PHP - Mysqli_num_rows Is Giving Me An Error

This is the form:

Code: [Select]
<form id="submit_form" method="post" action="">

       
<label for="knuffix_category"><h4>Choose a category</h4></label><br />
     
        <?php 
        require_once ($sort_category_func);
        $switch = 0;
        sort_category ($switch); 
        ?>


<br /><br />

       
        <label for="contribution_description"><h4>Enter your description here</h4></label><br />
        <textarea maxlength="150" type="text" name="contribution_description" value=""></textarea><br />
       
       
       
       
<label for="contribution"><h4>Enter your contribution here</h4></label><br />
<textarea maxlength="300" type="text" name="contribution" value=""></textarea><br />

       
       
<input type="submit" name="submit" value="Contribute" />
       
       
</form>



<?php
    include($submit_script);
?>



And this is the script:

Code: [Select]
<?php
if (isset($_POST['submit'])){

if (isset($user_name)) {
 // Connect to the database 
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
  
                  
                  
// Grab the score data from the POST
                
//$knuffix_name = strip_tags(trim($_POST['knuffix_name']));

                $contribution_name = uniqid();
                
                if (!empty($_POST['contribution_description']))
                    $contribution_description = $_POST['contribution_description'];
               // if (!empty($_POST['knuffix_contribution']))
                    $contribution = $_POST['contribution'];
                // if (!empty($_POST['cat'])) {
                    $contribution_category = strip_tags(trim($_POST['cat']));
               // }
                
                
                  // Check if the fields and variables are empty
                
  if (!empty($contribution_category) && !empty($contribution)) {
 
                      
                      
                      
                      // Check if the submission exists twice in the whole database
                      
                      $query_get = "SELECT contribution FROM con WHERE = '{$contribution}'";
                      
                      $query_run = mysqli_query($dbc, $query_get);
                      
                      $num_rows = mysqli_num_rows($query_run);
                      

                   
                      
                      if ($num_rows == 0) {

                      
                          
                          
                          
                          // Write the data into the database
                          
  $query = sprintf("INSERT INTO con (user_id, name, contribution, category, contribution_date, description) VALUES ('$user_id', '%s', '%s', '%s', now(), '%s')",
mysqli_real_escape_string($dbc, $contribution_name),
mysqli_real_escape_string($dbc, $contribution),
mysqli_real_escape_string($dbc, $contribution_category),
                                        mysqli_real_escape_string($dbc, $contribution_description));
 
                          
                            mysqli_query($dbc, $query) or die (mysqli_error($dbc));

                            mysqli_close($dbc);
   
  /* Set a cookie (later) to prevent duplicate submissions */
  // redirect the page to prevent duplicate submissions
// header ('Location: redirect.php');
echo 'Contributing was successful.';   
 
                                
                          
                                
                                
                      } else {
                          
                          
                          echo "This contribution already exists in the database. 
                              To keep this place clean we do not allow duplicate submissions.";
                          
                          
                          
                      }
                       
                      
                      
                      } else {



                            echo "Please enter all of the information to add your contribution.";



                            }

                        
                        
                      } else {



                            echo "You have to be signed-in to do a contribution.";



                            }
    }
?>


And this is the error message:

Code: [Select]
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\php_projects\myproject\controller\contribution\submit_script.php on line 38


As the error message states it is not becoming a result, and I do not see where the problem lies, everything looks right to me. Does somebody else see a mistake in there?

<!DOCTYPE html>
<html>
<head>
    <title>Registered Users</title>
    <link rel="stylesheet" href="includes/style.css" type="text/css" media="screen"/>
    <link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.25/css/jquery.dataTables.min.css">
</head>
<body>
    <?php
    include('includes/header.html');
    ?>
    <center>
        <table id="records">
            <thead>
                <tr>
                    <th>ID</th>
                    <th>First Name</th>
                    <th>Last Name</th>
                    <th>Email</th>
                </tr>>
            </thead>
            <tbody>
                <?php
                    require_once('mysqli_connect.php');
                    $query = "SELECT * FROM users";
                    $result = mysqli_query($dbc,$query);
                    
                    if(mysqli_num_rows($result) > 0) -------------------------------------------------Line 28

                    {
                        while($row = mysqli_fetch_assoc($result))
                        {?>
                            <tr>
                                <td><?php echo $row["id"];?></td>
                                <td><?php echo $row["First Name"];?></td>
                                <td><?php echo $row["Last Name"];?></td>
                                <td><?php echo $row["Email"];?></td>
                            </tr>
                            <?php
                        }
                    }
                    mysqli_close($dbc);
                ?>
            </tbody>
        </table>
    </center>
</body>
<?php include('includes/footer.html');?>
</html>

All,
I'm getting the following error when I work with a class:
Parse error: syntax error, unexpected T_STRING, expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or '}' in /webspace/httpdocs/offers/form_key.php on line 5

The complete code for this is class is:

<?php
class Form_Key
{

    protected $oldKey;

    public function __construct()
    {
        // Ensure we have an available session
        if ( NULL == session_id() )
        {
            session_start();
        }
        // Grab our former key for validation
        if ( isset( $_SESSION['form_key'] ) )
        {
            $this->oldKey = $_SESSION['form_key'];
        }
        // Assign the new key
        $_SESSION['form_key'] = md5( uniqid( mt_rand(), TRUE ) );
    }

    public function isValid()
    {
        return 'POST' == $_SERVER['REQUEST_METHOD']
            && isset( $_POST['form_key'] )
            && '' != trim( $_POST['form_key'] )
            && '' != trim( $this->oldKey )
            && $_POST['form_key'] === $this->oldKey;
    }

    public function getKey()
    {
        return $_SESSION['form_key'];
    }

    public function getOldKey()
    {
        return $this->oldKey;
    }

    public function render()
    {
        return '<input type="hidden" name="form_key" value="' . $_SESSION['form_key'] . '" />';
    }

    public function __toString()
    {
        return $this->render();
    }

} 
?>

The line that is giving the issue is this line:
protected $oldKey;

Is it ok just to remove this line?? If I comment it out it gives me the same message but for line 7, which is this one:
public function __construct()

Anyone have any ideas? Thanks in advance.

Hi,   I have a text area field on one of my forms in order for people to post articles, however, it doesn't work when I try and post HTML through it. It works if I post normal text. Also, PHP won't give me an error, it just doesn't insert it?

<?php

if ($_POST['add']) {
    $title   = addslashes($_POST['title']);
    $image   = htmlspecialchars($_POST['image']);
	$source   = mysql_real_escape_string($_POST['source']);
    $active  = $_POST['active'];
	$feature  = $_POST['feature'];
    $cat_id  = $_POST['cat_id'];
	$content = htmlspecialchars($_POST['content']);
    $months  = array(
        "",
        "January",
        "February",
        "March",
        "April",
        "May",
        "June",
        "July",
        "August",
        "September",
        "October",
        "November",
        "December"
    );
    $date    = date('d', time()) . ' ' . $months[date('n', time())] . ' ' . date('Y', time());
    if ($title == NULL || $image == NULL || $content == NULL) {
        echo '<br /><br /><center>Please, fill all inputs</center><br /><br />';
    } else {
        $add = "INSERT INTO `news` cat_id='$cat_id', title='$title', image='$image', content='$content', date='$date', author='".$user['admin']."', authorid='".$user['id']."', source='$source', active='$active', twitter='".$user['twitter']."', featured='$feature'" or die(mysql_error());
        $sql = mysql_query($add);
        $addgrowl = "INSERT INTO `growl` (toid, message) VALUES ('$id', 'Your article is now online!')";
        $sql = mysql_query($addgrowl);    
            echo '<script type="text/javascript">
window.location = "articles.php"
</script>
';
}
}
?>

Help :-(

hey guys. Right now upon form submission, I do an error check that puts all the errors into an array, then will display them at the top of the form. What I want to do though is break up the array, and be able to give each error msg in its indivdual table row in the html. Right now, my coding is
if($lname == '') {
$errmsg_arr[] = 'You must enter your last name';
$errflag = true;
}
if($email == '') {
$errmsg_arr[] = 'You must enter your email address';
$errflag = true;
}
if($city == '') {
$errmsg_arr[] = 'You must enter your city';
$errflag = true;
}

if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: signup_artist.php");
exit();
}


thats in form2.php

form 1.php has the following snipet of code

<?php
if( isset($_SESSION['ERRMSG_ARR']) && is_array($_SESSION['ERRMSG_ARR']) && count($_SESSION['ERRMSG_ARR']) >0 ) {
echo '<ul class="err">';
foreach($_SESSION['ERRMSG_ARR'] as $msg) {
echo '<li>',$msg,'</li>'; 
}
echo '</ul>';
unset($_SESSION['ERRMSG_ARR']);
}
?>

so basically would I would like to be able to do is identify each error as for instane $fname_error $lname_error etc and then load them into my html appropriately, as opposed to just an array listing them