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PHP - Simplexml_load_string Warnings. Is The Xml The Problem, Or Is It Me?

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Simplexml_load_string Warnings. Is The Xml The Problem, Or Is It Me?	View Content

Full Excel VBA Course - Beginner to Expert

Warnings???

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ok people sorry but its late, and I'm tired and I can't figure this out maybe I just need a break? I'm getting some warnings here, first here is my code:
Code: [Select]
<?php
echo"<?xml version=\1.0\"encoding=\UFT-8\"\x3f>";
echo"\n";
?>

<!DOCTYPE html PUBLIC"-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtmll/DTD/xhtmll-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml"
xml:lang="en"lang="en">

<head>
<title>Assignment 7: Making Space, Part 3</title>
<meta name="author" content="Jim Cozzy"/>
</head>
<body>
<h1> Assignment 7: Making Space, Part 3 </h1>
<?php
  echo "<h2>Today is ".date('l, F d, Y')."</h2>";
?>

<?php
  //calculate the area of an equilateral triangle
  function calcTriangle($cVal) {
    $halfSide = $cVal/2;
    $otherSide = sqrt(($cVal*$cVal) - ($halfSide*$halfSide));
    $areaValue = $otherSide * $halfSide;
    echo "<p>The area of an equilateral triangle whose sides are
    {$cVal} is {$areaValue}.</p>\n";
  }
$yeah = (isset($_POST['vfld']))?$_POST['vfld']:'';
    if (isset($_POST['submit'])) {
  if ($_POST['submit'] == "Clear") {
    $yeah = "";
   } elseif ($_POST['submit'] == "Calculate") {
      if (!eregi(square|circle|triangle, $yeah)) {
        echo "<p>Sorry I forgot to mention the the shape needs to be
                a Square, Circle, or a Triangle.</p>".
            "<p>Please return and try again.</p>";
     exit;
      }
}}

?>
<form method="post" action="makingspace3.php"
id="form1" name=form1">
<p>
Write down a shape and a number and I will compute the area for:
<input type="text" id="vfld" name="vfld"
value="<?php echo $yeah; ?>"/>
</p>
<p>
After you have done this click calcualte to see results
</p>
<p>
<input type="submit" name="submit" value="Calculate" />
<input type="submit" name="submit" value="Clear" />
</p>
</form>

<div id="footer">
<pre>This page written by: <cite>James Cozzy</cite> ©2011 and beyond</pre>
</div>

</body>

</html>

so when I enter something wrong into my form like ggggg I want it to print an error which it does, but it also says what am I missing????
Notice: Use of undefined constant square - assumed 'square' in /studfs1/homepage/CISS-225/makingspace3.php on line 49

Notice: Use of undefined constant circle - assumed 'circle' in /studfs1/homepage/CISS-225/makingspace3.php on line 49

Notice: Use of undefined constant triangle - assumed 'triangle' in /studfs1/homepage/CISS-225/makingspace3.php on line 49

Catch Warnings

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Is there any way to catch warnings without defining a warning handler?
Example:
try
{
$settings = parse_ini_file( 'main.ini' );
}
catch( ErrorException $e )
{
throw new ErrorException('No main configuration file found for webiste', '0003');
}

Only ends up throwing a warning if the file doesn't exist. I want to catch it and throw an error.

Array_map Warnings

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Howdy, this one is killing me! Hope some of the genius here will rub off on me.

Very simple example of what is happening:

Base.php
Code: [Select]
class Base_Model //class autoloaded
{
protected function scrub($value)
{
return trim(htmlspecialchars($value, ENT_QUOTES, "UTF-8"));
}
}

Foo.php
Code: [Select]
class Foo_Model extends Base_Model
{
if($_POST)
{
$_POST = array_map('parent::scrub', $_POST);
}
}

This works perfectly on the workstation (MAMP) server, but generates a warning ( Warning: array_map() [function.array-map]: The first argument, 'parent::scrub', should be either NULL or a valid callback in... ) when uploaded to the production server. Both are running the exact same version of php.

Any ideas would be appreciated on getting rid of the error, besides turning off warnings ( haha ).

William

Help With Coding And Warnings

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Okay I have been trying to work out this problem for 3 months I am not the type of person to give up but I have tried everything. I was first not able to connect to my database live with my form but I did a change from ($cxn,$sql) to ($sql,$link) since then I was able to download to phpmyadmin in Google Chrome and no other site. But in Google Chrome I get a message about global registry Warning: Unknown on line 0. all other sites give me a Warning line 116 MySql Link is

Here is my coding can someone help I have only been using PHP and I know someone is out there with much more knowledge than I have. This is the first page with the codes but it does not go to the second page which I will paste after this one. Any help would make my year better.
Code: [Select]
<php
@session_start();
switch (@$_POST['Button'])
{
case "Login1":
include "Connections/connect_to_mysql.php";
$sql = "SELECT id FROM Members
WHERE id= '$_POST[fusername]'";
$result = mysql_query($sql)
or die("Query died: fusername");
$num = mysql_num_rows($result);
if($num > 0) //login name was found
{
$sql = "SELECT id FROM Members
WHERE id='$_POST[fusername]'
AND password=md5('$_POST[fpassword])'";
$result2 = mysql_query($sql)
or die("Query died: fpassword");
$num2 = mysql_num_rows($result2);
if($num2 > 0) //password matches
{
$_SESSION['email1']="yes";
$_SESSION['id'] = $_POST['fusername'];
$sql = "INSERT INTO Members (id,loginTime)
VALUES ('$_SESSION[id]',NOW())";
$result = mysql_query($sql)
or die("Query died: insert");
header("Location: New_member.php");
}
else //password does not match
{
$message_1="The username, '$_POST[fusername]' exists, but you have not entered
the correct password! Please try again.";
$fusername=strip_tags(trim($_POST['fusername']));
include("login_form.php");
}
}
else //username not found
{
$message_1 = "The username you entered does not exist! Please try again.";
include("login_form.php");
}
break;

case "Register":
/* Check for blanks */
foreach($_POST as $field => $value)
{
if($field != "fax")
{
if(empty($value))
{
$blanks[] = $field;
}
else
{
$good_data[$field] = strip_tags(trim($value));
}
}
}
if(isset($blanks))
{
$message_2 = "The following fields are blank. Please enter the required information: ";
foreach($blanks as $value)
{
$message_2 .="$value, ";
}
extract($good_data);
include("login_form.php");
exit();
}
/* validate data */
foreach($_POST as $field => $value)
{
if(preg_match("/name/i",$field) and
!preg_match("/user/i",$field) and
!preg_match("/log/i",$field))
{
if (!preg_match("/^[A-Za-z' -]{1,50}$/",$value))
{
$errors[] = "$value is not a valid name. ";
}
}
if(preg_match("/email/i",$field))
{
if(!preg_match("/^.+@.+\\..+$/",$value))
{
$errors[]="$value is not a valid email addr.";
}
}
} //end if not empty
foreach($_POST as $field => $value)
{
$$field = strip_tags(trim($value));
}
if(@is_array($errors))
{
$message_2 = "";
foreach($errors as $value)
{
$message_2 .= $value." Please try again<br />";
}
include("login_form.php");
exit();
} // end if errors are found

/* check to see if username already exists */
include_once"Connections/connect_to_mysql.php";
$link = @mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
@mysql_select_db("$db_name") or die ("Query died:");
$sql = "SELECT id FROM Members
WHERE id='$id'";
$result = mysql_query($sql)
or die("Query died: id.");
$num = mysql_num_rows($result);
if($num > 0)
{
$message_2 = "$id already used. Select
another User Name.";
include("login_form.php");
exit();
} // end if username already exists
else // Add new member to database
{
$sql = "INSERT INTO Members (id,sign_up_date,username
password,firstName,lastName,email) VALUES
('$id',NOW(),,md5('$password'),
'$username','$firstName','$lastName','$email')";
mysql_query($sql);
$_SESSION['email1']="yes";
$_SESSION['id'] = $username;
/* send email to new Customer */
$emess = "You have successfully registered. ";
$emess .= "Your new username and password a ";
$emess .= "\n\n\t$username\n\t";
$emess .= "$password\n\n";
$emess .= "We appreciate your interest. \n\n";
$emess .= "If you have any questions or problems,";
$emess .= " email ebermylove@gmail.com";
$subj = "Your new customer registration";
# $mailsend=mail ("$email","$subj","$emess");

}
break;

default:
include("login_form.php");
}
?>
<php
session_start();

if (@$_SESSION['auth'] != "yes")
{
header("Location: login.php");
exit();
include("dogs.inc");
$link = mysqli_connect($host,$user,$password,$dbname)
or die("Couldn't connect to server.");
$sql = "SELECT firstName, lastName FROM `Members`
WHERE id='{$_SESSION['id']}'";
$result = mysqli_query($sql,$link)
or die("Couldn't execute query");
$row = mysqli_fetch_assoc($result);
extract($row);
}
echo "<html>
<head><title>New Member Welcome</title></head>
<body>
<h2 style='margin-top: .7in; text-align: center'>
Welcome $firstName </h2>\n";
?>
<p>Your new Member accounts lets you enter the members only section
of our web site. You'll find special discounts, a profile of matches,
live advise for experts, and much more.</p>
<p>Your new Member ID and password were emailed to you. Store them
carefully for future use.</p>
<div style="text-align: center">
<p style="margin-top: .5in; font-weight: bold">
Glad you could join us!</p>
<form action="member_page.php" method="post">
<input type="submit"
value="Enter the Members Only Section">
</form>
<form action="profile.php" method="post">
<input type="submit" value="Go to Main Page">
</form>
</div>
</body>
</html>

Undefined Argument Warnings

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Hey all. I have a function called displayForm in a form validation script. This function accepts a few arguments but these arguments are empty upon initial calling of the function. One of the arguments is and errors array that only gets populated if there were form errors.

So upon running my script initially I get missing argument Warnings.

Can these warnings simply be ignored? Would they cause problems throughout my script if I ignore these warnings? And how can I suppress these warnings if I would like to?

Thanks all for your help.

Cheers.

Suppress Foreach Error Warnings

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I use PHP's PDO extensively for data grabbing and inserting. When I am selecting data, the return is a PHP PDO object which in itself contains select statements. So I use foreach to parse through the object and build a table. For cells which I have data it's fine but for cells which do not get any data meaning that have no entry in the db, php throws out a error PHP Warning: Invalid argument supplied for foreach().
I tried using if(is_array) and also @ to suppress them but no luck . Can anyone point me to the right direction for suppressing these errors/warnings!!

Getting Errors/warnings On The Web Server But Not On Development Machine

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In php.ini on my local development server, error_reporting is set to
Quote

error_reporting = E_ALL | E_STRICT

And I get no errors.

However, on the remote web server I found a bunch of warnings in the error log. I fixed most of them, but this error/warning should have popped up on my development console:

Quote

[02-Dec-2011 17:47:22] PHP Warning: array_keys() [<a href='function.array-keys'>function.array-keys</a>]: The first argument should be an array in /home/bpslk/public_html/cart-checkout.php on line 177

[02-Dec-2011 17:47:22] PHP Warning: Invalid argument supplied for foreach() in /home/bpslk/public_html/cart-checkout.php on line 177

I can fix the code to eliminate the warning, no prob. But why am I NOT seeing this error on my development machine?

--Thanks

--HarryG

No Warnings Displayed If I Reference Arguments At Call Time For A Function

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Hello,

I was trying out some code to check out how some of the php settings work and when i tried allow_call_time_pass_reference to see if it works, it doesn't spew out any warning or error like the comments in the .ini file mention (or in the manual sites). Maybe i'm not using some other setting in order to see this? Here is my EXACT code.

<?php
ini_set('display_errors',1);
error_reporting(-1);

$variable = 'String';

function display($variable) {
return $variable;
}

echo display(&$variable);
?>

Outputs : String. No errors, no warnings.
I'm running PHP Version 5.3.2-1ubuntu4.5

Problem Fixed But Caused New Problem

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Guys thanks for helping me solve the problem i had but now i have another problem and i am lost. i have included the code below for you to have overview.

On the home page the url shows how it should be: http://localhost/board/italian.php?country=Italian
BUT
When you mouse over the NEXT link to go to next page you get this : http://localhost/board/?page=1&country=Italian and then an error 404 when you click on the link
page = 1 no matter if you are mousing over link for page 5 or 2

This is the code:

<table style="width:100%; margin-left:auto; margin-right:auto">
       <?php

           $items_per_page = 5;//how many records/rows to display per page

           $page_url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];

           $cleaned_current_page_country = htmlspecialchars($_GET["country"]);

           $page_number = (isset($_GET['page']))? filter_var($_GET['page']) : 1;
           //$page_number = filter_var($_GET['page']);
           //$page_number = $_GET['page'];

           $selected_restaurant_type = $_GET['country'];

           // find number of rows in the db
               $query = "SELECT * FROM menu_update WHERE special_of_theday = 1 AND restaurant_type = '".$selected_restaurant_type."' ";
               $query_result = mysqli_query($connection, $query);
               $row_count = mysqli_fetch_array($query_result);
               $total_rows = mysqli_num_rows($query_result);

           $expected_number_of_pages = ceil($total_rows/$items_per_page); //expected number of pages

           $page_position = (($page_number - 1) * $items_per_page); //get starting position to fetch record

           //fetch a group of records using sql LIMIT clause
               $query = "SELECT * FROM menu_update WHERE special_of_theDay = 1 AND restaurant_type = '".$selected_restaurant_type."' LIMIT " .$page_position. ',' .$items_per_page;
               $query_result = mysqli_query($connection, $query);
       ?>

           <table width="100%" border=0>
               <tr bgcolor = "#cccccc">
                   <td>Meal</td>
                   <td>Meal name</td>
                   <td>Meal Price</td>
                   <td>Order Now</td>
                   <td>Full Menu</td>
                   <td>Our Gallery</td>
                   <td>Find Us</td>
                   <td>Our Chef</td>
                   <td>Our Manager</td>
               </tr>
       <?php
               while($row_count = mysqli_fetch_array($query_result)){
                   echo "<tr>";
                       echo"
                           <td style ='width:200px; text-align:center; height:120px; background:#eaedf2'><a href='meal_fullPic.php'><img src= ".$row_count['img']." /></a></td>
                           <td style ='width:119px; text-align:center; height:120px; background:#eaedf2'> ".$row_count['meal_name']."</td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'>".$row_count['price']."</td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='order_processor.php'><img src= ".$row_count['order_now']." /></a></td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='full_menu.php'><img src= ".$row_count['full_menu_img']." /></a></td>

                           <td style ='width:170px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_gallery.php'><img src= ".$row_count['restaurant_pic']." /></a></td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_location.php'><img src= ".$row_count['location_pic']." /></a></td>
                           <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['chef_name']."</td>
                           <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['admin']."</td>
                       ";
                   echo "</tr>";
               }
               mysqli_close($connection);
       ?>
           </table>
       <div class="board_pagination">
       <?php
           //display the links to the pages
           for($current_page = 1; $current_page <= $expected_number_of_pages; $current_page++){
    if ($page_number != $expected_number_of_pages){
                   $data = array(
                       'page' => $page_number,
                       'country' => $cleaned_current_page_country,
                   );
                   $the_query = http_build_query($data);
                   $page_url .= '?' . $the_query;
                   echo '<li><a href=" '.$the_query.' ">'.$current_page.'</a></li> ';
               }
           }
       ?>
       </div>
   </table>

Php Problem.

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Hello everyone

Just found this forum and really need some help. I'm trying to create a popup contact form for my product pages. I managed to get a php form from a random site and i cant get the form to pop up.

http://www.e-sol.co.uk/printer-scanner-copier.php

If someone could help i would be very grateful.

Php/xml Problem.

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so i want to call my php file with a query string to select a record out of a XML file... im close but cant get the variable i pass to be the record selected..

example... index.php?auction=1

i highlighted in red, where it doesnt pick up the proper row

Code: [Select]
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);

$auction = $_GET{auction};

// parse the xml
$xmlFileData = file_get_contents("items.xml");
$xmlData = new SimpleXMLElement($xmlFileData);

[color=red]$location = $xmlData->item[$auction];[/color]

//Retrieving the content from the selected item
$title = $location->title;
$value = $location->value;
$itemnumber = $location->itemnumber;
$description = $location->description;
$donatedby = $location->donatedby;
$image1 = $location->image1;
$image2 = $location->image2;
$image3 = $location->image3;
$image4 = $location->image4;
$image5 = $location->image5;
$image6 = $location->image6;
$image7 = $location->image7;
$image8 = $location->image8;

?>

<style>
.itemimages img {width: 10px; height: 10px;}
img.large {width: 100px; height: 100px;}
</style>
<script language="JavaScript" src="http://code.jquery.com/jquery-1.5.min.js" type="text/javascript"></script>
<script>
$().ready(function() {

$(".itemimages img").click(
$("#largepic").attr("src","this");
});

});
</script>

<table border="1" cellpadding="10">
<tr>
<td valign="top">
<img src="<?php echo $image1 ?>" alt="" border="0" class="large" id="largepic"><br>
<div class="itemimages">
<?php
// image 1
if (trim(''.$image1) != '') {
echo "<img src='";
echo $image1;
echo "' alt='' border='0'/> ";
}
// image 2
if (trim(''.$image2) != '') {
echo "<img src='";
echo $image2;
echo "' alt='' border='0'/> ";
}
// image 3
if (trim(''.$image3) != '') {
echo "<img src='";
echo $image3;
echo "' alt='' border='0'/> ";
}
// image 4
if (trim(''.$image4) != '') {
echo "<img src='";
echo $image4;
echo "' alt='' border='0'/> ";
}
// image 5
if (trim(''.$image5) != '') {
echo "<img src='";
echo $image5;
echo "' alt='' border='0'/> ";
}
// image 6
if (trim(''.$image6) != '') {
echo "<img src='";
echo $image6;
echo "' alt='' border='0'/> ";
}
// image 7
if (trim(''.$image7) != '') {
echo "<img src='";
echo $image7;
echo "' alt='' border='0'/> ";
}
// image 8
if (trim(''.$image8) != '') {
echo "<img src='";
echo $image8;
echo "' alt='' border='0'/> ";
}
?>
</div>

</td>
<td valign="top">

<?php echo $title ?><br>
Estimated Value: <?php echo $value ?><br>
Item Number: <?php echo $itemnumber ?><br><br>

<?php echo $description ?><br><br>

Donated by: <?php echo $donatedby ?><br>
</td>
</tr>
</table>

Php Com Problem

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Hi,

I am trying to write in an existing excel file. But for some reason, I can't open the workbook (but I know I have made similar code work before). Here I am just pasting the code upto the error line

$filename="C:\\apache\\htdocs\\GAC2\\dload\\template.xls";
$excel = new COM("excel.application") or die("Unable to instanciate excel");
$excel->Visible = 1;
$excel->DisplayAlerts = 1;
try{
$wkb = $excel->Workbooks->Open($filename);
}catch (Exception $e) {
echo 'Caught exception: ', $e->getMessage(), "\n";
}

//........ Later part of the code

This returns the following error
Code: [Select]
Caught exception:
Source: Microsoft Office Excel
Description: Unable to get the Open property of the Workbooks class

For record, the I have already validated the file location. And I have tried the following line instead without any positive result.
Code: [Select]
$filename="C:\apache\htdocs\GAC2\dload\template.xls";
Looking for any help please.

-Abd

Oop Problem

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You may see a lot of things wrong with the code below, but please bring it to my attention! I'm trying to advance my knowledge on OOP, but I'm trying my best to get the basics down. The below code, after selecting the two characters to fight, the page returns nothing, just blank. I'm pretty sure the error lyes within class_li.php, but I'm not exactly sure.

Script is attached.

While / If Else Problem

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I am trying to use the following code snippet to return values from a database with a "," after each value apart form the last one. At the moment for testing I have 2 rows with values 1 and 3

$totalRows_Recordset3 equals 2 (I echoed this to check)

$i=1;
  while($row_Recordset3 = mysql_fetch_assoc($Recordset3))
  {
    if($i < $totalRows_Recordset3)
    {
      $str = $str.$row_Recordset3['reading'].",";
      $i=$i+1;
    }
    else
    {
      $str = $str.$row_Recordset3['reading'];
      $i=$i+1;
    }
  }

I get the number 3 which is the last value in the database, but I can't see why I don't get "1,3"

I am sure it is something simple but it just escapes me.

Thanks in advance for any help

Gordon

Php Problem

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The problem is whit the var $threadid .
It dosent work here(it dosen't display the data from database): $dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' ";
But if I assign a value before the that code the html page show correctly.

Code: [Select]
include("config.php");

$name = $_POST['name'];
$comment = $_POST['comment'];
$threadid = $_POST['threadid'];

if($name & $comment)
{
$dbh="INSERT INTO comments (name,comment,threadid) VALUES ('$name','$comment','$threadid') ";
mysql_query($dbh);

}

$dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' ";
$req = mysql_query($dbh);
while($row=mysql_fetch_array($req))
{
echo "<li>";
echo "<br>"."<b>".$row['name']."</b></br>"."<br>".$row['comment']."</br>";
echo "</li>";
}

Code: [Select]
$('.submit').click(function(){
location.reload();
var name = $("#name").val();
var comment = $("#comment").val();

var threadid = $("#v").val();

var dat = 'name='+name+'&comment='+comment+'&threadid='+threadid;

$.ajax({
type:"post",
url:"comment.php",
data:dat,
success:function(){
console.log("dat");
}
});

return false;

});

If...else Problem

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I have a simple text file called bio.txt:
Code: [Select]
wohlersr; Richie Wohlers; Intermediate; Four Aces
meyersg; Greg Meyers; Novice; Four Aces
I wrote a little script, to return the info for Greg Meyers if the user id is correct, or say "Member not found" if incorrect. For some reason, my output is giving me both. I assume it's a context error, but can't figure it out.

Here's my script:
Code: [Select]
<?php
$file=file("bio.txt");
$count=count($file);

$i=0;
while($i<=$count)
{
$row = explode(";", $file[$i]);

$id = $row[0];
$name = $row[1];
$class = $row[2];
$club = $row[3];

if($id=="meyersg"){
echo $name."</br>";
echo $class."</br>";
echo $club."</br>";
}else{
echo "Member not found.";
}

$i++;
}

?>
And here's the output I'm getting:
Code: [Select]
Member not found. Greg Meyers
Novice
Four Aces
Member not found.

Problem

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Little Php Problem

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Hi
I make 3 querys first Order by DESC, second RAND() and third ASC. Now I want that in the second query can't be the same ids (produgg_users.id) that already taken in the first. And in the third not the ids from first and second. Like I have it now the same users can show up in the RAND() and the ASC like in DESC. Here my Script so far:

$rs = mysql_query("select produgg_users.id, twitterUser, credits from produgg_users where twitterUser != '' and
                produgg_users.id IN (select concat_ws(',', id) from produgg_users where credits > 0 and id != ".$usersClass->userID().") and produgg_users.id NOT IN (select concat_ws(',', followedID) from produgg_activity where followerID = '".$usersClass->userID()."') and produgg_users.id NOT IN (select concat_ws(',', userid) from produgg_featured) ORDER BY credits DESC LIMIT 10") or die(mysql_error());

$nr = @mysql_num_rows($rs);

if($nr != 0) {

    print "<p>Simply click <strong>get 1 point</strong> and you will automatically FOLLOW the USER and earn the credit!</p>";

print "<p><font color=\"#0066FF\">Top Members</font></p>";

    while($row=@mysql_fetch_object($rs))
    {

        $divLeft = '<div id="thediv_'.$row->id.'"><div style="float:left;width:400px;">';
        $divRight = '<div style="float:left;width:150px;">';
        $clearDiv = '<div style="clear:both;"></div>';

        print "$divLeft<img src=\"http://api.twitter.com/1/users/profile_image/".strip_tags($row->twitterUser)."?size=mini\" alt=\"\" />  ".strip_tags($row->twitterUser)."  <font color=\"#999999\" size=\"1\">$row->credits Points</font></div>$divRight <a href='javascript:void(0);' id='vote_$row->id' class='getPoint'>Get 1 Point</a></div>$clearDiv<hr /></div>";
$z = 1;
    }
}

//Lucky Riders

$rs = mysql_query("select produgg_users.id, twitterUser, credits from produgg_users where twitterUser != '' and
                produgg_users.id IN (select concat_ws(',', id) from produgg_users where credits > 0 and id != ".$usersClass->userID().") and produgg_users.id NOT IN (select concat_ws(',', followedID) from produgg_activity where followerID = '".$usersClass->userID()."') and produgg_users.id NOT IN (select concat_ws(',', userid) from produgg_featured) ORDER BY RAND() LIMIT 5") or die(mysql_error());
$nr = @mysql_num_rows($rs);

if($nr != 0) {

print "<p><font color=\"#0066FF\">Lucky Riders</font></p>";

    while($row=@mysql_fetch_object($rs))
    {
        $divLeft = '<div id="thediv_'.$row->id.'"><div style="float:left;width:400px;">';
        $divRight = '<div style="float:left;width:150px;">';
        $clearDiv = '<div style="clear:both;"></div>';

        print "$divLeft<img src=\"http://api.twitter.com/1/users/profile_image/".strip_tags($row->twitterUser)."?size=mini\" alt=\"\" />  ".strip_tags($row->twitterUser)."  <font color=\"#999999\" size=\"1\">$row->credits Points</font></div>$divRight <a href='javascript:void(0);' id='vote_$row->id' class='getPoint'>Get 1 Point</a></div>$clearDiv<hr /></div>";
$z = 1;
    }
}

//Lucky Riders

$rs = mysql_query("select produgg_users.id, twitterUser, credits from produgg_users where twitterUser != '' and
                produgg_users.id IN (select concat_ws(',', id) from produgg_users where credits > 0 and id != ".$usersClass->userID().") and produgg_users.id NOT IN (select concat_ws(',', followedID) from produgg_activity where followerID = '".$usersClass->userID()."') and produgg_users.id NOT IN (select concat_ws(',', userid) from produgg_featured) ORDER BY credits ASC LIMIT 5") or die(mysql_error());
$nr = @mysql_num_rows($rs);

if($nr != 0) {

print "<p><font color=\"#0066FF\">Last Rides</font></p>";

    while($row=@mysql_fetch_object($rs))
    {
        $divLeft = '<div id="thediv_'.$row->id.'"><div style="float:left;width:400px;">';
        $divRight = '<div style="float:left;width:150px;">';
        $clearDiv = '<div style="clear:both;"></div>';

Php Sql Problem

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Hi there,

I have a map of America made in flash, where you click on a state and the page should display the SQL database information for that state in the HTML table - but instead all it shows is the first entry of the database regardless of which state you click and it doesn't display the 2 radio buttons.

My code is as follows

Flash Actionscript 3 (just showing one state)
Code: [Select]
function waClick(event:MouseEvent):void {
var waURL:URLRequest = new URLRequest("restaurants.php?state=Washington");
navigateToURL(waURL, "_self");
}
wa_btn.addEventListener(MouseEvent.CLICK, waClick);
PHP code

Code: [Select]
<?php
  include("mvfconnect.php");
  $theChoice = $_GET['state'];
  $query = "SELECT * FROM restaurants WHERE" .$theChoice;
  $result = @ mysql_query($query);
  if (!$result) {
   $message="Unfortunately we are having problems with this page, we promise to have it fixed as soon as possible";
    die($message);
  }
  $num = mysql_num_fields($result);
  $i=0;
  while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
   $show1=substr($row['state'],0,50)."...";
   $show2=substr($row['city'],0,50)."...";
   $show3=substr($row['rname'],0,50)."...";
   $show4=substr($row['address'],0,50)."...";
   $show5=substr($row['pnum'],0,50)."...";
   $show6=substr($row['web'],0,50)."...";
   $show7=substr($row['dishes'],0,50)."...";
   $show8=substr($row['dish_details'],0,50)."...";
   $show9=substr($row['challenges'],0,50)."...";
   $show10=substr($row['challenge_details'],0,50)."...";
   $show11=substr($row['youtube'],0,50)."...";
   $show12=substr($row['images'],0,50)."...";
   echo "<tr>".
          "<td>".$row['city']."</td>".
          "<td>".$row['rname']."</td>".
          "<td>".$row['address']."</td>".
          "<td>".$row['pnum']."</td>".
          "<td>".$row['web']."</td>".
          "<td>".$row['dishes']."</td>".
          "<td>".$row['challenges']."</td>".
              "<td id='state".$i."' style='display:none'>".$row['state']."</td>".
          "<td id='city".$i."' style='display:none'>".$row['city']."</td>".
          "<td id='rname".$i."' style='display:none'>".$row['rname']."</td>".
          "<td id='address".$i."' style='display:none'>".$row['address']."</td>".
          "<td id='pnum".$i."' style='display:none'>".$row['pnum']."</td>".
          "<td id='web".$i."' style='display:none'>".$row['web']."</td>".
          "<td id='dishes".$i."' style='display:none'>".$row['dishes']."</td>".
          "<td id='dish_details".$i."' style='display:none'>".$row['dish_details']."</td>".
          "<td id='challenges".$i."' style='display:none'>".$row['challenges']."</td>".
          "<td id='challenge_details".$i."' style='display:none'>".$row['challenge_details']."</td>".
          "<td id='youtube".$i."' style='display:none'>".$row['youtube']."</td>".
                  "<td id='images".$i."' style='display:none'>".$row['images']."</td>".
                  "<td><input type='radio' name='vid' id='vid".$i."' onclick='openVideo(".$i.")' /></td>".
          "<td><input type='radio' name='pic' id='pic".$i."' onclick='openImage(".$i.")' /></td>".
          "<td class='last'style='display:none'>".$show1." ".$show2." ".$show3." ".$show4." ".$show5." ".$show6."
                  ".$show7." ".$show8." ".$show9." ".$show10."</td>".
         "</tr>";
    $i++;
   }
?>

I have used this code before and it has worked fine, can someone please help me out!

Ehh.. I Have A Problem!

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Okay, so i'm using the basic php mysql insert into function but I have a strange (or maybe not so strange, I just don't get it..) problem..

I have two columns (do you call them columns? I don't really know for sure, whatever things that are in a table whatsoever).

Name and Trailer.

EDIT: they are TEXT's if that matters

Name is really nothing important, just the name.
But I want to embed trailers.. If I embed the Megavideo code below:

Code: [Select]
<object width="400" height="300"><param name="movie" value="http://www.megavideo.com/v/8UZFURZQ2b55ada04d73ae65c726a5383b9ee9d9"></param><param name="allowFullScreen" value="true"></param><embed src="http://www.megavideo.com/v/8UZFURZQ2b55ada04d73ae65c726a5383b9ee9d9" type="application/x-shockwave-flash" allowfullscreen="true" width="400" height="300"></embed></object>
Everything works fine, no problems and no bullshit.

But.. if I for an example try to embed this one (NovaMov):

Code: [Select]
<iframe style='overflow: hidden; border: 0; width: 400px; height: 300px' src='http://embed.novamov.com/embed.php?width=400&height=300&v=xngq7u8yejwbt' scrolling='no'></iframe>
It doesn't work. It just doesn't insert. Is inserting an "iframe" a problem or something? I just don't get why the hell it would do this?

Does anyone know anything about this? If so i'd be eager and very thankful to hear why and how to fix it.

thanks in advance