Question

我使用来自tizag的资源创建了一个上传脚本，该脚本返回了一个屏幕错误，但是apache日志中没有错误。

HTML表单

 <form action="upload.php" method="post" enctype="multipart/form-data">
 <input type="hidden" name="MAX_FILE_SIZE" value="90000000" />
  Select video to upload:
 Please choose a file: <input name="uploadedfile" type="file" /><br /> 
 <input type="submit" value="Upload File" /> 
   </form>

PHP代码

<?php
$target_path = "/var/www/html/upload/";
$target = $target_path . basename($_FILES['uploadedfile']['name'][0] );
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'] [0],        $target_path))
 { 
 echo "The file ". basename( $_FILES['uploadefile']['name'] [0]). " has been      uploaded"; 
  } 
  else {
 echo "Sorry, there was a problem uploading your file."; 
  }
  ?>

这不应该是非常难以实现的，但是在apache上没有错误，我很难进行故障排除。我的php知识有限，所以请记住这一点。

亲切的问候，

Mark Couto

Answer 1

这一行：

$target = $target_path . basename($_FILES['uploadedfile']['name'][0] );

就目前而言，$target只是一个杂散变量而没有在其他地方使用。

它应该是：

$target_path = $target_path . basename($_FILES['uploadedfile']['name'][0] );

＆＃34;我使用tizag＆＃34;
中的资源创建了一个上传脚本

您遵循的Tizag教程不会更改其变量。

他们的例子：

$target_path = "uploads/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
}

来源：http://www.tizag.com/phpT/fileupload.php

另外，['uploadefile']中的拼写错误应为['uploadedfile']

Answer 2

首先确保将php配置为允许文件上载。在你的＆＃34; php.ini＆＃34;文件搜索指令，并将其设置为ON，

file_uploads= ON

来源：http://www.w3schools.com/php/php_file_upload.asp

PHP上传脚本无法运行，并且apache日志没有错误

2 个答案: