displaying images from a database

blackwidow

hi guys,

I've managed to get a script working that will allow me to upload an image to a database, and it works fine...however, I have another script that should show the image, but it doesnt. I know I changed a couple of things in the upload script, but I'm not sure how to make the 'show_image' script match.

here is the upload script:

<?php
// code that will be executed if the form has been submitted:


if ((isset($_POST['submit']))&&(isset($_FILES['form_data']))) {
  upload();
} else {
  show_form();
}//end if

function upload() {

mysql_connect("xxxxx", "xxxxx", "xxxxx");
mysql_select_db("xxxxx");

//define variables
$image_file  = '';
$image_desc  = '';
$type        = '';
$size        = 0;
$name        = '';

$image_file = $_FILES['form_data']['tmp_name'];
$image_desc = $_POST['form_description'];
$type       = $_FILES['form_data']['type'];
$size       = $_FILES['form_data']['size'];
$name       = $_FILES['form_data']['name'];

$data = fread(fopen($image_file, "r"), filesize($image_file));
$data = addslashes($data);
$sql="INSERT INTO binary_data VALUES (NULL,'". $image_desc ."','". $data ."','". $name ."','". $size ."','". $type ."')"; 

@mysql_query($sql) or die(mysql_error());

//	"INSERT INTO tblimage VALUES(NULL,'". $imgtype ."','". $imgdata ."')";

$id= mysql_insert_id();
print "<p>This file has the following Database ID: <b>$id</b>";

//show the form again to load the next image
show_form();
}

function show_form()
{
// else show the form to submit new data:
?>

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype="multipart/form-data">
File Description:<br>
<input type="text" name="form_description" size="40">
<input type="hidden" name="MAX_FILE_SIZE" value="1000000">
<br>File to upload/store in database:<br>
<input type="file" name="form_data" size="40">
<p>
<input type="submit" name="submit" value="submit">
<input type="button" name="move" value="Show" onClick="javascript:window.location='show_image.php';">
</p>
</form>

<?php
}
?>

and here is the 'show_image' script:

<?php

require("conn.php");
//check to see if the id is passed
if(isset($_GET['id'])) {
    $id=$_GET['id'];

$query = "select bin_data, filetype from binary_data where id=$id";
//echo $query;
$result = connect($query);
$row = mysql_fetch_array($result);
  {
   $data = $row['bin_data'];
   $type = $row['filetype'];
  }
  if ($type=="pjpeg") $type = "jpeg";
  Header( "Content-type: $type");
  echo $data;
}
?>

Any help would be greatly appreciated as this is the main crux of things at the moment!

many thanks

Mark

reddrum

You may be better off just keeping the url of the image in the db and the images in a dir.

while ($row = mysql_fetch_array ($result)) {
   $photo=  $row['photo'];
   print"<img src='$photo' />";

 }

blackwidow

Hi,

I can see your point, but I deliberately wanted the images to be stored in database. I've spent the past three weeks with this little mini project at the back of my mind.

I have the images displaying on the screen now perfectly fine, but only when I append the 'id' to the url in the browser (/show_image.php?id=20) and it should be being passed straight onto it by the script so when the user clicks the 'show' button it does it automatically.........but it doesnt!

heres the modified upload code:

<?php

mysql_connect("xxxxx", "xxxxx", "xxxxx");
mysql_select_db("xxxxx");

// code that will be executed if the form has been submitted:


if ((isset($_POST['submit']))&&(isset($_FILES['form_data']))) {
  upload();
} else {
  show_form();
}//end if

function upload() {


//define variables
$image_file  = '';
$image_desc  = '';
$type        = '';
$size        = 0;
$name        = '';

$image_file = $_FILES['form_data']['tmp_name'];
$image_desc = $_POST['form_description'];
$type       = $_FILES['form_data']['type'];
$size       = $_FILES['form_data']['size'];
$name       = $_FILES['form_data']['name'];

$data = fread(fopen($image_file, "r"), filesize($image_file));
$data = addslashes($data);
$result="INSERT INTO binary_data VALUES (NULL,'". $image_desc ."','". $data ."','". $name ."','". $size ."','". $type ."')"; 

@mysql_query($result) or die(mysql_error());

$id= mysql_insert_id();
print "<p>This file has the following Database ID: <b>$id</b>";


//show the form again to load the next image
show_form();
}

function show_form()
{
// else show the form to submit new data:
?>

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype="multipart/form-data">
File Description:<br>
<input type="text" name="form_description" size="40">
<input type="hidden" name="MAX_FILE_SIZE" value="1000000">
<br>File to upload/store in database:<br>
<input type="file" name="form_data" size="40">
<p>
<input type="submit" name="submit" value="submit">
<input type="button" name="move" value="Show" onClick="javascript:window.location='show_image.php';">
<a href="show_image.php">view image</a> </p>
</form>

<?php
}
?>

and heres the modified show_image code:

<?php

mysql_connect("xxxxx", "xxxxx", "xxxxx");
mysql_select_db("xxxxx");

//check to see if the id is passed
if(isset($_GET['id'])) {
    $id=$_GET['id'];

$query = mysql_query("SELECT bin_data, filetype FROM binary_data WHERE id='". $id ."'");
//echo $query;
$result = ($query);
$row = mysql_fetch_array($result);
  {
   $data = $row['bin_data'];
   $type = $row['filetype'];
  }
  if ($type=="pjpeg") $type = "jpeg";
  Header( "Content-type: $type");
  echo $data;
}
?>

Sorry to be a pain, and appologise for not taking your advice about the location of the images.

Any ideas would be awsome!

many thanks

Mark

blackwidow

I've now resolved this........just in case anyone else has the same trouble, I removed the 'show' button in the form, and added:

echo "<br><br><a href=show_image.php?id=$id>view image</a>";

just beneath:

$id= mysql_insert_id();
    print "<p>This file has the following Database ID: <b>". $id ."</b>";

The link only shows up after the file has been uploaded, and takes you straight to it.

cheers

Mark