有人可以告诉我我做了什么错误的代码,我连接到本地到数据库称为视频。然后我插入表格位置并将其列位置的值$videoLocation
。将视频存储在数据库中,然后检索它
结果我得到的是,当我打开我的浏览器是
<?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 />";
?>
我想实现这个代码是针对视频在我的网页一旦用户提交它出现。
<?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";
}
?>
这是什么? INSERT INTO位置(位置) VALUES($ videoLocation) –
location是表的名称,括号内的位置是表的列名之一。 – user3448253
_“我得到的结果是当我打开我的浏览器是”_ - 是_what_?您可以看到您在此声明之后发布的PHP代码,未解析 - 或者是什么? – CBroe