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用php和mysql上传视频_Php_Mysql - Fatal编程技术网

用php和mysql上传视频

用php和mysql上传视频,php,mysql,Php,Mysql,我上传的文件不移动文件夹,而是在mysql中保存路径。我的代码有什么问题如何将视频保存在文件夹中,我已经创建了测试上传的文件夹名。获取文件夹中视频的exexct方法是什么 <?php error_reporting(1); $con=mysql_connect("localhost","root",""); mysql_select_db("ngo",$con); extract($_POST); $target_dir = "test_upload/"; $target_fi

我上传的文件不移动文件夹,而是在mysql中保存路径。我的代码有什么问题如何将视频保存在文件夹中,我已经创建了测试上传的文件夹名。获取文件夹中视频的exexct方法是什么

<?php

error_reporting(1);

$con=mysql_connect("localhost","root","");

mysql_select_db("ngo",$con);

extract($_POST);

$target_dir = "test_upload/";

$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 != "wmv")
{
    echo "File Format Not Suppoted";
} 

else
{

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

mysql_query("insert into video(video_name) values('$video_path')");

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

echo "uploaded ";

}

}

//display all uploaded video

if($disp)

{

$query=mysql_query("select * from video");

 while($all_video=mysql_fetch_array($query))

 {
?>

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

 <?php } } ?>

 <form method="post" enctype="multipart/form-data">

<table border="1" style="padding:10px">

<tr>

<Td>Upload  Video</td></tr>

<Tr><td><input type="file" name="fileToUpload"/></td></tr>

<tr><td>

<input type="submit" value="Uplaod Video" name="upd"/>

<input type="submit" value="Display Video" name="disp"/>

</td></tr>

</table>

</form>

调试代码以查找错误。大概是这样的:

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

if($result) {
  mysql_query("insert into video(video_name) values('$video_path')");
  echo "uploaded ";       
} else {
  echo $_FILES["file"]["error"];
}

停止使用不推荐的
mysql.*
API。将
mysqli.*
PDO
与准备好的语句一起使用准备好尝试,但结果相同#Jens@AbhishekShulka如果($imageFileType!=“mp4”&&&&$imageFileType!=“avi”&&&$imageFileType!=“mov”&&&$imageFileType!=“3gp”&&&&$imageFileType!=“wmv”){echo”文件格式不受支持,}else,您会收到什么错误消息{$video_path=$_FILES['fileToUpload']['name'];$result=move_uploaded_file($_FILES['fileToUpload'][“tmp_name”],$target_file);if($result){mysql_查询('insert-into-video(video_name)值('$video_path')))”);echo“uploaded”;}否则{echo$_FILES[“file”[“file”[“error”];}echo“uploaded”;}只有一个问题文件没有移动文件夹我不明白实际的问题是什么你能给出这个@Saili的解决方案吗