PHP - Why Doesn't It Go To The Next Page?

Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way.

First set up everything for DB:

<?php
//connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//create the main database if it doesn't already exist
$query = 'CREATE DATABASE IF NOT EXISTS moviesite';
mysql_query($query, $db) or die(mysql_error($db));

//make sure our recently created database is the active one
mysql_select_db('moviesite', $db) or die(mysql_error($db));

//create the movie table
$query = 'CREATE TABLE movie (
        movie_id        INTEGER UNSIGNED  NOT NULL AUTO_INCREMENT,
        movie_name      VARCHAR(255)      NOT NULL,
        movie_type      TINYINT           NOT NULL DEFAULT 0,
        movie_year      SMALLINT UNSIGNED NOT NULL DEFAULT 0,
        movie_leadactor INTEGER UNSIGNED  NOT NULL DEFAULT 0,
        movie_director  INTEGER UNSIGNED  NOT NULL DEFAULT 0,

        PRIMARY KEY (movie_id),
        KEY movie_type (movie_type, movie_year)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die (mysql_error($db));

//create the movietype table
$query = 'CREATE TABLE movietype (
        movietype_id    TINYINT UNSIGNED NOT NULL AUTO_INCREMENT,
        movietype_label VARCHAR(100)     NOT NULL,
        PRIMARY KEY (movietype_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

//create the people table
$query = 'CREATE TABLE people (
        people_id         INTEGER UNSIGNED    NOT NULL AUTO_INCREMENT,
        people_fullname   VARCHAR(255)        NOT NULL,
        people_isactor    TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,
        people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,

        PRIMARY KEY (people_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Movie database successfully created!';
?>



********************************************************************
***********************************************************************

second file to load info into DB:



<?php
// connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//make sure you're using the correct database
mysql_select_db('moviesite', $db) or die(mysql_error($db));

// insert data into the movie table
$query = 'INSERT INTO movie
        (movie_id, movie_name, movie_type, movie_year, movie_leadactor,
        movie_director)
    VALUES
        (1, "Bruce Almighty", 5, 2003, 1, 2),
        (2, "Office Space", 5, 1999, 5, 6),
        (3, "Grand Canyon", 2, 1991, 4, 3)';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the movietype table
$query = 'INSERT INTO movietype
        (movietype_id, movietype_label)
    VALUES
        (1,"Sci Fi"),
        (2, "Drama"),
        (3, "Adventure"),
        (4, "War"),
        (5, "Comedy"),
        (6, "Horror"),
        (7, "Action"),
        (8, "Kids")';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the people table
$query  = 'INSERT INTO people
        (people_id, people_fullname, people_isactor, people_isdirector)
    VALUES
        (1, "Jim Carrey", 1, 0),
        (2, "Tom Shadyac", 0, 1),
        (3, "Lawrence Kasdan", 0, 1),
        (4, "Kevin Kline", 1, 0),
        (5, "Ron Livingston", 1, 0),
        (6, "Mike Judge", 0, 1)';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Data inserted successfully!';
?>



**************************************************************
****************************************************************

MY CODE FOR THE QUESTION:


<?php
$db = mysql_connect('localhost', 'root', '000') or
die ('Unable to connect. Check your connection parameters.');

mysql_select_db('moviesite', $db) or die(mysql_error($db));

//get our starting point for the query from the URL
if (isset($_GET['offset'])) {
   $offset = $_GET['offset'];
} else {
   $offset = 0;
}

//get the movie
$query  = 'SELECT
   movie_name, movie_year
FROM
   movie
ORDER BY
   movie_name
LIMIT ' . $offset . ' , 1';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_assoc($result);
?>
<html>
<head>
<title><?php echo $row['movie_name']; ?></title>
</head>
<body>
<table border = "1">
<tr>
<th>Movie Name</th>
<th>Year</th>
</tr><tr>
<td><?php echo $row['movie_name']; ?></td>
<td><?php echo $row['movie_year']; ?></td>
</tr>
</table>
<p>
<a href="page.php?offset=0">Page 1</a>,
<a href="page.php?offset=1">Page 2</a>,
<a href="page.php?offset=2">Page 3</a>
</p>
</body>
</html>

Some code from my pages ,   Page1 ( Redirecting page )

<html>
<title>login_redirect.</title>
body>

<form name="redirect" action="http://mysite/page2.php" method="post">
<input type="hidden" name="mac" value="$(mac)">
</form>

<script language="JavaScript">
<!--
document.redirect.submit();
//-->
</script>
</body>
</html>

Page 2 ( select product )

<?php


session_start();

ini_set('display_errors',1);
error_reporting(E_ALL); 

include '../lib/config.php';
include '../lib/opendb.php';

// get user mac adres from redirect post page1
$_SESSION['macid'] = $_POST['mac'];
// set $macid for other use ( maybe not needed, am learning )
$macid = $_SESSION['macid'];

// echo $macid does show mac adress, so variable is not empty here


if (!empty($_POST["submit"]))

{
$product_choice = $_POST['accounttype'];
$query= "SELECT AccountIndex, AccountCost, AccountName FROM AccountTypes WHERE AccountIndex='$product_choice'";
$result = mysql_query($query) or die('Query failed. ' . mysql_error()); 
while($row = mysql_fetch_array($result))
{
$_SESSION['AccountIndex'] = $row['AccountIndex'];
$_SESSION['AccountCost'] = $row['AccountCost'];
$_SESSION['AccountName'] = $row['AccountName'];
}


header('Location: page3.php');

}

// did leave out the other/html/form stuff here

Page 3 ( show Session variables )

<?php


ini_set('display_errors',1);
error_reporting(E_ALL); 

session_start();

print_r($_SESSION);

?>

Now, on page 3 i do see the right session varables, only the "macid" is empty.   why ?    

Never had this one before.

Here's a string from a URL:
$urltext = Product_Name_'with_single_quotes'_"_B
Code: [Select]
$name = str_replace( "_", " ", $urltext );
echo 'raw: ' . $name . "<br>";
$name = mysql_real_escape_string( $name );
echo 'mysql_real_escaped: ' . $name . '<br>';

Doesn't seem possible, but both of the "echos" return the same string. My query fails because there are no backslashes in the SQL statement. What's going on here?

raw: Product Name 'with single quotes' " B
mysql_real_escaped: Product Name 'with single quotes' " B

I've joined querys because I'm making my own forums. However, when it gets to the echoin out part, it shows the category and only ONE of the sub categorys.

Code: [Select]
<div id="forumContainer">
<?php
$lastboard = '';

$forumQ = mysql_query("SELECT

   f1.cat_name as catName,
   f1.cat_id as catID,
   f2.cat_id as subCatID,
   f2.sub_id as subID,
   f2.sub_name as subName,
   f2.sub_desc as subDesc
   
   FROM `forum_cats` as f1
    LEFT JOIN `forum_sub` as f2
    ON f1.cat_id = f2.cat_id
   GROUP BY f2.cat_id
   ");
  
while($forumF = mysql_fetch_assoc($forumQ))
{
if($forumF['catName'] != $lastboard)
{
echo '<div class="forumCat">'.$forumF['catName'].'</div>';
$lastboard = $forumF['catName'];
}
echo '<div class="forumSub">'.$forumF['subName'].'</div>';
}
?>
</div>

I'm a complete newb at programming so please bear with me
i'm trying to check to make sure my sql query

          $result=mysql_query("SELECT screenname FROM users WHERE privacy='0'");

shows the correct results so i did a print_r($result); die();

but when i reload the page, it shows a syntax error after the die();

isn't the script suppose to completely stop so syntax errors after die() shouldn't matter?

also i'm expecting the $result to be an array of names and i wanted to do a query to another table to grab comments from each of those screennames, is the only way to do a for loop of mysql queries?

Thanks

Is there something wrong with this query?


elseif($_POST["titlee"] && $_POST["contente"])
{

$titlee = $_POST['titlee'];
$contente = $_POST['contente'];
$to = $_POST['edit'];

mysql_query("UPDATE custom_pages SET title='$titlee' AND content='$contente' WHERE id='$to'");
echo  '<div class="post">
<div class="postheader"><h1>Updated</h1></div>
<div class="postcontent">
<p>Your custom page has been updated.</p>
</div>
<div class="postfooter"></div>
</div>
';

}

Hi, can anybody check on this code? I don't have any errors in it, except that the supposedly alert after I deleted an information is not working. Thank you very much.

**This is a modified code provided by a friend. It worked originally (I just lost the original file since I sold my old computer xD

Here's the code:

<?php
include("dbconnection.php");



if(mysql_num_rows($result) > 0)
{
$query = "SELECT * FROM records WHERE id = '".$_GET["id"]."'";
$result = mysql_query($query, $connection);

$status = mysql_result($result, 0, "status");
$territory = mysql_result($result, 0, "territory");
$job_title = mysql_result($result, 0, "job_title");
$area_of_work = mysql_result($result, 0, "area_of_work");
$employer = mysql_result($result, 0, "employer");
$department = mysql_result($result, 0, "department");
$location = mysql_result($result, 0, "location");
$date_posted = mysql_result($result, 0, "date_posted");
$closing_date = mysql_result($result, 0, "closing_date");
$gender = mysql_result($result, 0, "gender");
$first_name = mysql_result($result, 0, "first_name");
$last_name = mysql_result($result, 0, "last_name");
$telephone_number = mysql_result($result, 0, "telephone_number");
$title = mysql_result($result, 0, "title");
$address_1 = mysql_result($result, 0, "address_1");
$address_2 = mysql_result($result, 0, "address_2");
$address_3 = mysql_result($result, 0, "address_3");
$city = mysql_result($result, 0, "city");
$country = mysql_result($result, 0, "country");
$postal_code = mysql_result($result, 0, "postal_code");
$website = mysql_result($result, 0, "website");
$email_address = mysql_result($result, 0, "email_address");
$cg_comment = mysql_result($result, 0, "cg_comment");
$date_emailed = mysql_result($result, 0, "date_emailed");
$mailing_comments= mysql_result($result, 0, "mailing_comments");
$telesales_comments= mysql_result($result, 0, "telesales_comments");
}

if(isset($_POST["btnSubmit"]))
{
$id = $_POST["id"];

$query = "DELETE FROM records WHERE id = '".$id."'";
mysql_query($query) or die(mysql_error());

echo "<script>
alert('You have successfully deleted a record');
window.location = 'view_client.php';
</script>";

}

?>

$Row[2] doesn't work but if I use $Row["linkcat"] than it works fine.
Is there somewhere to activate in the php.ini file ?


Code: [Select]
// I got 9 fields to in the table tbl_link !  $Row[2] should give me a number.

$Verbinding = mysql_connect($db_host, $db_user, $db_passw);
mysql_select_db($db_name);
$sql = "Select * from tbl_link where userid=1";
$ResultShow = mysql_query($sql);

while ($Row = mysql_fetch_assoc($ResultShow)){
 
    echo $Row[2];

}

I've written a script where it counts the page impressions on my website.

When a visitor lands on the page, it counts their impression and then sets a $_SESSION so that if they reload or return to the page it doesn't count it again. But the problem is it sometimes does.

Does this mean that the $_SESSION isn't being set probably? That's what it seems to be. And if so, what would be the reason for that?

Here's my basic code:

Code: [Select]
if(isset($_SESSION['userid'])){
     $userid = $_SESSION['userid'];
}
else{
     $userid = rand(1000, 10000000);
     $_SESSION['userid'] = $userid;
}

$date = date('Y-m-d');

$counters = mysql_query("SELECT * FROM counters
WHERE userid='$userid'
AND impressions='1'
AND date='$date'")
or die(mysql_error());

if(!mysql_num_rows($counters)){
     mysql_query("INSERT INTO counters
     (userid, impressions, date)
     VALUES('$userid', '1', current_date)")
     or die(mysql_error());
}

Hi guys, me again.

I have been using mail to send emails ( With your great help ) successfully in about three websites. For some reason, my email doesn't want to work with the site I am busy with now. This is my code :

Code: [Select]
<html>
      <head>
            <title>Practice</title>
      </head>
      <body>
<table border="1">
       <tr>
           <td>
<?php
      $emailto="hannes@ncc-cla.com";
      $emailsubject="confirmation";
      $emailfrom="footloose.sa@gmail.com;

                  if(!empty($_POST['uname'])){
                  $yourname=$_POST['uname'];
                  $yoursurname=$_POST['usurname'];
                  echo "hello
            </td>
            <td>"
                ." " . "$yourname" . " " . "$yoursurname
            </td>
         </tr>";

         $emailbody="hello
            </td>
            <td>"
                ." " . "$yourname" . " " . "$yoursurname
            </td>
         </tr>";
}
 else{
   echo"<tr>
            <td>
                enter your name
            </td>
        </tr>";

        $emailbody .="<tr>
            <td>
                enter your name
            </td>
        </tr>";
}
   if(isset($_POST['gender'])){      /*isset is what it is set to can be used */
   $gender=$_POST['gender'];
   echo "<tr>
             <td>
                 You are a
             </td>
             <td>
                 " . " " . "$gender
             </td>
   </tr>";
          $emailbody .="<tr>
             <td>
                 You are a
             </td>
             <td>
                 " . " " . "$gender
             </td>
   </tr>";
 }
       else{
             echo "<tr>
                       <td>
                           Select Gender
                       </td>
                    </tr>";
             $emailbody .="<tr>
                       <td>
                           Select Gender
                       </td>
                    </tr>";
       }
 if(isset($_POST['plang'])){
   $plang=$_POST['plang'];
   echo "<tr>
             <td>
                 You are interested in
             </td>
             <td>
                 " . " " . "$plang
             </td>
         </tr>";
         $emailbody .="<tr>
             <td>
                 You are interested in
             </td>
             <td>
                 " . " " . "$plang
             </td>
         </tr>";
 }
   else{
     echo "<tr>
               <td>
                   Select Programming language
               </td>
           </tr>";
           $emailbody .="<tr>
               <td>
                   Select Programming language
               </td>
           </tr>";
   }
 print "<tr>
            <td>
                Your qualification is
            </td>
            <td>";
            $emailbody .= "<tr>
            <td>
                Your qualification is
            </td>
            <td>";
 foreach ($_POST['qlevel'] as $level){
 print "$level<BR>";
 $emailbody .="$level<BR>";
 }
 $headers='From:'. $emailfrom;
                
                 mail($emailto,$emailsubject,$emailbody,$headers);
?>
             </td>
         </tr>
</table>
</body>
</html>

I just started with php and made an image. But when I try to open it in my browser, it shows a broken image icon. I have put the font in the same folder as the .php file, but it doesn't work. Can somebody help me solve this problem?

Here is the code:
Code: [Select]
<?php
$font_path = "nakedchick.TTF";
$font_size = 30;

$img_number = imagecreate(300,50);
$backcolor = imagecolorallocate($img_number,0,162,255);
$textcolor = imagecolorallocate($img_number,255,255,255);

imagefill($img_number,0,0,$backcolor);
$number = " So your ip is $_SERVER[REMOTE_ADDR]? ";

ImageTTFText($img_number, $font_size,10,5,5,$number,$font_path,$textcolor);

header("Content-type: image/png");
imagepng($img_number);
?>

And also the files: (.php in attachment)
http://falcogerritsjans.nl/u/signature2.php
http://falcogerritsjans.nl/u/nakedchick.ttf

Hi, I am trying to get the previous id for the table xdocument, but browser outputs this warning:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given

for this code:
Code: [Select]
<?php
//HELPER FCN used by FCN: writeXMLtoDBviaDOM
function mysql_next_id($table) {
    $result = mysql_query('SHOW TABLE STATUS LIKE "'.$table.'"');
    $rows = mysql_fetch_assoc($result);
    return $rows['Auto_increment'];
}

//TEST RUN:
$_filePath="C:\dir\email.xml";
$node=basename($_filePath);
$dom=new DOMDocument();
$dom->load($node);
$labelPath=array();
mysql_connect("localhost","root");
mysql_select_db("dummydpev");
$isXdocExist=mysql_query("SELECT (file_Path,file_Name) FROM xdocument WHERE file_Path='$_filePath' AND file_Name='$node'");
$docId=0;

if(mysql_num_rows($isXdocExist)==0) {
mysql_query("INSERT INTO xdocument (file_Path,file_Name) VALUES ('$_filePath','$node')");
$docId=mysql_next_id("xdocument")-1;
}
else
$docId=mysql_next_id("xdocument")-1;

writeXMLtoDBViaDOM($dom->documentElement,$labelPath,$docId,$_filePath);
?>
So I think my problem is with the multiple WHERE clause conditions, can I do it like what I have above?

Any help much appreciated!

  Why doesn't this code update two mysql databases (on two servers)? What am I doing wrong?

//SERVER 1 
$link = mysql_connect("localhost","usern1","pw1"); 
mysql_select_db("db_one1"); 

//SERVER 2 
$link = mysql_connect("xxx.xxx.xx.xxx","usern2","pw2"); 
mysql_select_db("db_one2"); 

$query = "INSERT INTO db1(subject, search, News, img) VALUES('$hsubject','$key','$news','$img')"; 
$result = mysql_query($query); 

$query = "INSERT INTO db2(subject, search, News, img) VALUES('$hsubject','$key','$news','$img')"; 
$result = mysql_query($query); 

$sentOk = "The data has been added to the database."; 
echo "sentOk=" . $sentOk; 

I'm trying to do a simple PHP form with Securimage Captcha and whenever I add this variable "$securimage = new Securimage();" the page displays nothing.  Could someone please help me with this?  My php file is already in the securimage directory so that's why I'm pointing directly to the file.

Here is my code so far (I say it's PDO but it pretty much isn't. What is the object here? The database connection? LOL!);
 

$sql = 
	"
	SELECT 
	SUM(IF(`sent` != 'N' , 1 , 0 )) as 'Emails Sent',	
	SUM(IF(`completed` NOT IN('N','paper') , 1 , 0 )) as 'Completed Electronically', 	
	SUM(IF(`completed` = 'paper' , 1 , 0 )) as 'Completed Manually', 
	SUM(IF(`completed` != 'N' , 1 , 0 )) as 'Total Number Completed',
	SUM(IF(`remindercount` = '1' , 1 , 0 )) as 'Reminder Sent Once',
	SUM(IF(`remindercount` = '2' , 1 , 0 )) as 'Reminder Sent Twice',
	SUM(IF(`remindercount` = '3' , 1 , 0 )) as 'Reminder Sent Thrice'
	FROM `tokens_$survey_id`
	";
				
$statement = $dbh->prepare($sql);
$statement->execute();

$result = $statement->fetch(PDO::FETCH_OBJ);
				
foreach($result as $key => $value) {
	 echo "<tr>
		<td>$key</td>
		<td>$value</td>
	       </tr>";
}

This is all well and good if the tokens_$survey_id table is actually there. Sometimes, for a genuine reason, there won't be a tokens table for that particular survey. How do I account for this? At the moment I get an error..
 

Warning: Invalid argument supplied for foreach() in /var/www/html/index.php on line 149 

I tried this but I am not satisfied this is correct;
 

if(!$result) {
die();
}

I don't want the code to die! If I take out the die() statement then this if is ignored for some reason I don't understand.