Question

如何上传视频？我用图像文件做了这个，但没有使用只显示＆＃34;错误＆＃34;

的视频

    <?php 
 session_start();

 if( !isset($_SESSION["username"]) ){




         echo "<p id='errors'>Por favor, loguese, si no es redireccionado haga click <a href='log.php'>aqui</a></p>";
 header("location: log.php");


 }else{
 };
$img = $_GET["fileToUploadv"];

$target_dir = "video/";
$target_file = $target_dir . basename($_FILES["fileToUploadv"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["fileToUploadv"]["tmp_name"]);
    if($check !== false) {
        $uploadOk = 1;
    } else {
        echo "The file is not an image!.";
        $uploadOk = 0;
    }
}
// Check if file already exists

// Check file size

// Allow certain file formats
if($imageFileType != "mp4" && $imageFileType != "mp4" ) {
    echo "Only use a .JPG file.";
    $uploadOk = 0;
}
if (file_exists($target_file)) {

    unlink("video/video.mp4");
    unlink("video/video.avi");

}

// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
    echo "Error.";
// if everything is ok, try to upload file
} else {
    if (move_uploaded_file($_FILES["fileToUploadv"]["tmp_name"], $target_file)) {





        echo "The file ". basename( $_FILES["fileToUploadv"]["name"]). " has been uploaded.";
        rename("video/".basename( $_FILES["fileToUploadv"]["name"]), "video/video.mp4");
        rename("video/".basename( $_FILES["fileToUploadv"]["name"]), "video/video.avi");

    } else {
        echo "Error.";
    }
}
?>

这个脚本有filetouploadv，它有一个表单输入来上传mp4或avi是html

<form id="formimg" method="post" action="uploadvideo.php" enctype="multipart/form-data"> 

    <label for="uploadimgg">
        <p id="selectimg">Please select a MP4 or AVI file to upload.</p>
        <i class="fas fa-upload"></i>
    </label>
    <input id="uploadimgg" type="file" name="fileToUploadv" accept="video/mp4,video/avi">

  <input class="hideshowsub"  id="submitpho" type="submit">
</form>

应该将名称更改为视频并上传到视频/但是不起作用... move_uploaded_file因为该reutrn false并执行＆＃34;错误＆＃34;消息

Answer 1

<?php 
   require("config.php");

   $sql = "SELECT * FROM inventario";
   $result = mysqli_query($link, $sql);
   echo("<table>");
   echo("<tr>");
   echo ("<th class='midtable'>"."Username"."</th>");
   echo ("<th class='midtable'>"."Password"."</th>");
   echo ("<th class='midtable'>"."</th>");
   echo ( "</tr>");

   while ($row = mysqli_fetch_array($result))  
        {
          echo("<tr>");
          echo("<td><div class='data'>".$row['username']."</div><div class='edit'><input type='text' name='usrname' value='".$row['username']."'></div></td>"."<td><div class='data'".$row['password']."<div class='edit'><input type='text' name='password' value='".$row['password']."'></div></td>"."<td>"."<i class='fas fa-edit'></i>"."<i class='fas fa-trash-alt'></i>"."</td>"  );
          echo("</tr>") ;
         }; ?>

然后使用jQuery隐藏类数据并显示行中的类编辑元素和提交按钮以更新数据库中的值...并且可能只更新该行...

用php和html上传视频

1 个答案: