claro aqui lo tienes
Esto tengo yo y me funciona de maravillas, solo necesito mostrar todas las imagenes de un usuario x o categoria
formulario
Código PHP:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<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" />
</form>
</body>
</html>
Insertar
Código PHP:
// Create MySQL login values and
// set them to your login information.
$username = "root";
$password = "";
$host = "localhost";
$database = "candente";
// 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 testblob ";
$query .= "(image) VALUES ('$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);
ver la foto
Código PHP:
//header('Content-type: image/jpg');
$username = "root";
$password = "";
$host = "localhost";
$database = "candente";
@mysql_connect($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 = $_GET['image_ctgy'];
if(!isset($id) || empty($id)){
die("Please select your image!");
}else{
$query = mysql_query("SELECT * FROM testblob WHERE image_ctgy='".$id."'");
$row = mysql_fetch_array($query)
//while($row = mysql_fetch_array($query)){
$content = $row['image'];
//}
header('Content-type: image/jpg');
echo $content;
}
Espero te ayude por que funciona perfecto, y espero que alguien me ayude a como mostrar mas de una foto...
Saludos