有人可以告诉我这段代码我做错了什么,我将localhost连接到名为video的数据库。然后我插入表位置及其列位置值$videoLocation
。
我得到的结果是当我打开浏览器时
<?php
echo "Upload: " . $_FILES["file"]["name"] . "<br />";
echo "Type: " . $_FILES["file"]["type"] . "<br />";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";
?>
我想通过此代码尝试实现的目的是,一旦用户提交了该视频,该视频就会显示在我的网页上。
<?php
$allowedExts = array("jpg", "jpeg", "gif", "png", "mp3", "mp4", "wma");
$allowType = array("video/mp4","audio/mp3","audio/wma","image/png","image/gif","image/jpeg");
$maxSize = 20000000000000;
$extension = pathinfo($_FILES['file']['name'], PATHINFO_EXTENSION);
$pathToUpload = 'upload/';
if( in_array($_FILES["file"]["type"], $allowType) && in_array($extension, $allowedExts) && $_FILES["file"]["size"] <= $maxSize)
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br />";
echo "Type: " . $_FILES["file"]["type"] . "<br />";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";
if (file_exists($pathToUpload . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"], $pathToUpload . $_FILES["file"] ["name"]);
$videoLocation = "upload/".$_FILES['file']['name'];
// now insert $videoLocation into a database table
$db = new mysqli('127.0.0.1', 'root', '', 'video');
INSERT INTO location (location)
VALUES ($videoLocation)
//so you can fetch it on whatever page you feel like
}
}
}
else
{
echo "Invalid file";
}
?>
答案 0 :(得分:0)
试试这个
$db = new mysqli('127.0.0.1', 'root', '', 'video');
$stmt = $db->prepare("INSERT INTO location (location) VALUES (?)");
$stmt->bind_param('s', $videoLocation);
$stmt->execute();
//so you can fetch it on whatever page you feel like
答案 1 :(得分:0)
我最近为想要将图像文件存储在数据库中的其他人回答了一个问题。以下是如何做到这一点:
1)数据库列必须定义为longblob。
接下来,代码使用PDO而不是mysqli。
// database connection...
$dsn = 'mysql:host=localhost;dbname=testmysql';
$username = 'test';
$password = 'test';
$options = array(
PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8',
);
$connection = new PDO($dsn, $username, $password, $options);
$connection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
然后你需要一个打开的文件句柄,如下所示:
$fileVideo = fopen($videoLocation,'rb');
然后准备语句并绑定文件:
$stmt = $connection->prepare("INSERT INTO location (location) VALUES (?)");
$stmt->bindParam(1, $fileVideo, PDO::PARAM_LOB);
$connection->beginTransaction();
$stmt->execute();
$connection->commit();
fclose($fileVideo);
我还没有对您的文件进行上述测试,但原始代码与上面的内容相似,但名称更改工作正常。