POST['q8[bookmarks]'];
$q9 = $_POST['q9[search]'];
$q10 = $_POST['q10[homepage]'];
$q11 = $_POST['q11[5topsites]'];
$q12 = $_POST['q12[state]'];

if ($_POST['fin'] == "complete") 
{
  
     $dbc = mysqli_connect('localhost', 'root', 'password', 'database')
     or die('Could not connect');
     $query = "INSERT INTO user_data (id, user, email, password, paypal, q1[favsite], q2[isp], q21[bill], email_services, ebay, amazon, q6[purchase], q7[social], q8[bookmarks],             q9[search], q10[homepage], q11[5topsites], q12[state]) VALUES ('$user', '$email', '$password', '$paypal', '$q1', '$q2', '$q21', '$email_services', '$ebay', '$amazon', '$q6', '$q7', '$q8', '$q9', '$q10', '$q11', '$q12')";       
     
      mysqli_query($dbc,$query)
      or die('Error querying database');
      
    include_once("../phpmailer/class.phpmailer.php");
$mail = new PHPMailer;
$mail->ClearAddresses();
$mail->AddAddress('', '');
$mail->From = '';
$mail->FromName = '';
$mail->Subject = 'Thanks for finishing the survey!';
$mail->Body = "Hello, $user. This is a reminder that you have finished the survey and your credit is currently 
being processed. Please login to your account at ../../ to view the status of your credit & cash out.
";
if ($mail->Send()) {
    echo "<center>Mail Sent.</center>";
} else {
    echo $mail->ErrorInfo;
}
echo "<center><h2>Thanks for completing the survey! Please <a href='login.php'>login</a> to your account to view the status of your credit & cash out.</h2></center>";
}
?>

It has nothing to do with PHPMailer, I of course edited the variables just now so all my info wouldnt be public, but everything is fine untill you press submit & I get the or die() error message "Error querying database".

What the hell did I do wrong? Is it possible that I cant name variables in the format I used with most of them ($var1 = $_POST['var[desc]']; ?

The below is constantly loading the password page no matter what is enter in the password input on the form (on another page). I have checked the $login variable by echoing it and it is carrying the value from the field.


<?php
$off="yes";
$login=$_POST['login'];
if($off=="yes" && (!isset($_SESSION['login']) || $_SESSION['login']!="pass")){
    if($login=="" || $_SESSION['login']!="pass"){
    header('Location: ./index2.html');
    exit();
    }
    elseif(!isset($_SESSION['login'])){
    session_start();
    $_SESSION['login']=$login;
    header('Location: ./index.php');
    exit();
    }
}
elseif($off=="no" || $_SESSION['login']=="pass"){
content goes here.
}
?> 

How to write 2 "if" statements together?

if($userResult==$myResult) and ($userResult<0)
{

Hey Guys,

I am creating an array and I have a need to put a if statement inside it. Is this even possible?

Here is where I am at so far, its just not working.


 function load_permissions($gid)
    {
     $data = array();
     $this->db->select('`groups`.`name` AS `permission_name`, `permissions`.``');
$query = $this->db->get('`groups`,`permissions`');
if ($query->num_rows() > 0){
foreach ($query->result_array() as $row){
  $data[] = array(
  "name" => $row['permission_name'],
  "read" => if($row['level'] >= 1)
   { 
    img("img/true.gif")
   } else { 
    img("img/false.gif")
   },
 
  );
  }
 }
 
 $query->free_result();
return $data;



Thanks,
Peter

Hello, I am learning PDO SQL statements, I have mananged to connect using PDO:


Code: [Select]
try
{
$this->link = $dbh = new PDO('mysql:host='.$this->dbhost.';dbname='.$this->dbname.'', $this->dbuser, $this->dbpass);
}
catch (PDOException $e)
{
echo 'Connection failed: ' . $e->getMessage();
}


I am not getting any error, so I guess that is a good start. I then tried the PDO SQL statement out. My old mySQL query is commented out in the


Code: [Select]
if (isset($_SESSION['id']) && isset($_SESSION['password']))
{
$_SESSION['id'] = ( isset( $_SESSION['id'] ) ) ? $_SESSION['id'] : FALSE;
$_SESSION['password'] = ( isset( $_SESSION['password'] ) ) ? $_SESSION['password'] : FALSE;


//$logged = mysql_query("SELECT * FROM `db_members` WHERE `id`='".$_SESSION['id']."' AND `password` = '".$_SESSION['password']."'");
//$logged = mysql_fetch_array( $logged );

// the new pdo statement
$ff = $dbh->prepare('SELECT * FROM db_members WHERE id = '.$_SESSION['id'].' AND password = '.$_SESSION['password'].'');

$ff->execute();
$logged = $ff->fetchAll();

echo $logged['username'];
}

I am trying to assign the session to logged variable. So all I am asking is go into db_members and check the id and password that is the same as session id and password and collect the rows data such as username.

My login script and everything works perfectly, even session id and password is valid when echo'd but I cannot assign it to the logged variable like my old sql statements.


Code: [Select]
$logged = mysql_query("SELECT * FROM `db_members` WHERE `id`='".$_SESSION['id']."' AND `password` = '".$_SESSION['password']."'");
$logged = mysql_fetch_array( $logged );

I used other PDO statements and it works perfectly but I just don't understand why this is not working.... can I please get some help if you have any solution to this?

Anything wrong with this because I got a parse error ONLY after adding to create this.

<?php
                            $visible = array('Yes', 'No');
                            foreach($visible as $visible):
                            $visible2 = array('yes', 'no');
                            foreach($visible2 as $visible2):
                        ?>      

Hi Guys,
Newbie trying to understand PHP conditionals.
In the below Code: [Select]
[[*id]] is a global variable which equates to my current page id so it could have the value of 1 for example. I tested it using echo and the correct values pass to it in the cms I'm using. What I cant achieve is for the conditionals below to work correctly so perhaps you can see a syntax error or show me away of doing the same using an array or something perhaps? THANKS

Code: [Select]
<?php
$id='[[*id]]'

if ($id=="1" || $id=="2")
  echo "we are in 1 or 2";
elseif ($id=="3" || $id=="4")
  echo "we are in 3 or 4";
else
  echo "not in 1 to 4";
?>

I am trying to write an if statement that checks to make sure the information put in the text box is an imbedded file. Does anyone know how to do that?

Hey guys,

is there any way to have a LIKE condition on an IF statement?
Something like:
Code: [Select]
if $GET_['WTV'] LIKE %LOVE% {
$category=romance
Any clues?
Thanks in advance.