Tutorial: Storing Images in MySQL with PHP / Part II / Display your images

[tonymorrison39]

well written thanks

[helena]

HI, I tried this script in my webpage the upload works fine but i couldnt display the image it says "please select the image" when i check my mysql it shows 1 row affected..
could u please help me resolve this problem

[Jaan]

Did you edit that script?
If yes please post it here.. then I can help you

[helena]

Did you edit that script?
If yes please post it here.. then I can help you

HI , thanks for your reply..
if you could sorld out this issue today , it ll be very helpful to me..
insert.php

<?php
// Create MySQL login values and 
// set them to your login information.
$username = "user1";
$password = "password1";
$host = "mysql2.ten4onehost.com";
$database = "michaels";

// 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);
?>

show.php

<?php

$username = "user1";
$password = "password1";
$host = "mysql2.ten4onehost.com";
$database = "michaels";
@mysql_connect($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 = $_GET['id'];
$query = mysql_query("SELECT * FROM tbl_images WHERE id=1);
$row = mysql_fetch_array($query);
$content = $row['image'];

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

thank you

[Jaan]

Change this:

$query = mysql_query("SELECT * FROM tbl_images WHERE id=1); 

to this:

$query = mysql_query("SELECT * FROM tbl_images WHERE id='1'"); 

[k1net1cs]

I think this line in show.php needs to be changed :

$row = mysql_fetch_array($query); 

into

$row = mysql_fetch_assoc($query); 

[Orjan]

I think this line in show.php needs to be changed :

$row = mysql_fetch_array($query); 

into

$row = mysql_fetch_assoc($query); 

Not really. mysql_fetch_array() gives back the combined answer that mysql_fetch_assoc() and mysql_fetch_row() would give, so either one is fine.

[phb50530]

Works perfect, thank you!

[k1net1cs]

Not really. mysql_fetch_array() gives back the combined answer that mysql_fetch_assoc() and mysql_fetch_row() would give, so either one is fine.

Ah, right, my bad. =/
Playing MW2 after quite a few hours straight with code analyzing don't really mix well...

Anyway, I can't see where in the codes that could make it says "please select the image" anywhere...?

[Orjan]

I'd say that your line shall be:

$query = mysql_query("SELECT * FROM tbl_images WHERE id='$id'"); 

and the line above, is even better if it is:

$id = mysql_real_escape_string($_GET['id']);