PHPでデータベースに画像を保存するプログラム

1.テーブルを作成
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>
更新後イメージ
</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 “エラー: " . $_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=startnews24_pictest’;
echo '<pre>’;
try{
$pdo = new PDO($dsn,’root’,’root’);
$pdo->exec(“INSERT INTO `upload`(`type`,`data`) values ('$type’,0x$data)");
$id = $pdo->lastInsertId();
echo '成功にアップロード!<a href="view.php?id=’.$id.'">View</a>’;
$pdo = null;
}catch (PDOException $e){
echo $e->getMessage();
}
echo '</pre>’;
fclose($fp);
}

4.view.php

<?php
$id = $_GET['id’];
if(is_numeric($id)){
$dsn=’mysql:host=localhost;dbname=startnews24_pictest’;
try{
$pdo = new PDO($dsn,’root’,’root’);
$rs = $pdo->query('select * from `upload` where `id`=’.$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();
}

PHP

Posted by arkgame