PHP - Warning: Mysql_pconnect() [function.mysql-pconnect]:

Please help me with this code. It keep appear in my website.

"Warning: preg_match() [function.preg-match]: No ending delimiter '.' found in /home/kengsite/public_html/wp-content/plugins/statpress/statpress.php on line 1184"

And this is how the line look like.

if (!preg_match(".ico$", $urlRequested)) { return ''; }

Hi All,

I am trying to create a PHP code that will run a form, however i keep getting the error message

Warning: preg_match() [function.preg-match]: No ending delimiter '^' found in D:\xampp\htdocs\HFCC\Scripts\mailprocess.php on line 55

Invalid email address entered. Go Back and try again

attached is the complete copy of the code i am using, i have an email text box and label with the id 'email' also my method is POST and my action is Scripts/mailprocess.php.

Please let me know if you need any additional details. Any help will be much appreciated.

Thanks

some please help - basically have a simple password change php page

$sql="UPDATE t_members SET Password = '$New_Password' WHERE Username = '$My_Username' AND  Password = '$Old_Password'";
   
   echo $sql;
   
   $result=mysql_query($sql) or die(mysql_error());
   $firstcount = mysql_num_rows($result);

and i get this annoying error Warning: mysql_num_rows(): supplied argument is not a valid

the sql statement works as i see it updating the database correctly

 I am simply trying to display the last 8 images from a mysql database and i am getting this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/aretheyh/public_html/nealeweb.com/gnew2/recent.php on line 8

Can somebody please tell me what i am doing wrong.

Code: [Select]
<?php
$i = 1;
echo '<table border="0" cellpadding="0" cellspacing="0">
<tr>';
include("config.inc.php"); 

        $result =  mysql_query("SELECT * FROM gallery_photos ORDER BY id DESC LIMIT 8"); 
        while($row = mysql_fetch_assoc($result)) { // Start of the loop 
        $img = $row['".$images_dir."/tb_']; 

echo "<td><img src=$img></td>"; 

if ($i == 4) {
echo '</tr><tr>';
}

$i++;
}
echo '</tr></table>'; 

Hi, first post, and yes it is a question.

I am stuck, and not by choice, this error has given me more headaches than I care to admit.

I have a script that I am attempting here, that is very simple for now, all I want to show is the Added_By field to show it is accessing the database correctly and the right row/line all together.

The page to test this at is he

http://kaboomlabs.com/PDI/@dm!n/viewncmr.php?id=2

The Error is this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pawz/public_html/kaboomlabs.com/PDI/@dm!n/viewncmr.php on line 18

This is row 18:     $row = mysql_fetch_array($data);

Now I know this is not a secure form yet, I am working on getting the basic functions down then I'll secure it, so please let me worry about that when the time comes.

Here is the script, can anyone see a blatant issue or not? Thanks in advance.

Code: [Select]
<?php
  require_once('../connectvars.php');

        // Connect to the database
        $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

  // Grab the profile data from the database
  if (!isset($_GET['id'])) {
    $query = "SELECT * FROM ncmr WHERE id = '" . $_SESSION['id'] . "'";
  }
  else {
    $query = "SELECT * FROM ncmr WHERE id = '" . $_GET['id'] . "'";

  $data = mysqli_query($dbc, $query);
  }
  if (mysqli_num_rows($data) == 1) {
    // The user row was found so display the user data
    $row = mysql_fetch_array($data);

echo '<fieldset>';

        if (!empty($row['Added_By'])) {
        echo '<div id ="added"><label>Added By:</label>' . $row['Added_By'] . '</div></fieldset>';
        }
}
  // End of check for a single row of user results
  else {
    echo '<p class="error">There was a problem accessing your profile.</p>';
  }

  mysqli_close($dbc);
?>

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource on line 14.

<?PHP
require_once("dbcheck.php");
require_once("check.php");
$Username=$_COOKIE['Username'];
$Password=$_COOKIE['Password'];
$Nickname=$_COOKIE['Nickname'];
$Username=mysql_real_escape_string($Username);
$Password=mysql_real_escape_string($Password);

date_default_timezone_set('Asia/Calcutta');

$action = $_GET['do'];
$Result=mysql_query("SELECT * FROM Checks WHERE UserName='$Username'");
$Rowexist=mysql_num_rows($Result);
if($Rowexist!=0){
while($Rows=mysql_fetch_array($Result))
{
$Serial    = $Rows['Serial'];
$Banned    = $Rows['Banned'];
$IP       = $Rows['IP'];
$Used       = $Rows['Used'];
$First       = $Rows['First'];
$Duration    = $Rows['Duration'];
$Total       = $Rows['Total'];
$Time       = $Rows['Time'];
}
if($action=="" || $action=="Banlist"){
mysql_query("UPDATE Checks SET NickName='$Nickname' WHERE UserName='$Username'");
?>

Hi,
I am having problem with mysql and php.... when i test the database on my local host it didt produce any error but when i put it on my web host it gave me error messages... "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource" Please help me as i am new to php.


Quote

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
        <link rel="stylesheet" type="text/css" href="http://www.hingloong.com/pricesearch/style.css" />
        <title>Price Search Query</title>
    </head>
    <body>
        <form method=GET action="search.php">
            <div id=container>
               
                <select name=general>
                    <option value = all>Select</option>
                    <option value = laptop>Laptop</option>
                    <option value = mobile>Mobile</option>
                    <option value = games>Games</option>
                    <option value = camera>Camera</option>
                </select>
               
                <label><b>Brand:</b></label>
                <input type=text name=brand>
                <label><b>Model:</b></label>
                <input type=text name=model>
                <input type=submit name=search value=Search>
               
               
        </form>
        <br><br>
        <input type=button name=gamelist value=View&nbsp;Games onclick="location.href='gamelist.php'">
        <input type=button name=mobilelist value=View&nbsp;Mobile onclick="location.href='mobilelist.php'">
        <input type=button name=cameralist value=View&nbsp;Cameras onclick="location.href='cameralist.php'">
        <input type=button name=laptoplist value=View&nbsp;Laptops onclick="location.href='laptoplist.php'">
        </div>
    </body>
</html>

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

$general = $_GET['general'];
$brand = $_GET['brand'];
$model = $_GET['model'];




if($general == 'mobile')
{
   
    echo "<br>";
    echo "<h3>Mobile Phone Prices</h3>";
    echo "<div id=container>";
    echo "<table>";
    echo "<tr>";
    echo "<td class=head>Phone Brand</td>";
    echo "<td class=head>Phone Model</td>";
    echo "<td class=head>Phone Loan</td>";
    echo "<td class=head>Phone Buy</td>";
    echo "<td class=head>Phone Sell</td>";
    echo "<td class=head>Phone Rrp</td>";
    echo "</tr>";
   
   
$query = "select * from mobilephones where phone_brand LIKE ('$brand%') AND phone_model LIKE ('$model%') ";
$temp = mysql_query($query) or die(mysql_error());





}


while ($row = mysql_fetch_array($temp))
{

    echo "<tr>";
    echo "<td>" .$row['phone_brand']. "</td>";
    echo "<td>" .$row['phone_model']. "</td>";
    echo "<td>" .$row['phone_loan']. "</td>";
    echo "<td>" .$row['phone_buy']. "</td>";
    echo "<td>" .$row['phone_sell']. "</td>";
    echo "<td>" .$row['phone_rrp']. "</td>";


   
   
}

    echo "</table>";
    echo "</div>";



?>

Hi,

After moving server my script started showing this error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/user/public_html/blog.php on line 107

Here is the code on line 107:
if(mysql_num_rows($ExeqryComm)!=0){

Any help please?

Thank you all

Hi, I am having an issue with a piece of code. I am getting the following error, any ideas please?

Quote

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ...


Code: [Select]
<?php
include "scripts/connect.php";

$last = $_GET['ref']; // This will be the ID of the page

echo "the ref number is $last";

$query = mysql_query("SELECT * FROM news WHERE `ref` = '$last'");

if(mysql_num_rows($query)==0){

die("Something went wrong, contact Administrator!");

  }else{
 
while($info = mysql_fetch_array($query)){
$title = $info ['title'];

echo "<p><a href=\"..news.php\" target=\"_blank\">$title >></a></p>";

}
}
?>

Hi all, complete newbie here.

i have copied this example php code to insert data and a picture into a mysql database. however i get this error.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

can be seen here : http://wycombedjs.co.uk/mix/sqltest.php

Any help would be much appreciated.. thanks


Code: [Select]
<?php
$db_host = 'xxx'; // don't forget to change 
$db_user = 'xxx'; 
$db_pwd = 'xxx';

$database = 'mixdb';
$table = 'ae_gallery';
// use the same name as SQL table

$password = '123';
// simple upload restriction,
// to disallow uploading to everyone


if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

// This function makes usage of
// $_GET, $_POST, etc... variables
// completly safe in SQL queries
function sql_safe($s)
{
    if (get_magic_quotes_gpc())
        $s = stripslashes($s);

    return mysql_real_escape_string($s);
}

// If user pressed submit in one of the forms
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
    // cleaning title field
    $title = trim(sql_safe($_POST['title']));

    if ($title == '') // if title is not set
        $title = '(empty title)';// use (empty title) string

    if ($_POST['password'] != $password)  // cheking passwors
        $msg = 'Error: wrong upload password';
    else
    {
        if (isset($_FILES['photo']))
        {
            @list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']);
            // Get image type.
            // We use @ to omit errors

            if ($imtype == 3) // cheking image type
                $ext="png";   // to use it later in HTTP headers
            elseif ($imtype == 2)
                $ext="jpeg";
            elseif ($imtype == 1)
                $ext="gif";
            else
                $msg = 'Error: unknown file format, .png, .jpeg, .gif only!';

            if (!isset($msg)) // If there was no error
            {
                $data = file_get_contents($_FILES['photo']['tmp_name']);
                $data = mysql_real_escape_string($data);
                // Preparing data to be used in MySQL query

                mysql_query("INSERT INTO {$table}
                                SET ext='$ext', title='$title',
                                    data='$data'");

                $msg = 'Success: image uploaded to Mix db';
            }
        }
        elseif (isset($_GET['title']))      // isset(..title) needed
            $msg = 'Error: file not loaded';// to make sure we've using
                                            // upload form, not form
                                            // for deletion


        if (isset($_POST['del'])) // If used selected some photo to delete
        {                         // in 'uploaded images form';
            $id = intval($_POST['del']);
            mysql_query("DELETE FROM {$table} WHERE id=$id");
            $msg = 'Photo deleted';
        }
    }
}
elseif (isset($_GET['show']))
{
    $id = intval($_GET['show']);

    $result = mysql_query("SELECT ext, UNIX_TIMESTAMP(image_time), data
                             FROM {$table}
                            WHERE id=$id LIMIT 1");

    if (mysql_num_rows($result) == 0)
        die('no image');

    list($ext, $image_time, $data) = mysql_fetch_row($result);

    $send_304 = false;
    if (php_sapi_name() == 'apache') {
        // if our web server is apache
        // we get check HTTP
        // If-Modified-Since header
        // and do not send image
        // if there is a cached version

        $ar = apache_request_headers();
        if (isset($ar['If-Modified-Since']) && // If-Modified-Since should exists
            ($ar['If-Modified-Since'] != '') && // not empty
            (strtotime($ar['If-Modified-Since']) >= $image_time)) // and grater than
            $send_304 = true;                                     // image_time
    }


    if ($send_304)
    {
        // Sending 304 response to browser
        // "Browser, your cached version of image is OK
        // we're not sending anything new to you"
        header('Last-Modified: '.gmdate('D, d M Y H:i:s', $ts).' GMT', true, 304);

        exit(); // bye-bye
    }

    // outputing Last-Modified header
    header('Last-Modified: '.gmdate('D, d M Y H:i:s', $image_time).' GMT',
            true, 200);

    // Set expiration time +1 year
    // We do not have any photo re-uploading
    // so, browser may cache this photo for quite a long time
    header('Expires: '.gmdate('D, d M Y H:i:s',  $image_time + 86400*365).' GMT',
            true, 200);

    // outputing HTTP headers
    header('Content-Length: '.strlen($data));
    header("Content-type: image/{$ext}");

    // outputing image
    echo $data;
    exit();
}
?>
<html><head>
<title>MySQL Blob Image Gallery Example</title>
</head>
<body>
<?php
if (isset($msg)) // this is special section for
                 // outputing message
{
?>
<p style="font-weight: bold;"><?=$msg?>
<br>
<a href="<?=$PHP_SELF?>">reload page</a>
<!-- I've added reloading link, because
     refreshing POST queries is not good idea -->
</p>
<?php
}
?>
<h1>Blob image gallery</h1>
<h2>Uploaded images:</h2>
<form action="<?=$PHP_SELF?>" method="post">
<!-- This form is used for image deletion -->

<?php
$result = mysql_query("SELECT id, image_time, title FROM {$table} ORDER BY id DESC");
if (mysql_num_rows($result) == 0) // table is empty
    echo '<ul><li>No images loaded</li></ul>';
else
{
    echo '<ul>';
    while(list($id, $image_time, $title) = mysql_fetch_row($result))
    {
        // outputing list
        echo "<li><input type='radio' name='del' value='{$id}'>";
        echo "<a href='{$PHP_SELF}?show={$id}'>{$title}</a> &ndash; ";
        echo "<small>{$image_time}</small></li>";
    }

    echo '</ul>';

    echo '<label for="password">Password:</label><br>';
    echo '<input type="password" name="password" id="password"><br><br>';

    echo '<input type="submit" value="Delete selected">';
}
?>

</form>
<h2>Upload new image:</h2>
<form action="<?=$PHP_SELF?>" method="POST" enctype="multipart/form-data">
<label for="title">Title:</label><br>
<input type="text" name="title" id="title" size="64"><br><br>

<label for="photo">Photo:</label><br>
<input type="file" name="photo" id="photo"><br><br>

<label for="password">Password:</label><br>
<input type="password" name="password" id="password"><br><br>

<input type="submit" value="upload">
</form>
</body>
</html>

In the following code, at the mysql query immediately following the <!--end accordianButton div-->,

Some of the rows will echo out the content, while some of the rows will throw the Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in... error.  Any idea why?

Go here to see what I mean:  http://www.chalmerscommunitychurch.com/P2P_archives.php


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Pastor to People Blog Archives</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"> </script>
<script type="text/javascript" src="js/javascript.js"> </script>
<style>
.accordionButton {   
   width: 100%;
   height:30px;
   float: left;
   background: url(../images/button.png);
   border-bottom: 1px solid #FFFFFF;
   cursor: pointer;
   }
 
.accordionContent {   
   width: 100%;
   float: left;
   display: none;
   
   }
</style>
</head>

<body>
<?php
require("include.php");
$con = mysql_connect("$db_host","$db_username","$db_pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("chalmers_db", $con);


           
            $result = mysql_query("SELECT title, date FROM blog");
            while($row = mysql_fetch_array($result))
          {
           $title=$row['title'];
           $sqldate=$row['date'];
           $date=date('m-d-Y',strtotime($sqldate));
            ?>
            <!--start accordionButton div-->
            <div class="accordionButton"><?php echo $title;?>, <?php echo $date;?>
            </div>
            <!--end accordianButton div-->
               
                   
                    <?php
               
                        $query = mysql_query("SELECT content FROM blog WHERE title = '".$title."' ORDER BY date DESC");
                        while ($row = mysql_fetch_array($query)){
                           $content = $row['content'];
                           
                        ?>   
                           <!--start accordionContent div-->
                           <div class="accordionContent" align="justify">
                              <?php echo $content;?>                     

                           </div>
                           <!--end accordionContent div-->
                        <?php
                        }
                        ?>                                                
                          
               
               <?php } ?>   
       <p><p><p><p><a href="http://www.chalmerscommunitychurch.com"><h3>Back to Chalmers Community Church</h3></a>
       </body>
</html>

Hi guys, for the life of me, what am I doing wrong, I cannot figure out this one, getting this error :

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ......... search.php on line 36

my code :

Code: [Select]

$start = ($page-1)*$per_page;
$sql = "SELECT productId, productCode, image, name, price, stock_level FROM inventory WHERE productCode LIKE '%".$searchp."%' OR name LIKE  '%".$searchp."%' AND Seller_ID = '" . $_SESSION['SESS_SELL_ID'] . "' order by name limit $start,$per_page";
$rsd = mysql_query($sql);


<?php


while($row = mysql_fetch_array($rsd))  // ERROR OCCURS HERE : line 36
{

$idpc=$row['productId'];
$idc=$row['productCode'];
$idi=$row['image'];
$idn=$row['name'];
$idp=$row['price'];
$ids=$row['stock_level'];

?>


all help appreciated as always.