PHP - Resource Id #51

Hi,

I have the following code which I have been able to put together with alot of the brilliant help on this forum.  When I run this code it just came up with a blank screen and the CSS however I added another error print "echo "fetchdata: $fetchdata<br>Failed with error: " . mysql_error()  . '<br>';" and the result now is "fetchdata: Resource id #2".

However, when search for this on Google many of the responses data back from between 2002-2006.  Does it just mean that there is an error with the second if query or does  Resource id #2 refer to a specific error? Im really puzzled why I cant find anything more modern to this error on Google.

<?php
if( isset($_GET['id']) && ctype_digit($_GET['id']) ) { // validate that $_GET['id'] is set, and contains only numeric characters
$id = (int) $_GET['id']; // cast value as an integer, and assign to $id
$query = "SELECT * FROM productfeed WHERE id = $id";
if( !$fetchdata = mysql_query($query) ) { // numeric values shouldn't be quoted in query strings.
echo "query: $query<br>Failed with error: " . mysql_error()  . '<br>';
} else {
while($row = mysql_fetch_array($fetchdata))

$id = $row['id'];
$image = $row['awImage'];
$link = $row['link'];
$description = $row['description'];
$fulldescription = $row['fulldescription'];
$price = $row['price'];




echo "<div class=\"productdisplayshell\"> <div class=\"productdisplayoutline\"> <div class=\"productborder\"><center> <a href=\"$link\" target=\"_blank\" ><img src=\"$image\" /></a> </center> </div></div> <div class=\"productdescriptionoutline\"><div class=\"productdescriptionbox\"> <a href=\"$link\" target=\"_blank\" >$description</a> </div><div class=\"productfulldescriptionbox\"> $fulldescription </div></div> <div class=\"productpriceoutline\"> <div class=\"productpricebox\"><center>&#38;#163; $price</center></div> <div class=\"productbuybutton\"><center><a href=\"$link\" target=\"_blank\" ><img src=/images/buybutton.png /></a></center></div></div></div>";

echo "fetchdata: $fetchdata<br>Failed with error: " . mysql_error()  . '<br>';

}
} else {
echo 'Product is not available.  Please visit our <a href="http://www.ukhomefurniture.co.uk">Homepage</a>';
exit();
} 

?>

I'm trying to show my friend my website and it's not letting him or me view it.

I am using my own IP-address. (dashed out for security, but it is correct)
http://--.---.--.---/index-1.php
When he and I type this into our browser, we can an error called "Resource Not Found".

But, when I use localhost address, it works fine.
http://localhost/index-1.php

Does anyone know what is wrong?
Do I need to open a specific port?

USING XAMPP.

Here is what is echoing the string "Resource id #1". However I do not know if it is the php or javascript that is outputting this. Can anyone tell me why this is showing and how to make it not show? 

PHP:
Code: [Select]
$directory = "Images/items/$product/";
 
//get all image files with a .jpg extension.
$images = glob($directory . "*.jpg");

$imgone = $images[0];



$gallery = '<tr><td valign="top" align="center">';
foreach($images as $image)
{
$tn = explode("/", $image);
$tnname = $tn[3];

$gallery .= '<a href="#" rel="'.$image.'" class="image" alt="Images/items/'.$product.'/large/'.$tnname.'"><img src="Images/items/'.$product.'/thumbs/'.$tnname.'" class="thumb" border="1" style="margin-bottom:7px;"/></a> ';
}


if(is_dir("Images/items/".$product))
$gallery .= "</td></tr><tr><td width='300'>".$link."<div id='image' class='bigimg' align='left'>";

if(is_dir('Images/items/'.$product))
$gallery .= '<img src="'.$imgone.'" border="0"/></div></a></td></tr>';

JS (jQuery)
Code: [Select]
$(function() {
$(".image").click(function() {
var image = $(this).attr("rel");
var large = $(this).attr("alt");
$('#image').hide();
$('#image').fadeIn('slow');
$('#image').html('<a href="' + large + '" ><img src="' + image + '"/></a>');
return false;
    });
});

I am currently trying create a sales system where it checks the user's username against the database to check whether they are in the list of buyers.

The mysql query returns "Resource id #35", I need it to return the actual username (which I manually inserted into the database to test).


PHP code that fetches from database:


<?php
$con = mysql_connect("x","x","x");
if (!$con)
   {
   die('Could not connect: ' . mysql_error());
   }

mysql_select_db("x", $con);

$check_buyer = "SELECT * FROM Buyers WHERE Buyer='x'";

        $buyer = mysql_query($check_buyer);
?>



Product page:

<?php
include("/home/x/public_html/scripts/buyer.php");

if ($user->data['user_id'] == ANONYMOUS)
{
echo 'To use ' . $product . 'you must be logged in!';
    echo '<br /><a href="http://x/forum/ucp.php?mode=register">Register</a>';
echo ' or ';
     echo '<a href="http://x/forum/ucp.php?mode=login">Sign In</a>';
}

elseif ($user->data['username_clean'] == $buyer)
{
echo "<h3>Welcome to x</h3>";
}

else
{
    echo "You need to buy this product to use it!";

echo $user->data['username_clean']; //test whether username is outputted correctly - which it did
echo $buyer; //Fetched from mysql - returned "Resource id #35", not the desired username
}
?>


Any help would be great!


Thanks,

otester

Hi, I was in the process of making thumnails for avatars It used to work before but now the files are being saved as resource ID and not in an image format. I can not point out what the issue is and also may I add when I try to echo nothing is happening.

Code: [Select]
$filename = $_FILES['myfile']['tmp_name'];

if ($_POST['cpic'])

{


// Set a maximum height and width
$width = 100;
$height = 100;


// Get new dimensions
list($width_orig, $height_orig) = getimagesize($filename);

$ratio_orig = $width_orig/$height_orig;
$ratio_orig;


if ($width/$height > $ratio_orig) {
   $width = $height*$ratio_orig;
} else {
   $height = $width/$ratio_orig;
}


// Resample
$image_p = imagecreatetruecolor($width, $height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $width, $height, $width_orig, $height_orig);

$location = "cid/$myid/$filename";
imagejpeg($image_p, $location, 100);


$query =mysql_query ("UPDATE clp SET avatar ='$location' WHERE cid='1'");
header ("LOCATION: editprofile.php");
}

If $result contains the result of mysql_query, a select count distinct query, how do I access the count?  I've tried $result[0] to no avail and $result only returns the resource #.

I keep getting the mysql resource id #4 on a query that I am running and I have tried everything that I have read to fix it and nothing is working. I tried using the mysql_fetch_array, mysql_fetch_accoc, and the mysql_fetch_row functions

I would appreciate any help that can be given to me.

Here is my code as it stands now
Code: [Select]
<?php

include('includes/config.php');

$last = $_GET['l_name'];
$first = $_GET['f_name'];

$sql = 'SELECT * FROM `ttmautos` WHERE `l_name` LIKE \'$last\' AND `f_name` LIKE \'$first\''; 
$autographs = mysql_query($sql, $connection)or die(mysql_error()); 
$row = mysql_fetch_array($autographs); 

echo $sql;
echo $autographs;

$l_name = $row['l_name'];
$f_name = $row['f_name'];
$sent = $row['date_sent'];
$return = $row['date_return'];
$address = $row['address'];
$isent = $row['item_sent'];
$ireturn = $row['item_return'];
$project = $row['project'];
$team = $row['team'];

$address = stripslashes($address);

?>

How can I fix this error?

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

Code: [Select]
function fetch_most_recent_fans($ctag)
{
$sql = "SELECT
`company_fans`.`company_id`,
`company_fans`.`user_id`,
`company_fans`.`fan_date`,
`companies`.`companytag`,
`users`.`firstname`,
`users`.`lastname`,
`users`.`username`
FROM `company_fans`
LEFT JOIN `companies`
ON `companies`.`companyid` = `company_fans`.`company_id`
LEFT JOIN `users`
ON `users`.`id` = `company_fans`.`user_id`
WHERE `companies`.`companytag` = {ctag}
ORDER BY `company_fans`.`fan_date` DESC
LIMIT 10";

$query = mysql_query($sql);

$return = array();

while (($row = mysql_fetch_assoc($query)) !== false)
{
$return[] = $row;
}

return $return;
}

Hi

I changed my connection in my application to odbc mssql
-> using adodb.org libraries
-> It connects fine
-> The update query below works
-> But when I pass the select query below SQL odbc_exec does not return a resource id for my query
-> None of my select queries returns a resource id

Below is the details that I logged.
-> Any idea how to fix
-> If I paste the select sql in MS Sql management studio it executes fine.

Please assist in what could be the cause? 

Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql = SET CONCAT_NULL_YIELDS_NULL OFF
odbc_exec = Resource id #89
Time->2020-07-30 19:09:14adodb-odbc.inc.php Connection Arguments => DSN=amphibian;UID=amphibian;PWD=speedster;DATABASE=crmapm;,,
Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql = Update pwd_incorrect set record_status_id = 2 where date <> '2020-07-30'
odbc_exec = Resource id #92
Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql = SET CONCAT_NULL_YIELDS_NULL OFF
odbc_exec = Resource id #95
Time->2020-07-30 19:09:14adodb-odbc.inc.php Connection Arguments => DSN=amphibian;UID=amphibian;PWD=speedster;DATABASE=crmapm;,,
Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql = SET CONCAT_NULL_YIELDS_NULL OFF
odbc_exec = Resource id #98
Time->2020-07-30 19:09:14adodb-odbc.inc.php Connection Arguments => DSN=amphibian;UID=amphibian;PWD=speedster;DATABASE=crmapm;,,
Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql = select c.name,t.name,c.length,c.isnullable, c.status,
(case when c.xusertype=61 then 0 else c.xprec end),
(case when c.xusertype=61 then 0 else c.xscale end)
from syscolumns c join systypes t on t.xusertype=c.xusertype join sysobjects o on o.id=c.id where o.name='pwd_user_deny'
odbc_exec = Resource id #101
Time->2020-07-30 19:09:14 adodb-odbc.inc.php sql =
SELECT user_name_used,password_used,ip_address,date,ref_no,record_status_id FROM crmapm.dbo.pwd_user_deny WHERE record_status_id = 1 and ip_address = '::1'
odbc_exec =

Okay, I have a table with two columns..'annoyed' and 'ignored'. I poulated 'annoyed' with Bob and 'ignored' with Sally. I ran the following test due to problems I was having in my main script and I get the old MySQL "Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/testing_ig.php on line 16

Here is the code:


<?php
$DBhost = "localhost";
$DBuser = "xxxxx";
$DBpass = "xxxxx";
$DBName = "xxxxx";
$table = "ignore";

mysql_connect($DBhost,$DBuser,$DBpass) or die();
@mysql_select_db("$DBName") or die();

$ignorequery = "SELECT * FROM $table WHERE annoyed='Bob' AND ignored='Sally'";
$checkignore = mysql_query($ignorequery);
$ifignored = mysql_num_rows($checkignore); //<--- This is line 16

if ($ifignored >= 1) {
echo "exists";
} else {
echo "doesn't exist";
}
?> 


I must be really tired because I'm not seeing the problem...?

I created this function to update my tour system.  The query is working and is updating one row in the table, but I get a resource boolean error in the return section of the function.  Any idea why?

Code: [Select]
function update_tour($uid, $step)
{
$step = (int)$step;
$uid = (int)$uid;

$sql = "UPDATE `users`
SET
`tour_step` = ${step}
WHERE `id` = '${uid}'";

$q = mysql_query($sql) or die(mysql_error());

return (mysql_num_rows($q) === 1) ? mysql_result($q, 0): false;
}

I always use my queries like below
mysql_query("SELECT id FROM someTable", $cn);

but some times i see in my error_log says: mysql_fetch_array(): supplied argument is not a valid MySQL result resource. Why is this happening?
I mean, there is no problem with my sites and all they are working as they should be but why php generates this error?

Hi!

I need help understanding how to access resource files using ResourceBundle Class and what should be ResourceBundle file format.

I have an example of usage in the form of:

$r = ResourceBundle::create( 'en', $fileName);

but what is the correct format of Resource file?

Can it be a format of property file in the form of
key=value
key=value?

or can it be a .txt file of format?:

root:table {
  myName:string { "Here goes my name" }
}


Can someone share a very simple example of ready to use Resource file with a plain text in it?

Thank you!

I am trying to make a thumbnail from an FLV using ffmpeg and ffmpeg-php. I got them all installed.

This is my script:

<?php
$extension = "ffmpeg";
$extension_soname = $extension . "." . PHP_SHLIB_SUFFIX;
$extension_fullname = PHP_EXTENSION_DIR . "/" . $extension_soname;

// load extension
if(!extension_loaded($extension))
{
    dl($extension_soname) or die("Can't load extension $extension_fullname\n");
}

include("../connect.php");

header("Content-type: image/jpeg");
$id  = $_GET['id'];
$query  = mysql_query("SELECT filename FROM video WHERE id = '$id'");
$result  = mysql_fetch_array($query);
$moviefile  = "../../../../armsmedia/videos/".$result['filename'];
$mov  = new ffmpeg_movie($moviefile,false);
$img  = $mov->getFrame(1);
$showImg  = $img->toGDImage();
$mkNewImg  = new ffmpeg_frame($showImg);
$maxWid  = 150;
$oldWid  = $mkNewImg->getWidth();
if($oldWid > $maxWid) {
$newWid = $maxWid;
}
$newHgt  = $newWid / $movRatio;
$mkNewImg->resize($newWid,$newHgt);
$newImg  = $mkNewImg->toGDImage();
imagejpeg($newImg,$mkThumbFile,40);
imagedestroy($newImg);
?>


I get a 500 Internal server error so I checked my apache logs and it is saying:

Unable to locate ffmpeg_frame resource in this object. in /var/www/html/inc/video/thumbnail.php on line 24 which is pointing to:
Code: [Select]
$oldWid = $mkNewImg->getWidth();

If I do a print_r($showImg) it returns: Resouce id #7
Any ideas?

Hi

I'm having a bit of bother with my login.  I created a login using this tutorial  http://www.phpeasystep.com/phptu/6.html and it works perfectly. 

So i have attempted to change it to meet my own database. So basically i've changed the database, table names etc to meet my own. I haven't changed any other lines. When i run it i get an error message:

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\checklogin.php on line 26

The code is below:

Code: [Select]
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="final year project"; // Database name
$tbl_name="tbl_user"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// username and password sent from form
$mem_username=$_POST['mem_username'];
$mem_password=$_POST['mem_password'];

// To protect MySQL injection (more detail about MySQL injection)
$mem_username = stripslashes($mem_username);
$mem_password = stripslashes($mem_password);
$mem_username = mysql_real_escape_string($mem_username);
$mem_password = mysql_real_escape_string($mem_password);

$sql="SELECT * FROM $tbl_name WHERE username='$mem_username' and password='$mem_password'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $mem_username and $mem_password, table row must be 1 row

if($count==1){
// Register $mem_username, $mem_password and redirect to file "login_success.php"
session_register("mem_username");
session_register("mem_password");
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
?>

Line 26 is $count=mysql_num_rows($result);


I'm baffled as to why the test database worked. I tried another test database but got the same error. baffled.com

Hope someone can help

MOD EDIT: [code] . . . [/code] tags added.