2011-04-06 106 views
4

我想上传服务器上的图像,下面是我在互联网上找到的脚本,并在本地工作,当我部署代码和数据库是给我“未能打开流:权限被拒绝“错误。php复制功能,未能打开流:权限被拒绝

<?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>"; 
}*/ 

?> 

我看到一些计算器应答者喜欢answers1answer2,但我不知道该怎么做?还有其他建议吗?

谢谢。

+0

你应该张贴** **齐全错误消息,不仅存根 – 2011-04-06 20:06:20

+0

您还应该使用'move_uploaded_file()'而不是'copy()'。 – prodigitalson 2011-04-06 20:06:37

+0

我只是写了一些排序来覆盖波斯尼亚 – eomeroff 2011-04-06 20:12:18

回答

17

确保保存文件(tmp)的文件夹权限设置为777。 类型chmod -R 777 path终端

+0

如何做到这一点? – eomeroff 2011-04-06 20:07:33

+0

在您的ftp客户端(或基于Web的文件管理器)中,您应该能够右键单击该文件夹并查看属性或权限的选项。从那里,777权限仅仅意味着所有者,组和公众都可以读取,写入和执行。请确保只更改'tmp'文件夹中的此权限,因为您不想授予每个文件夹的写权限。 – 65Fbef05 2011-04-06 20:14:51

+0

@eomeroff chmod -R 777终端上的路径 – 2016-02-29 21:12:37

0

你copy命令似乎有语法错误:

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

如果你想这样做

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

它不会反正工作。 PHP解析器不gready,并认为这是

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

,并尝试做

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

在任何情况下,你应该使用move_uploaded_file()以处理移动上传的文件,而不是copy() 。 m_u_l有额外的检查,以确保在上传完成和您的脚本尝试移动文件之间没有人篡改文件。

+0

所以这行应该去$ copied = copy(“Array ['tmp_name']”,$ newname); ? – eomeroff 2011-04-06 20:10:22

+0

我只是试过了,它不起作用。它说:警告:复制(/tmp/someImg.jpg)[function.copy]:未能打开流:没有这样的文件或目录在... – eomeroff 2011-04-06 20:16:18

+1

他意味着你可以省略第一撇号像这样'$ copied =复制($ _ FILES ['image'] ['tmp_name'],$ newname);' – 65Fbef05 2011-04-06 20:17:44

1

的文件夹,您试图复制你的文件必须具有相同的权限作为你的PHP用户上。(Apache用户如果您的服务器是Apache)的

./
drwxrwxr-X根根应用
drwxrwxr-X阿帕奇阿帕奇FilesystemDir

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