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

[Paul NIcolai Sunga]

here's the code for retrieving the info in the pic..but the info in filedata only showing a set of strings..

 
<?php 
$Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
    if(mysqli_connect_errno()){
        die("Mysql connection failed:".mysqli_connect_errno());
        }
//$result1 = mysqli_query($Link, "SELECT filedata FROM filestorage ORDER BY form_no DESC LIMIT 1");
$result1 = mysqli_query($Link, "SELECT fileid, filename, filemime, filesize, created, filedata from filestorage");
if($result1)
{
if(mysqli_num_rows($result1) == 0){
echo "<p> there are no files in the database</p>";
}
else
{
echo "<table width = '100%'><tr>";
echo "<td><b>Name</b></td>";
echo "<td><b>Mime</b></td>";
echo "<td><b>Size(bytes)</b></td>";
echo "<td><b>Created</b></td>";
echo "<td><b>Filedata</b></td>";
 
while($row = mysqli_fetch_array($result1))
{
echo "<tr><td>".$row['filename']."</b></td>";
echo "<td>".$row['filemime']."</b></td>";
echo "<td>".$row['filesize']."</b></td>";
echo "<td>".$row['created']."</b></td>";
echo "<td><input type=\"image\" src=".$row['filedata']."/></b></td></tr>";
}
 
echo "</table>";
}
mysqli_free_result($result1);
}
else
{
echo "error! sql query failed";
echo "<pre>".$link->error."</pre>";
}
mysqli_close($Link);
 
 
?>
 

[Dormilich]

as for any image in html, the src attribute specifies an URI to the image, not the image data itself.

[Paul NIcolai Sunga]

how could i convert the strings into an image? what function should i use? to convert it?



thanks

[Dormilich]

you're not getting the point here. the image data belong into a separate file.

// html
<input type="image" src="path/to/image.php"/>

// image.php
// ...
echo $row['filedata'];

if you closely examine Atli's code samples, you'll notice that he doesn't mix (binary) image data with html code.

[Paul NIcolai Sunga]

i see.. it should be in a different page? ..so that it will be a valid code in src?

[Dormilich]

@Paul NIcolai Sunga
the URI of that page

[Canabeez]

Here, save this as image.php

header("Content-Type: image/jpeg");
 
$fileid = isset($_REQUEST['fileid']) ? $_REQUEST['fileid'] : false;
 
if($fileid)
{
   $Link = mysqli_connect("localhost","root","tupi", "amyak_maleh");
   if(mysqli_connect_errno())
   {
      die("Mysql connection failed:".mysqli_connect_errno());
   }
 
   $result = mysqli_query($Link, "SELECT filedata from filestorage where fileid={$fileid}");
 
   if(is_resource($result))
   {
      $row = mysql_fetch_row($result);
      echo $row['filedata'];
   }
}

Then call:

<input type="image" src="image.php?fileid=YOUR_FILE_ID">

[Paul NIcolai Sunga]

the result is

Submit Query

why is that?


thanks

[Paul NIcolai Sunga]

here's the code, and it appears only a this string: Submit Query

<input type="image" src="image.php?fileid=<?Php $form_no; ?>" />
 

thanks

[Paul NIcolai Sunga]

any other suggestions please?


thanks

[Dormilich]

where does "Submit Query" come from?

[Paul NIcolai Sunga]

from the image.php.



thanks

[Paul NIcolai Sunga]

whats this?when i click the submit query which is the one appeared insted of the image, it jumps to the image.php page then this is what appears:

http://localhost/image.php?x=&y=

[Dormilich]

@Dormilich
I mean, where dou you define such a string (which line of code causes that to appear)?

[Paul NIcolai Sunga]

this is the code..

<form action="image.php" method="post">
 
<input type="image" src="image.php?form_no=<?Php echo $form_no; ?>" /> 
      </form>

[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]

@Paul NIcolai Sunga
.....................


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]

@Canabeez
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]

@Paul NIcolai Sunga
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">
        <img src="img.php?iid=sas091" 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']))
        {
            header('Content-Type: ' . $this->DBimg['img_mime']);
            header('Content-Length: ' . $this->DBimg['img_size']);
            echo $this->DBimg['img_data'];
        }
    }
 
    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]

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]

@Paul NIcolai Sunga
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..

[Dormilich]

@Paul NIcolai Sunga
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]

@Paul NIcolai Sunga
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?