如何使用PHP从MySQL数据库存储和检索图像？

首先，您创建一个MySQL表来存储图像，例如：

create table testblob ( image_id tinyint(3) not null default '0', image_type varchar(25) not null default '', image blob not null, image_size varchar(25) not null default '', image_ctgy varchar(25) not null default '', image_name varchar(50) not null default '' ); 然后，您可以将图像写入数据库，如下所示：

/*** * All of the below MySQL_ commands can be easily * translated to MySQLi_ with the additions as commented / $imgData = file_get_contents($filename); $size = getimagesize($filename); mysql_connect("localhost", "$username", "$password"); mysql_select_db ("$dbname"); // mysqli // $link = mysqli_connect("localhost", $username, $password,$dbname); $sql = sprintf("INSERT INTO testblob (image_type, image, image_size, image_name) VALUES ('%s', '%s', '%d', '%s')", / * For all mysqli_ functions below, the syntax is: * mysqli_whartever($link, $functionContents); ***/ mysql_real_escape_string($size['mime']), mysql_real_escape_string($imgData), $size[3], mysql_real_escape_string($_FILES['userfile']['name']) ); mysql_query($sql); 您可以使用以下方法在网页中显示数据库中的图像：

$link = mysql_connect("localhost", "username", "password"); mysql_select_db("testblob"); $sql = "SELECT image FROM testblob WHERE image_id=0"; $result = mysql_query("$sql"); header("Content-type: image/jpeg"); echo mysql_result($result, 0); mysql_close($link);来源：stack overflow