Tutorial: Storing Images in MySQL with PHP

**eXcellion** · 06-01-2009, 03:18 AM

A question about uploading an image to my database.
(same question as in other topic, but I need to finish it within 12 hours..)

I used the tutorial recommended above, but it seems as if I'm to plain dumb to get it to work, I followed the instructions, but it's not working, I get errors about these lines:

Code:


      $fp = fopen($tmpName, 'r');

      $data = fread($fp, filesize($tmpName));

      $data = addslashes($data);

      fclose($fp);

Warning: fread(): supplied argument is not a valid stream resource in /home/vhosting/21/sintpietersgent.be/www/leerlingenraad/seminarie/insert.php on line 26

Warning: fclose(): supplied argument is not a valid stream resource in /home/vhosting/21/sintpietersgent.be/www/leerlingenraad/seminarie/insert.php on line 28

This is the entire code:

bestellingenCP.php

HTML Code:

<table>
<form enctype="multipart/form-data" action="insert.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="102400" type="hidden">
<tr><td>Afbeelding:</td><td><input name="image" accept="image/jpeg" type="file"></td></tr>
<tr><td>Naam:</td><td><input name="naam" type="text" /></td></tr>
<tr><td>Prijs:</td><td><input name="prijs" type="text" /></td></tr>
<tr><td>Omschrijving:</td><td><textarea name="omschrijving" ></textarea></td></tr>
<tr><td></td><td><input value="Submit" type="submit"></td></tr>
</form>
</table>

insert.php (the problem..)

Code:


<?php

$host="localhost";

$name = "correct";

$pass = "correct";

$dbname = "correct";



$dbi = mysql_connect($host, $name,$pass) or

die("Kan niet verbinden met de database. Error :" . mysql_error());

mysql_select_db($dbname,$dbi);



if (isset($_FILES['image'])) { 





      $tmpName  = $_FILES['image']['tmp_name'];  

       



      $fp = fopen($tmpName, 'r');

      $data = fread($fp, filesize($tmpName));

      $data = addslashes($data);

      fclose($fp);

      $naam = $_POST['naam'];

      $prijs = $_POST['prijs'];

      $omschrijving = $_POST['omschrijving'];

      



        $query = "INSERT INTO bestellingen (naam,prijs,beschrijving,image) VALUES

('$naam','$prijs','$omschrijving','$data')"; 

      $results = mysql_query($query, $dbi);

      

}

else {

   print "No image selected/uploaded";

}

mysql_close($dbi);

?>

**ssSuave** · 06-01-2009, 05:27 PM

A note referring to my previous post:

I learned that an upploaded file ( $_FILES['userfile']['tmp_name'] ) has permission 600 and that I have to use chmod() to change permission.

Do you know where and how i would apply it?

Thank you in advance for your replies! I really appreciate it

Jaan · 06-02-2009, 06:17 AM

Use this function PHP: chmod - Manual when you have uploaded this image.. just after move_uploaded_file() function..

**ssSuave** · 06-02-2009, 08:01 PM

Thanks a Lot buddy!

**plyswthsqurles** · 07-26-2009, 08:40 PM

great tutorial, just what i needed

arslan220 · 07-27-2009, 01:35 AM

thanks for this tutorial.

**hardinera** · 07-29-2009, 11:28 PM

thanks for the tutorial really nice hehe

um.. is it posible that pic_id can hold 2 or more picture at the same time?

table would be :

pic_id : int
pic_1: blob
pic_2: blob
pic_3: blob
....etc...

then if i call pic_id all pictures will appear? kinda like that hehe

or updating the pics?
just wondering hihihi.. is there a simple/short code for that? >.> like this tutorial?

thanks,
hardinera

**Filips** · 07-30-2009, 05:35 PM

Helllo Jordon,
I followed your instruction -

DB and Table
DATABASE Name : binary;

TAble Name : tbl_images (
> id tinyint(3) unsigned NOT NULL auto_increment,
> image blob NOT NULL,
> PRIMARY KEY (id)
> );

HTML Code:

<html>
<head>
</head>
<body>
<h1>Upload Your Image</h1>
<form enctype="multipart/form-data" action="insert.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">
</body>
</html>

insert.php

Code:


<html>
<head></head>
<body>
<?php
// Create MySQL login values and 
// set them to your login information.
$username = "root";
$password = "1234";
$host = "localhost";
$database = "binary";

// Make the connect to MySQL or die
// and display an error.
$link = mysql_connect($host, $username, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

// Select your database
mysql_select_db ($database);  

// 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 tbl_images ";
      $query .= "(image) VALUES ('$data')";
      $results = mysql_query($query, $link);
      
      // Print results
      print "Thank you, your file has been uploaded.";
}
else {
   print "No image selected/uploaded";
}
// Close our MySQL Link
mysql_close($link);
?>  
</body>
</html>

End of insert.php

view.php

Code:


<html>
<head> 
</head>
<body>
<?php

$username = "root";
$password = "1234";
$host = "localhost";
$database = "binary";

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

@mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$id = 1;
//$id = $_GET['id'];

if(!isset($id) || empty($id)){
die("Please select your image!");
}else{

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];

header('Content-type: image/jpg');
echo $content;
}
?> 
</body>
</html>

end of view.php

Problem 1: I can add 1 image,even if I tried to insert another image, it shows successfull,but when I open the mysql talbe, I see only one record

Problem 2: view.php open my Dreamweaver and shows garbage data.

Can you please help me?

**hardinera** · 07-30-2009, 06:34 PM

if i replace insert with update

$query = "UPDATE members SET image='$data' WHERE id=1";

but nothing happens >.>

am i missing sumthing?

**Orjan** · 07-31-2009, 12:37 AM

Originally Posted by Filips

view.php

Code:


<html>
<head> 
</head>
<body>
<?php

$username = "root";
$password = "1234";
$host = "localhost";
$database = "binary";

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

@mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$id = 1;
//$id = $_GET['id'];

if(!isset($id) || empty($id)){
die("Please select your image!");
}else{

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['image'];

header('Content-type: image/jpg');
echo $content;
}
?> 
</body>
</html>

end of view.php

The view.php shall NOT have any html tags! Remove them and it will work better to view the images.

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