Question

这是我从数据库中检索视频的代码，在上传时我已将视频复制到mysql的htdocs文件夹并将路径存储到数据库中，而我尝试检索视频时发出错误mime type not matched，这个是我的数据库it has only two column，this is my htdocs folder contains of video

代码：

<?php

error_reporting(1);
$servername="localhost";
$username="root";
$password="";
$dbname="demo1";
$conn=new mysqli($servername,$username,$password,$dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

extract($_POST);

$target_dir = "uploads/";

$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);

if($upd)
{
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

if($imageFileType != "mp4" && $imageFileType != "avi" && $imageFileType != "mov" && $imageFileType != "3gp" && $imageFileType != "mpeg")
{
    echo "File Format Not Suppoted";
}

else
{

$video_path=$_FILES['fileToUpload']['name'];

    $sql = "INSERT INTO video (path) VALUES('$video_path')";
if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

move_uploaded_file($_FILES["fileToUpload"]["tmp_name"],$target_file);

echo "uploaded ";


}

}

//display all uploaded video

if($_POST['disp'])

{
  //  echo "came in";

$query=mysqli_query($conn,"select * from video");

    while($all_video=mysqli_fetch_array($query))

    {
//echo "test_upload/uploads/".$all_video['path'];

?>


<h1>test_upload/uploads/<?php echo $all_video['path']; ?></h1>
   <video width="300" height="200" controls>
     <source src="test_upload/uploads/<?php echo $all_video['path']; ?>" type="video/mp4">
</video>



    <?php } } ?>

Answer 1

您需要尝试调试代码。就像尝试添加php生成的标签并尝试在新标签中打开它一样。

<a href="test_upload/uploads/raj.mp4" >test link put in `raj.php` file</a>

我相信你忘记添加开始/所以试试这个链接

/test_upload/uploads/raj.mp4

或使用此链接：

uploads/raj.mp4

我认为在你的情况下它只是正确的路径。

PHP-从Database Mysql中搜索视频

1 个答案: