I know, I know, this sounds soo easy. But I can't seem to find the correct answer on the Internet.
我知道,我知道,这听起来很容易。但是我似乎在网上找不到正确的答案。
One of the solution I found was to use is_dir
.
我发现的一个解决方案是使用is_dir。
if(is_dir($dir))
echo 'directory exists';
else
echo 'drectory not exist';
But this is wrong-- All this function does is to check whether the $dir
is a directory, it doesn't check whether the directory exists, or not. In other words if I put:
但是这是错误的——这个函数所做的就是检查$dir是否是一个目录,它不检查该目录是否存在。换句话说,如果我说
$rootDir = "C:\\Documents and Settings\\test\\My Documents\\Image Directory\\Me Dog\\";
then the function will return a true, even though you can find no such directory on your web server.
然后该函数将返回一个true,即使您在web服务器上找不到这样的目录。
Any ideas?
什么好主意吗?
4 个解决方案
#1
54
Should work correctly. From is_dir()
documentation:
应该正常工作。文档的作用是:判断给定文件名是否是:
Returns TRUE if the filename exists and is a directory, FALSE otherwise.
如果文件名存在且为目录,则返回TRUE,否则返回FALSE。
Well, anyway if it doesn't try this:
好吧,不管怎样,如果它不尝试这个:
if(file_exists($dir) && is_dir($dir))
BTW. results of these functions are cached in stat cache. Use clearstatcache() to clean that cache.
顺便说一句。这些函数的结果缓存在stat缓存中。使用clearstatcache()清理缓存。
#2
3
You'll probably want to use opendir() after is_dir() agrees that the path (could) be a directory.
在is_dir()同意路径(可能)是一个目录之后,您可能希望使用opendir()。
If the resource returned by opendir() is valid, you know you have a directory, and already have a handle to read it.
如果opendir()返回的资源是有效的,则您知道您有一个目录,并且已经有一个读取它的句柄。
Just be sure to call closedir(), either way, if a valid handle is returned.
只要确保在返回有效句柄时调用closedir()即可。
Edit:
编辑:
This answer assumes that you'll be opening the directory either way. If you just need to ensure a path is sane / valid, file_exists() is much cheaper.
这个答案假设您将以任何一种方式打开这个目录。如果只需要确保路径是健全/有效的,file_exists()要便宜得多。
#3
1
You could try this:
你可以试试这个:
if(is_dir($dir) && is_writeable($dir))
{
// ...
}
#4
1
bool file_exists ( string $filename )
Checks whether a file or directory exists.
检查文件或目录是否存在。
http://php.net/manual/en/function.file-exists.php
http://php.net/manual/en/function.file-exists.php
#1
54
Should work correctly. From is_dir()
documentation:
应该正常工作。文档的作用是:判断给定文件名是否是:
Returns TRUE if the filename exists and is a directory, FALSE otherwise.
如果文件名存在且为目录,则返回TRUE,否则返回FALSE。
Well, anyway if it doesn't try this:
好吧,不管怎样,如果它不尝试这个:
if(file_exists($dir) && is_dir($dir))
BTW. results of these functions are cached in stat cache. Use clearstatcache() to clean that cache.
顺便说一句。这些函数的结果缓存在stat缓存中。使用clearstatcache()清理缓存。
#2
3
You'll probably want to use opendir() after is_dir() agrees that the path (could) be a directory.
在is_dir()同意路径(可能)是一个目录之后,您可能希望使用opendir()。
If the resource returned by opendir() is valid, you know you have a directory, and already have a handle to read it.
如果opendir()返回的资源是有效的,则您知道您有一个目录,并且已经有一个读取它的句柄。
Just be sure to call closedir(), either way, if a valid handle is returned.
只要确保在返回有效句柄时调用closedir()即可。
Edit:
编辑:
This answer assumes that you'll be opening the directory either way. If you just need to ensure a path is sane / valid, file_exists() is much cheaper.
这个答案假设您将以任何一种方式打开这个目录。如果只需要确保路径是健全/有效的,file_exists()要便宜得多。
#3
1
You could try this:
你可以试试这个:
if(is_dir($dir) && is_writeable($dir))
{
// ...
}
#4
1
bool file_exists ( string $filename )
Checks whether a file or directory exists.
检查文件或目录是否存在。
http://php.net/manual/en/function.file-exists.php
http://php.net/manual/en/function.file-exists.php