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

[Dormilich]

well, it does not explain where the string comes from, but it explains the strange submit.

obviously image.php is designed to be an image (representation ), thus if you put it in your form's action it will be executed onsubmit (say, if you submit the form, you'll be returned an image (which is possible, but doesn't make sense)). solution: put the form handling script in the action attribute.

[Paul NIcolai Sunga]

solution: put the form handling script in the action attribute.

what exactly i will put in the action attribute?

[Dormilich]

what exactly i will put in the action attribute?

............... ......


the script (resp. its file name) that processes the form data (may be the same as the current page)

[Paul NIcolai Sunga]

what shall i do? it is still not working..

[Dormilich]

what is not working, the form or the image?

[Paul NIcolai Sunga]

the image is not showing.. i tried to show it through the src itself and its okei but when the image that came from the database it is not showing,.. maybe the problem is in the database or in the code, but there is no sign error in the codes.

[Dormilich]

ok, then you have to make sure that the picture is saved correctly in the DB . there are some MySQL GUIs out, but unless you have a Mac I can’t recommend anything.

(If that’s possible I could try connecting to your DB and look myself)

[Canabeez]

Here's one I use... SQLYog

[Dormilich]

Here's one I use... SQLYog

does it show you any pictures?

[Paul NIcolai Sunga]

i tried another code:
here's the code. it only displays a blank photo. i mean an x that is usually seen in a blank photo..

<?php

$Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
			if (!$Link)
  			{
  			trigger_error("Could not connect", E_USER_ERROR );
  			}

$result=mysqli_query($Link, "SELECT * FROM pix");
while($row=mysqli_fetch_assoc($result))
{
 header("Content-type: image/jpeg");
  echo "<img src=\"".$row['imgdata']."\" alt=\"".$row['title']."\" width=\"100\" length=\"100\"/>";
}
?>

[Dormilich]

header("Content-type: image/jpeg");
  echo "<img [U]src=\"".$row['imgdata']."\"[/U] alt=\"".$row['title']."\" width=\"100\" length=\"100\"/>";

this is completely wrong, the src attribute has to be an url!

[Dormilich]

I’ll give you a working example

HTML

    <h4>
        <a class="hook" id="mord">Mord an Bord</a>
    </h4>

	<p>11. Januar 2009, SachsenSonntag</p>
	<div class="mitte">
		[U]<img src="img.php?iid=sas091"[/U] width="150" height="99" alt="Bild des Zeitungsartikels" title="Wohl bekomm’s! Ein Schlückchen zur Begrüßung der Passagiere" id="img_sas091"/>
	</div>

img.php

<?php
	require 'load.img.php';
	$img = new ArticleIMG;
	$img->display();
?>

methods

	public function display()
	{
		if (!$this->requestImage())
		{
			exit;
		}
		
		if (!empty($this->DBimg['img_mime']) and !empty($this->DBimg['img_size']) and !empty($this->DBimg['img_data']))
		{
			[B]header('Content-Type: ' . $this->DBimg['img_mime']);
			header('Content-Length: ' . $this->DBimg['img_size']);
			echo $this->DBimg['img_data'];[/B]
		}
	}

	protected function requestImage()
	{
		$this->load();
		
		if ($this->checkID($this->IID))
		{
			return $this->getImageData();
		}
		return false;
	}

	private function getImageData()
	{
		try 
		{			 
			$PS = aDB::execute('img', array(':id' => $this->IID));
			$this->DBimg = $PS->fetch(PDO::FETCH_ASSOC);
		}
		catch (PDOException $pdo)
		{
			ErrorLog::logException($pdo, __CLASS__, __METHOD__);
			return false;
		}
		return true;
	}

[Paul NIcolai Sunga]

yes

is this correct?

 
$pic = $_FILES['Browse'];

$formid = mysqli_query($Link, "Select form_no from registration_form where last_insert_id() = form_no");

$formnum = mysqli_fetch_row($formid);

	
$querypic = mysqli_query($Link, "INSERT INTO pix(pid, imgdata) values('$formnum', '$pic')");

i am not certain with this code..

[Dormilich]

is this correct?

$formid = mysqli_query($Link, "Select form_no from registration_form where last_insert_id() = form_no");

completely unnecessary (have a close look at what you want to do here… ;))

otherwise I can’t tell (too less info)

[Paul NIcolai Sunga]

okei.. im trying a lot of codes that is why i am so confused.. the issue about the displaying picture from mysql is still unsolved..