How Can I Store and Retrieve Images Using MySQL and PHP?-PHP Tutorial-php.cn

How Can I Store and Retrieve Images Using MySQL and PHP?

Mary-Kate Olsen

Release： 2024-12-17 08:56:24

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How Can I Store and Retrieve Images Using MySQL and PHP?

Storing and Retrieving Images Using MySQL and PHP

Inserting an image into a MySQL database involves creating a table to store the images and then executing an SQL query to insert the image data. To start, a MySQL table called "testblob" is created with columns for storing the image, its type, size, category, and name.

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 ''
);

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Inserting an image into this table is achieved using an SQL query that specifies the image type, actual image data, image size, and image name.

$imgData = file_get_contents($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);

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Retrieving an image from the database can be done using another SQL query that selects the image data from the appropriate table.

$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);

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