php复制功能,未能打开流:权限被拒绝
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answers1和answer2,但我不知道该怎么做?还有其他建议吗?
谢谢。
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
$copied = copy("Array['tmp_name']" ....);
在任何情况下,你应该使用move_uploaded_file()以处理移动上传的文件,而不是copy()
。 m_u_l有额外的检查,以确保在上传完成和您的脚本尝试移动文件之间没有人篡改文件。
所以这行应该去$ copied = copy(“Array ['tmp_name']”,$ newname); ? – eomeroff 2011-04-06 20:10:22
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
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
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
,如果你还不能创建在目标文件夹中的文件,并且你已经把权限就可以了755,请检查以下内容:
如果您的文件是:/path/to/test-in.txt
你应该有X权限上:
- /路径
- /路径/到
- 和读权限/路径/到/测试in.txt
检查更多的细节在这里 fopen() fails to open stream: permission denied, yet permissions should be valid
你应该张贴** **齐全错误消息,不仅存根 – 2011-04-06 20:06:20
您还应该使用'move_uploaded_file()'而不是'copy()'。 – prodigitalson 2011-04-06 20:06:37
我只是写了一些排序来覆盖波斯尼亚 – eomeroff 2011-04-06 20:12:18