无法显示我存储在数据库中的图像

Question

i want to display all the images stored in the database.. then my problem is that only small square will appear.. i need help..

my maincontent.php

<?php
   error_reporting(E_ALL);
   $link = mysql_connect("localhost", "root", "") or die("Could not connect: " . mysql_error());
   mysql_select_db("login") or die(mysql_error());
   $sql = "SELECT id FROM products";
   $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
    while($row=mysql_fetch_assoc($result)){
    echo '<img src="t2.php?id='.$row['id'].'"/>';
    }
    mysql_close($link);
        ?>

my t2.php

<?php
error_reporting(E_ALL);
if(isset($_GET['id']) && is_numeric($_GET['id'])) {

$link = mysql_connect("localhost", "root", "") or die("Could not&amp;nbsp;connect: " .  mysql_error());

mysql_select_db("login") or die(mysql_error());

$sql = "SELECT photo FROM products WHERE id={$_GET['id']}";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

header("Content-type: image/jpeg");
echo mysql_result($result, 0);
mysql_close($link);
}else{
echo 'Please use&;nbsp;a real id number';

}
?>

i hope someone can help me regarding this problem.

Answer 1

I do not know how you saved your image in your database(direct images or image path). but it seems that you have to be changed those:

Change this line :

 $result = mysql_query($sql) or die("Invalid query: " . mysql_error());

to:

 $result = mysql_query($sql) or die("Invalid query: " . mysql_error());

and your t2.php

Change :

$result = mysql_query($sql) or die("Invalid query: " . mysql_error());

to :

$result = mysql_query($sql) or die("Invalid query: " . mysql_error());

you need to use mysqli_* or PDO . mysql_* all functions are deprecated.

Answer 2

Change to this. your t2.php

 <?php
    $imageId = intval($_GET["id"]);

    $query = mysql_query("SELECT img FROM images WHERE id = ". $imageId);
    $row = mysql_fetch_array($query);

    $mime = null;
    // place $type init. here
    if ($type=="pjpeg") // <<< where do you get $type btw?
        $mime = "image/jpeg";

    $b64Src = "data:".$mime.";base64," . base64_encode($row["img"]);
    echo '<img src="'.$b64Src.'" alt="" />';
    ?>