PHP - Retrieving Image From Database (blob)? Just Getting Binary.

I'm planning on storing RIPEMD-160 passphrase hash's raw binary data in MySQL as BINARY(20).  Right now, I'm not sure if one of the following INSERT approach is more preferrable and more efficient:

Approach (1) - Using mysqli::real_escape_string()
=================================================

Code: [Select]
$binhash = $mysqli->real_escape_string(hash('ripemd160', $passphrase, true));
$mysqli->query("INSERT INTO testtbl (Passphrase) VALUES ('" . $binhash . "')");
This approach saves hash() from converting raw binary data to hexadecimal format, but need to process and escape the binary string.

Approach (2) - Using UNHEX() on MySQL
=====================================

Code: [Select]
$hexhash = hash('ripemd160', $passphrase);
$mysqli->query("INSERT INTO testtbl (Passphrase) values (UNHEX('" . $hexhash . "'))");
This approach needs hash() to convert binary data to hexadecimal but no need for escaping the string and it put extra load on MySQL to convert the hexadecimal back to binary.

Hi everyone, i am just trying to learn php for a bit of fun really and started making a sort of 'facebook' website. I am having trouble however trying to display different users images, for example when trying to find a correct 'friend' only the image of the last result is being shown for all people with the same name... here is my code below, if anyone can help me out that would be great

file 1

$count=1; 

while ($numids>=$count){
  echo "<form method=\"post\" action=\"friendadded.php\">";
  $frienduserid=$_SESSION["passedid[$ii]"];
  $friendfirstname=$_SESSION["passedfirstname[$ff]"];
  $friendlastname=$_SESSION["passedlastname[$ll]"];
  $_SESSION['friendsuserpicid'] = $frienduserid;
  echo "<table width=\"700\" height=\"50\" border=\"1\" align=\"center\">";
  echo "<tr>";
  echo "<th></th>";
  echo "<th>First Name</th>";
  echo "<th>Last Name</th>";
  echo "</tr>";
  echo "<tr>";
  echo "<td><center>";
  echo "<img border=\'0\' src=\"frienduserpic.php\" width=\"80\" height=\"80\" align=\"middle\"/>";
  echo "</center></td>";
  echo "<td><center>";
  echo $friendfirstname;
  echo "</center></td>";
  echo "<td><center>";
  echo $friendlastname;
  echo "</center></td>";
  echo "</tr>";
  echo "</table>";
  echo "<center><input type=\"submit\" value=\"Add this friend\" name=\"Add Friend\"></center><br/>";  

  $ii=$ii+1;
  $ff=$ff+1;
  $ll=$ll+1;
  $count=$count+1; 
  
  echo "</form>";
}




file 2

session_start();
$passeduserid=$_SESSION['friendsuserpicid'];
$timespost=$_SESSION['postednum'];
$host= 
$username=
$password=
$db_name= 
$tbl_name=
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$query = mysql_query("SELECT * FROM $tbl_name WHERE picid='".$passeduserid."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header("Content-type: image/jpeg");
echo $content;


Thanks in advance

Hi!

I am building a little PHP/MySQL application where pictures are uploaded and stored in MySQL in a longblob. These are special circumstances and storing images in the database in an absolute must.

Uploading is working fine. The upload script inserts the following into the field `image_data`:
base64_encode(file_get_contents($_FILES['image']['tmp_name']))

The problem is displaying the images. I cannot for the life of me make it work. I've tried the code on multiple systems with various versions of PHP and MySQL.

I have two files: one called view.php and one called show.php.

# view.php
# It gets $info['id'] from a query getting the ID of the recent-most image uploaded. 

echo '<img src="show.php?id='.$info['id'].'" alt="" />';

# show.php
# $id is determined by $_GET['id'] and passes through security checks I've omitted here.
# There is zero (not even a whitespace) output before the header()s are sent.

$query = "SELECT `image_data`,`image_mime`,`image_size` FROM `upload`.`files` WHERE `id` = '$id';";

$sql = mysql_query($query) or die(mysql_error());

$image = mysql_fetch_assoc($sql);

header('Content-Type: ' . $image['image_mime']);
header('Content-Length: ' . $image['image_size']);

echo base64_decode($image['image_data']);

The problem is that no image is displayed either in view.php or when I call show.php directly with a valid ID. I have verified that $image['image_mime'] and $image['image_size'] contain the right data.

However, if I download show.php and change extension to for example .jpg, the image is there. So the image is stored correctly in the database and $image['image_data'] is outputting the right data.

I even compared checksums for the image before and after and they're identical, so I would conclude that the error is in the outputting of the image - but I can't figure out what.

Error_report is set to E_ALL but there's nothing useful coming out.

Any ideas?

Code: [Select]
$query = mysql_query("SELECT a.*, b.*
         FROM friendlist a
INNER JOIN friendlist b
ON (a.friendemail=b.friendemail)
INNER JOIN users c
ON (b.friendemail = c.EmailAddress)
WHERE a.email = 'asdf@gmail.com'
AND c.Username LIKE '%carol%'
GROUP BY a.id
ORDER BY count(*) DESC");
Code: [Select]
while ($showfriends = mysql_fetch_array($query)) {
     echo $showfriends['Username'];
}
and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries?

Hi everyone,

I've read lots of tutorials on this, but something is not clicking in my brain with it.  I think my coding is close, but, as of right now, all I get is a red x for the image when I try to display the image.  Let me share my code as a starting point - I would truly appreciate any helpful comments or blatant errors that are pointed out or shared with me.

Here is the upload image form and code (so they clicked the item name and then go into this):

if($_REQUEST['modifyfeatured'])
{
$id=$_REQUEST['modid'];

$sqlid="SELECT * FROM product WHERE id='$id'";
$resultid=mysql_query($sqlid, $dbh);
$idrow = mysql_fetch_array($resultid);

echo "<p>";
echo "Fill out the form below to add a short text line (i.e. 20% Off!) and/or an image (like the new note).<br>";
echo "You are modifying the following item: <p><b>";
echo $idrow['product_id'];
echo " ";
echo $idrow['title'];
echo "</b><p>";

?>
<table border="0" cellpadding="2" cellspacing="0">
<tr>
<td>
<form method="post" enctype="multipart/form-data">
<input type="hidden" name="id" value="<? echo $id; ?>">
Short Text Message:
</td><td>
<input type="Text" name="message"></td></tr>
<tr><td>Small Image:</td><td>
<input type="hidden" name="MAX_FILE_SIZE" value="2000000">
<input name="userfile" type="file" id="userfile"></td></tr>
<tr><td>&nbsp;</td><td>
<input name="upload" type="submit" class="box" id="upload" value="Submit">
</td></tr></table>
</form>
<?

}
//then when the click the upload link, here is the code for that:

if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$message=$_REQUEST['message'];
$id=$_REQUEST['id'];

$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}

include 'library/config.php';
include 'library/opendb.php';

//$query = "INSERT INTO featured_prods (name, size, type, image, message ) ".
//"VALUES ('$fileName', '$fileSize', '$fileType', '$content', '$message')";

$query="UPDATE featured_prods SET name='$fileName', size='$fileSize', type='$fileType', image='$content', message='$message' WHERE id='$id'";

//$sqldone="UPDATE product SET product_id='$product_id', title='$title', description='$description', regular_price='$regular_price', sale_price='$sale_price', stat='$stat', weight='$weight', close_out='$close', additional='$additional', additionalpix='$additionalpix_name' WHERE  id='$id'";

//$resultdone=mysql_query($sqldone, $dbh);

mysql_query($query) or die('Error, query failed'); 
include 'library/closedb.php';

echo "<br>File $fileName uploaded<br>";
} //end if upload is hit


Okay, now here is the code trying to display the image:

<? $featuredquery="SELECT * FROM featured_prods";
$featuredresult=mysql_query($featuredquery, $dbh);
while($featuredrow=mysql_fetch_array($featuredresult)){

$id=$featuredrow['id'];
$prodquery="SELECT * FROM product WHERE id='$id'";
$prodresult=mysql_query($prodquery, $dbh);
$prodrow=mysql_fetch_array($prodresult);

echo $prodrow['title'];
echo "<br>";
echo $featuredrow['message'];
echo "<br>";
?>
<img src="getimage.php?id=<?echo $id;?>" alt="cover" />
<?
}


And here is the code for getimage.php:

<?
$id=$_GET["id"];
$result = mysql_query("select image from featured_prods where id='$id'"); 
$row = mysql_fetch_row($result); 
         
$data = base64_decode($row[0]); 
         
$im = imagecreatefromstring($row[0]); 
imagejpeg($im);

header('Content-type: ' . image/jpeg); // 'image/jpeg' for JPEG images
echo $data;

?>


Again, any help would be appreciated.  I'm a real rookie here, and I appreciate everyone's time and effort to assist me very much.