PHP - Reference Stored Image

Hi   I am winning I think   I have got the records displayed for the current user logged in so basically they can only see their submitted listings and just working on the edit of them so the current user logged in can update their listing and is all working apart from the update of the images   the images are stored by the file name on the database and then gets moved onto the server so the actual images are not stored on the database only the file names are and the images are moved onto the server, hope that makes sense   what I can't do at the mo is work out how to update the file names of the images on the database and update on the server   I get the following error   Notice: Undefined index: photo on line 209 and line 211   the coding for them lines are below

$pic1= basename($_FILES['photo']['name'][0]);

$pic2= basename($_FILES['photo']['name'][1]);

The rest of the coding below that is

if(!empty($_FILES['photo']['tmp_name']))
{
	
	// Number of uploaded files
	 $num_files = count($_FILES['photo']['tmp_name']);

	 /** loop through the array of files ***/
	 for($i=0; $i < $num_files;$i++)
	 {
		 // check if there is a file in the array
		 if(!is_uploaded_file($_FILES['photo']['tmp_name'][$i]))
		 {
			 $messages[] = 'No file uploaded';
		 }
		 else
		 {
			 // move the file to the specified dir
			 if(move_uploaded_file($_FILES['photo']['tmp_name'][$i],$target.'/'.$_FILES['photo']['name'][$i]))
			 {
			
				 $messages[] = $_FILES['photo']['name'][$i].' uploaded';
				
			 }
			 else
			 {
				 // an error message
				 $messages[] = 'Uploading '.$_FILES['photo']['name'][$i].' Failed';
			 }
		 }
	 }

The update query is below

// save the data to the database
 mysql_query("UPDATE privatelistings SET listingtitle='$listingtitle', make='$model', model='$model', exteriorcolour='$exteriorcolour', enginesize='$enginesize', fueltype='$fueltype', yearregistered='$yearregistered', transmission='$transmission', mileage='$mileage', nodoors='$nodoors', bodystyle='$bodystyle', price='$price', photo='$pic1', photo1='$pic2' WHERE id='$id'")
 or die(mysql_error()); 

I am going to be updating to mysqli just want to get it working first   My HTML form coding is below

<form action="" method="post" enctype="multipart/form-data">
 <input type="hidden" name="id" value="<?php echo $id; ?>"/>
 <div>
 <strong>Listing Title: *</strong> <input type="text" name="listingtitle" value="<?php echo $listingtitle; ?>"/>
 <br/>
 <strong>Make: *</strong> <input type="text" name="make" value="<?php echo $make; ?>"/>
 <br/>
 <strong>Model: *</strong> <input type="text" name="model" value="<?php echo $model; ?>"/>
 <br/>
 <strong>Exterior Colour: *</strong> <input type="text" name="exteriorcolour" value="<?php echo $exteriorcolour; ?>"/>
 <br/>
 <strong>Engine Size: *</strong> <input type="text" name="enginesize" value="<?php echo $enginesize; ?>"/>
 <br/>
 <strong>Fuel Type: *</strong> <input type="text" name="fueltype" value="<?php echo $fueltype; ?>"/>
 <br/>
 <strong>Year Registered: *</strong> <input type="text" name="yearregistered" value="<?php echo $yearregistered; ?>"/>
 <br/>
 <strong>Transmission: *</strong> <input type="text" name="transmission" value="<?php echo $transmission; ?>"/>
 <br/>
 <strong>Mileage: *</strong> <input type="text" name="mileage" value="<?php echo $mileage; ?>"/>
 <br/>
 <strong>Number of Doors: *</strong> <input type="text" name="nodoors" value="<?php echo $nodoors; ?>"/>
 <br/>
 <strong>Body Style: *</strong> <input type="text" name="bodystyle" value="<?php echo $bodystyle; ?>"/>
 <br/>
 <strong>Price: *</strong> <input type="text" name="price" value="<?php echo $price; ?>"/>
 <br/>
 <strong>Photo One:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'>
 <br>
 <strong>Photo Two:</strong> <input type='hidden' name='size' value='350000'><input type='file' name='photo[]'>
 <br>
 <input type="submit" name="submit" value="Submit">
 </div>
 </form>

sorry was not sure what other I need to show, so you guys get the idea?

I put together the following blocs of code for uploading pictures into a database and displaying them on a webpage.  The pictures are supposed to be displayed on the member's only page of a website I'm working on, upon logging in, and they are supposed to be the member's uploaded picture.  I created several members and and used one of my existing member accounts to test the uploading process.  The picture upload process appeared to have been successful when I checked on myphpadmin.  Yet,  when I login with this account, no picture is displayed, instead, a tiny jpg icon is displayed at the top left corner of the box in which the picture was supposed to be displayed.  Same thing when I login with the other accounts with which I haven't yet uploaded a picture.

I'll start with the code that installs the table in the database

$query = "CREATE TABLE images (

image_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
member_id INT UNSIGNED,
like_id INT UNSIGNED,
image LONGBLOB NOT NULL,
image_name varchar(255)  NOT NULL,
image_type varchar(4) NOT NULL,
image_size int(8) NOT NULL,
image_cartegory VARCHAR(20) NOT NULL,
image_path VARCHAR(300),
image_date DATE  

          )";




Then here is the code  which allows the member to upload his pictu
<form enctype="multipart/form-data" action="insert_image.php" method="post" name="changer">
                         <input name="MAX_FILE_SIZE" value="102400" type="hidden">
                         <input name="image" accept="image/jpeg" type="file">
                         <input value="Submit" type="submit">
                        </form>


 And here is the insertimage.php which inserts the image into our database:  Note that I have to authenticate the user in order to register his session id which is used later on in the select query  to identify him and select the right image that corresponds to him.


<?php

//This file inserts the main image into the images table.

//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);





//authenticate user
//Start session

session_start();

       
        //Connect to database
require ('config.php');

//Check whether the session variable id is present or not. If not, deny access.

if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) {

header("location: access_denied.php");

exit();

}

        else{ 
// Make sure the user actually 
// selected and uploaded a file
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { 

      // Temporary file name stored on the server
      $tmpName  = $_FILES['image']['tmp_name'];  
       
      // Read the file 
      $fp      = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);
      

      // Create the query and insert
      // into our database.
      $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')";
      $results = mysql_query($query);
      
      // Print results
      print "Thank you, your file has been uploaded.";
      
}
else {
   print "No image selected/uploaded";
}

// Close our MySQL Link
mysql_close();


} //End of if statmemnt.
 


?>  



On the page which is supposed to display the image upon login in, I inserted the following html code in the div that's supposed to contain the image:


                    <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border-           color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; ">
                 
                         <img src=picscript.php?imname=potwoods>
                             
                    </div>

                   
And finally, the picscript.php  contained the select query:
<?php

                             //address error handling

                             ini_set ('display_errors', 1);
                             error_reporting (E_ALL & ~E_NOTICE);

                             //include the config file
                              require('config.php');

                             $image = stripslashes($_REQUEST[imname]);
                             $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND cartegoty = 'main' ");
                             $row = mysql_fetch_assoc($rs);
                             $imagebytes = $row[image];
                             header("Content-type: image/jpeg");
                             print $imagebytes;

                            
 ?>



Now the million dollar question is, "What is preventing the picture from getting displayed?"  I know this is a very lengthy and laborious problem to follow but I'm sure there is someone out there who can point out where I'm not getting it.  Thanks.

Hi Everyone,

I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two  images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening.....

Here is MY code..........



<?php
      session_start();
      $sid = $_SESSION['id'];
      $_SESSION['imageDispCnt'] = 0;
$myQuery = "SELECT * from image";
$conn = mysql_connect("localhost","User","Passwd");
mysql_select_db("database_Name",$conn);
$result = mysql_query($myQuery);
$img =Array();
$id =Array();
$i =0;
$imagepath = 'http://localhost/images/';
while ($row = mysql_fetch_array($result))
      {
      $img[$i] =  $imagepath.$row['img_name'];
      $id[$i] =  $row['imageid'];
      $i = $i + 1;
       }
 ?>
      </head>
      <script language="JavaScript1.2">
var scrollerwidth='800px';
var scrollerheight='600px';
var scrollerbgcolor='white';
var pausebetweenimages=3200;
var s;
var sec;
var d;
var j;
var imgid;
var milisec = 0;
var seconds = 0;
var flag = 1;
var ses_id = '<?php echo $sid;?>';
var count = 0;
var i = 0;
var imgname;
var imgid;
var k =0;
var slideimages=new Array();
var img_id = new Array();
var index;

<?php
  $l =0;   
$count = array();
   $j = rand(0,199);
   while($l < 200)
   {
    while(in_array($j, $count))
        {
           $j = rand(0,199);
        }
        $count[$l] =  $j;
        $l++;
       
     }?>
    <?php
  $k = 0;
  for($k = 0;$k<count($count);$k++){
  ?>
  index = <?php echo $k;?>;
  <?php
        $indx = $count[$k];?>
  if(index == 0){
     slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>';
     img_id[0] = '<input type="hidden" value="0" id="imgId" />';
  }
 
  else if(index > 0)
  {
    slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>';
     img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />';
  }
<?php
}
?>


Can Any one plese help me

Appreciate your help...


Thanks

Okay, so here is the deal.

Have a table which stores image as blob files. Now i want to read the image width and height directly from the blob field.
Is this possible and if yes, how?

Things i tried so far;
list($size[0],$size[1],$type, $attr) = getimagesize('image.php?i=26ddd45b02859e836d13d4b9fde34281');      
print_r($size);

$img = 'image.php?i=26ddd45b02859e836d13d4b9fde34281';
echo imagesy($img);

image.php grabs the image from DB and show's it with header("Content-type: image/jpg");   
It works for just showing the images with the <img> tag.

Any ideas of help would be great!

 wrote a stored procedure this morning and i don’t know how to get the values out of it through a class function in php or phpmyadmin.

here is what i wrote :

public function totalProcedures($friend_name,$session_id) 
{ /* *query to fetch stored procedure */ 
try { 
//executing the stored procedure 
$sql_sp="CALL timeline (:friend, :session,@updates, @group_posts)";
$stmt_sp= $this->_db->prepare($sql_sp); 
$stmt_sp->bindValue(":friend",$friend_name);
$stmt_sp->bindValue(":session",$session_id); 
$stmt_sp->execute();
$rows=$stmt_sp->fetch(PDO::FETCH_ASSOC); 
$stmt_sp->closeCursor(); // closing the stored procedure 
//trying to get values from OUT parameters.
$stmt_sp_2=$this->_db->prepare("select @updates,@group_posts");
$stmt_sp_2->execute();
return $stmt_sp_2->fetch(PDO::FETCH_ASSOC);
} catch (PDOException $ei)
{
echo $ei->getMessage(); } 
}

can someone helpme how to get results.

here is the storedprocedu

DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `timeline`(IN `friend` VARCHAR(255), IN `session_id` VARCHAR(255), OUT `updates` VARCHAR(62555), OUT `group_posts` VARCHAR(62555))
BEGIN


select * FROM updates where author in (friend,session_id)  order by time desc limit 5;



select * FROM group_posts where author_gp in (friend,session_id) order by pdate desc limit 5;
END$$
DELIMITER ;

i get the result in php myadmin as follows:

 how do i do this inside a php class function.

CALL timeline('shan2batman','aboutthecreator', @updates, @group_posts);