i tried atli's manual and the picture is not appearing?

[Dormilich]

the issue about the displaying picture from mysql is still unsolved..

so, did you verify that the picture is in the DB (and not corrupted)?

[Paul NIcolai Sunga]

yeah.. in atli's code there were files in the DB but when i tried to check my own, there was no files.. im going to start again.. hope i can do it.. i'll try to fix the codes.. the insert query maybe has an error..
here's now my code:

 
<?php 
error_reporting (E_ALL ^ E_NOTICE);

$pid = base64_decode($_GET['id']);

if(isset($_FILES['uploaded_file']))
{
if($_FILES['uploaded_file']['error'] == 0)
	{$Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
	if(mysqli_connect_errno()){
		die("Mysql connection failed:".mysqli_connect_errno());
		}
		$name = mysqli_real_escape_string($Link, $_FILES['uploaded_file']['name']);
		$data = mysqli_real_escape_string($Link, file_get_contents($_FILES ['uploaded_file']['tmp_name']));
		
		$query = "INSERT INTO pix(pid, title, imgdata) values('$pid', '$name', '$data')";
		
		$result = mysqli_query($Link, $query);
		if($result)
		{
$asd = "Succes! YOur file was successfully added!";
$alert = base64_encode($asd);
header("Location:registration.php?alert=$alert");
exit();
		}
		else
		{
		$asd = "Error! An error occurred while the file was being uploaded.
		Error code:".$_FILES['uploaded_file']['error'];
$alert = base64_encode($asd);
header("Location:picul.php?alert=$alert");
exit();
		}
		mysqli_close($Link);
}

it is not inserting..

[Paul NIcolai Sunga]

yeah.. in atli's code there were files in the DB but when i tried to check my own, there was no files.. im going to start again.. hope i can do it.. i'll try to fix the codes.. the insert query maybe has an error..
here's now my code:

 
<?php 
error_reporting (E_ALL ^ E_NOTICE);

$pid = base64_decode($_GET['id']);

if(isset($_FILES['uploaded_file']))
{
if($_FILES['uploaded_file']['error'] == 0)
	{$Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
	if(mysqli_connect_errno()){
		die("Mysql connection failed:".mysqli_connect_errno());
		}
		$name = mysqli_real_escape_string($Link, $_FILES['uploaded_file']['name']);
		$data = mysqli_real_escape_string($Link, file_get_contents($_FILES ['uploaded_file']['tmp_name']));
		
		$query = "INSERT INTO pix(pid, title, imgdata) values('$pid', '$name', '$data')";
		
		$result = mysqli_query($Link, $query);
		if($result)
		{
$asd = "Succes! YOur file was successfully added!";
$alert = base64_encode($asd);
header("Location:registration.php?alert=$alert");
exit();
		}
		else
		{
		$asd = "Error! An error occurred while the file was being uploaded.
		Error code:".$_FILES['uploaded_file']['error'];
$alert = base64_encode($asd);
header("Location:picul.php?alert=$alert");
exit();
		}
		mysqli_close($Link);
}

it is not inserting..

with this code the data has inserted.. but now im having a problem retrieving the photo..

[Paul NIcolai Sunga]

i used this:

<input name="image" type="image" src="asdf.php?imgid=<?Php echo $form_no; ?>" width="110" height="100"/>

asdf.php where the image come from.

error_reporting (E_ALL ^ E_NOTICE);

$id = $_GET['imgid'];

$Link = mysqli_connect("localhost","root","***", "***");
			if (!$Link)
  			{
  			trigger_error("Could not connect", E_USER_ERROR );
  			}
if(!isset($id) || empty($id)){
die("Please select your image!");
}else{

$query = mysqli_query($Link, "SELECT * FROM pix WHERE pid='$id' ");
$row = mysqli_fetch_assoc($query);
$content = $row['imgdata'];

header('Content-type: image/jpg');
echo $content;

}

and i tried this one: it still not showing:

  header('Content-Type: image/jpeg');
    $id = $_GET['imgid'];
   $id = isset($_REQUEST['imgid']) ? $_REQUEST['imgid'] : false;
    
   if($id)
    {
      $Link = mysqli_connect("localhost","root","***", "***");
      if(mysqli_connect_errno())
      {
         die("Mysql connection failed:".mysqli_connect_errno());
    }
    
     $result = mysqli_query($Link, "SELECT imgdata from pix where pid = '$id' ");
   
      if(is_resource($result))
      {
         $row = mysql_fetch_row($result);
		  header('Content-Type: image/jpeg');
        echo $row['imgdata'];
		
      }
  }

the photo is not showing..which do you think the good code should i use and enhance? so that i will not get confused...


thanks a lot...

[Paul NIcolai Sunga]

help. which code should i use?