【PHP】mysqlに画像を保存する

1.テーブル構造
CREATE TABLE `upload` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`type` varchar(20) NOT NULL,
`data` mediumblob NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;

2.index.html

<!doctype html>
<html>
<head>
<title>
Post-Image
</title>
</head>
<body>
<form action="post.php" method="post" enctype="multipart/form-data">
<input type="file" name="file" id="file"/>
<input type="submit" value="OK"/>
</form>
</body>
</html>

3.post.php

<?php
if ($_FILES["file"]["error"] > 0)
{
    echo "Error: " . $_FILES["file"]["error"] . "<br />";
}
else
{
    $type = $_FILES["file"]["type"];
    $size = $_FILES['file']['size'];
    $tmp=$_FILES["file"]["tmp_name"];
    $fp = fopen($tmp,'rb');
    $data = bin2hex(fread($fp,$size));
    $dsn='mysql:host=localhost;dbname=test';
    echo '<pre>';
    try{
        $pdo = new PDO($dsn,'root','root');
        $pdo->exec("INSERT INTO `upload`(`type`,`data`) values ('$type',0x$data)");
        $id = $pdo->lastInsertId();
        echo 'upload success!<a href="view.php?id='.$id.'">View</a>';
        $pdo = null;
    }catch (PDOException $e){
        echo $e->getMessage();
    }
    echo '</pre>';
    fclose($fp);
}

4.view.php
query('select * from `upload` where `id`=’.$id);
$row = $rs->fetchAll();
$data = $row[0];
header(“Content-Type:${data['type’]}");
echo $data['data’];
$pdo = null;
}catch (PDOException $e){
echo $e->getMessage();
}
}else{
exit();
}

Source

Posted by arkgame