php複製功能，未能打開流：權限被拒絕

Question

我想上傳服務器上的圖像，下面是我在互聯網上找到的腳本，並在本地工作，當我部署代碼和數據庫是給我「未能打開流：權限被拒絕「錯誤。

<?php 
//define a maxim size for the uploaded images in Kb 
define ("MAX_SIZE","5000"); 

//This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. 
function getExtension($str) { 
     $i = strrpos($str,"."); 
     if (!$i) { return ""; } 
     $l = strlen($str) - $i; 
     $ext = substr($str,$i+1,$l); 
     return $ext; 
} 

//This variable is used as a flag. The value is initialized with 0 (meaning no error found) 
//and it will be changed to 1 if an errro occures. 
//If the error occures the file will not be uploaded. 
$errors=0; 

    //reads the name of the file the user submitted for uploading 
    $image=$_FILES['image']['name']; 

    //if it is not empty 
    if ($image) 
    { 
    //get the original name of the file from the clients machine 
     $filename = stripslashes($_FILES['image']['name']); 
    //get the extension of the file in a lower case format 
     $extension = getExtension($filename); 
     $extension = strtolower($extension); 
    //if it is not a known extension, we will suppose it is an error and will not upload the file, 
    //otherwise we will do more tests 
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png")) 
     { 
     //print error message 
      echo '<h1>Nepoznata vrsta fajla!</h1>'; 
      $errors=1; 
     } 
     else 
     { 
//get the size of the image in bytes 
//$_FILES['image']['tmp_name'] is the temporary filename of the file 
//in which the uploaded file was stored on the server 
$size=filesize($_FILES['image']['tmp_name']); 

//compare the size with the maxim size we defined and print error if bigger 
if ($size > MAX_SIZE*1024) 
{ 
    echo '<h1>To large file!</h1>'; 
    $errors=1; 
} 

//we will give an unique name, for example the time in unix time format 
$image_name=time().'.'.$extension; 
//the new name will be containing the full path where will be stored (images folder) 
$newname="Content/Images/".$image_name; 
//we verify if the image has been uploaded, and print error instead 
//$copied = copy($_FILES['image']['tmp_name'], $newname); 

$copied = copy('$_FILES['image']['tmp_name'], $newname); 

//echo $_FILES['image']['tmp_name'].'<br/>'; 
//echo $_FILES['image']['name']; 

if (!$copied) 
{ 
    echo '<h1>Error occurred!</h1>'; 
    $errors=1; 
}}} 


//If no errors registred, print the success message 
/*if(isset($_POST['Submit']) && !$errors) 
    { 
    echo "<h1>You have successfully uploaded image.</h1>"; 
}*/ 

?> 

我看到一些計算器應答者喜歡answers1和answer2，但我不知道該怎麼做？還有其他建議嗎？

謝謝。

Answer 1

確保保存文件（tmp）的文件夾權限設置爲777。 類型chmod -R 777 path終端

Answer 2

你copy命令似乎有語法錯誤：

$copied = copy('$_FILES['image']['tmp_name'], $newname); 
       ^--- extra quote? 

如果你想這樣做

$copied = copy("$_FILES['image']['tmp_name']", $newname); 

它不會反正工作。 PHP解析器不gready，並認爲這是

$_FILES['image'] -> array 
['tmp_name'] -> string 

，並嘗試做

$copied = copy("Array['tmp_name']" ....); 

在任何情況下，你應該使用move_uploaded_file（）以處理移動上傳的文件，而不是copy() 。 m_u_l有額外的檢查，以確保在上傳完成和您的腳本嘗試移動文件之間沒有人篡改文件。

Answer 3

的文件夾，您試圖複製你的文件必須具有相同的權限作爲你的PHP用戶上。（Apache用戶如果您的服務器是Apache）的

./
drwxrwxr-X根根應用
drwxrwxr-X阿帕奇阿帕奇FilesystemDir

Answer 4

，如果你還不能創建在目標文件夾中的文件，並且你已經把權限就可以了755，請檢查以下內容：

如果您的文件是：/path/to/test-in.txt

你應該有X權限上：

• /路徑
• /路徑/到
• 和讀權限/路徑/到/測試in.txt

檢查更多的細節在這裏 fopen() fails to open stream: permission denied, yet permissions should be valid