我试图在用户请求时在浏览器中显示存储在mysql数据库中的PDF文件。
MySQL表的结构:
CREATE TABLE `file` (
`id` Int Unsigned Not Null Auto_Increment,
`name` VarChar(255) Not Null Default 'Untitled.txt',
`mime` VarChar(50) Not Null Default 'text/plain',
`size` BigInt Unsigned Not Null Default 0,
`data` MediumBlob Not Null,
`created` DateTime Not Null,
PRIMARY KEY (`id`)
)
php代码:
$query = "
SELECT `type`, `name`, `size`, `data`, `mime`
FROM `file`
WHERE `id` = {$id}";
$result = $dbLink->query($query);
if($result) {
// Make sure the result is valid
if($result->num_rows == 1) {
// Get the row
$row = mysqli_fetch_assoc($result);
// Print headers
header('Accept-Ranges: bytes');
header('Content-Transfer-Encoding: binary');
header("Content-Type: ".$row['mime']);
header("Content-Length: ".$row['size']);
header("Content-Disposition: inline; filename=".$row['name']);
echo $row['data'];
}
}
将文件保存到数据库的代码:
if(isset($_FILES['uploaded_file'])) {
// Make sure the file was sent without errors
if($_FILES['uploaded_file']['error'] == 0) {
// Connect to the database
$dbLink = new mysqli('localhost', 'REDACTED',"REDACTED", 'pdfs');
if(mysqli_connect_errno()) {
die("MySQL connection failed: ".mysqli_connect_error());
}
// get all required data
$name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']);
$type = $dbLink->real_escape_string($_FILES['uploaded_file']['type']);
$data = $dbLink->real_escape_string(file_get_contents($_FILES ['uploaded_file']['name']));
$size = intval($_FILES['uploaded_file']['size']);
// Create the SQL query
$query = "
INSERT INTO `file` (
`name`, `type`, `size`, `data`, `created`
)
VALUES (
'{$name}', '{$type}', {$size}, '{$data}', NOW()
)";
// Execute the query
$result = $dbLink->query($query);
// Check if it was successfull
if($result) {
echo 'Success! Your file was successfully added!';
}
else {
echo 'Error! Failed to insert the file'
. "<pre>{$dbLink->error}</pre>";
}
但是,当我尝试查看pdf时,adobe reader开始加载,然后我收到此错误消息:
文件已损坏且无法修复。 我做错了什么?
答案 0 :(得分:1)
无论你想做什么,都有更好的方法。
header
标签就足够了。答案 1 :(得分:0)
调用die();在echo $ row ['data']之后;确保数据后没有添加其他输出。不确定这是否是问题,但它将消除它作为一种可能性。