PHP - 警告：复制（..）：无法打开流：中没有这样的文件或目录
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PHP - Warning: copy(..): failed to open stream: No such file or directory in
提问by Brian
I'm fairly new to PHP and was trying to create a simple PHP file upload system.
我对 PHP 还很陌生，正在尝试创建一个简单的 PHP 文件上传系统。
I followed a tutorial from (http://www.phpeasystep.com/phptu/2.html). I only altered the $HTTP_POST_FILES, as it was giving me errors, and from what I read it's old in PHP.
我遵循了 (http://www.phpeasystep.com/phptu/2.html) 中的教程。我只更改了$HTTP_POST_FILES，因为它给了我错误，并且从我读到的内容来看，它在 PHP 中很旧。
I got less error messages but I am getting an error in the copy()function, with these given error messages:
我收到的错误消息较少，但在 copy()函数中出现错误，并显示以下错误消息：
Warning: copy(Task2/uploads/anonymous.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\Task2\upload.php on line 13
Warning: copy(Task2/uploads/DSCF4639.JPG): failed to open stream: No such file or directory in C:\xampp\htdocs\Task2\upload.php on line 14
Warning: copy(Task2/uploads/jien maroon.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\Task2\upload.php on line 15
I thought it was a problem with permission (read/write permissions in Windows 7), but from a quick google search it seems that XAMPP is set by default to deal with permission on Win 7.
我认为这是权限问题（Windows 7 中的读/写权限），但从谷歌快速搜索来看，XAMPP 似乎默认设置为处理 Win 7 上的权限。
This is the code:
这是代码：
<?php
//set where you want to store files
//in this example we keep file in folder upload
//$_FILES['ufile']['name']; = upload file name
//for example upload file name cartoon.gif . $path will be upload/cartoon.gif
$path1= "Task2/uploads/".$_FILES['ufile']['name'][0];
$path2= "Task2/uploads/".$_FILES['ufile']['name'][1];
$path3= "Task2/uploads/".$_FILES['ufile']['name'][2];
//copy file to where you want to store file
copy($_FILES['ufile']['tmp_name'][0], $path1);
copy($_FILES['ufile']['tmp_name'][1], $path2);
copy($_FILES['ufile']['tmp_name'][2], $path3);
//$_FILES['ufile']['name'] = file name
//$_FILES['ufile']['size'] = file size
//$_FILES['ufile']['type'] = type of file
echo "File Name :".$_FILES['ufile']['name'][0]."<BR/>";
echo "File Size :".$_FILES['ufile']['size'][0]."<BR/>";
echo "File Type :".$_FILES['ufile']['type'][0]."<BR/>";
echo "<img src=\"$path1\" width=\"150\" height=\"150\">";
echo "<P>";
echo "File Name :".$_FILES['ufile']['name'][1]."<BR/>";
echo "File Size :".$_FILES['ufile']['size'][1]."<BR/>";
echo "File Type :".$_FILES['ufile']['type'][1]."<BR/>";
echo "<img src=\"$path2\" width=\"150\" height=\"150\">";
echo "<P>";
echo "File Name :".$_FILES['ufile']['name'][2]."<BR/>";
echo "File Size :".$_FILES['ufile']['size'][2]."<BR/>";
echo "File Type :".$_FILES['ufile']['type'][2]."<BR/>";
echo "<img src=\"$path3\" width=\"150\" height=\"150\">";
///////////////////////////////////////////////////////
// Use this code to display the error or success.
$filesize1=$_FILES['ufile']['size'][0];
$filesize2=$_FILES['ufile']['size'][1];
$filesize3=$_FILES['ufile']['size'][2];
if($filesize1 && $filesize2 && $filesize3 != 0)
{
echo "We have recieved your files";
}
else {
echo "ERROR.....";
}
//////////////////////////////////////////////
// What files that have a problem? (if found)
if($filesize1==0) {
echo "There're something error in your first file";
echo "<BR />";
}
if($filesize2==0) {
echo "There're something error in your second file";
echo "<BR />";
}
if($filesize3==0) {
echo "There're something error in your third file";
echo "<BR />";
}
?>
Any help would be appreciated !
任何帮助，将不胜感激 ！
Thanks !
谢谢 ！
回答by Naftali aka Neal
Do notuse copy, use move_uploaded_file(...)
千万不能使用copy，使用move_uploaded_file(...)
回答by Sebastian Viereck
make sure that the destination directory exists, copy will not create directories for you. The 3rd parameter will also create directories recursive.
确保目标目录存在，复制不会为您创建目录。第三个参数也将创建递归目录。
if (!is_dir($directory)) {
mkdir($directory, 0777, true);
}
回答by Brian
Solved the problem !
解决了问题！
The problem was with the Path.. instead of Task2/uploads/I had to put ../Task2/uploads/.
问题出在 Path .. 而不是Task2/uploads/我必须把../Task2/uploads/.
Thanks!
谢谢！
