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问题:
I have a folder structure as follows:
mydomain.com
->Folder-A
->Folder-B
I have a string from Database that is '../Folder-B/image1.jpg', which points to an image in Folder-B.
Inside a script in Folder-A, I am using dirname(FILE) to fetch the filename and I get
mydomain.com/Folder-A
. Inside this script, I need to get a string that says 'mydomain.com/Folder-B/image1.jpg
. I tried
$path=dirname(__FILE__).'/'.'../Folder-B/image1.jpg';
This shows up as mydomain.com%2FFolder-A%2F..%2FFolder-B%2Fimage1.jpg
This is for a facebook share button, and this fails to fetch the correct image. Anyone know how to get the path correctly?
Edit: I hope to get a url >>>mydomain.com%2FFolder-B%2Fimage1.jpg
回答1:
For PHP < 5.3 use:
$upOne = realpath(dirname(__FILE__) . '/..');
In PHP 5.3 to 5.6 use:
$upOne = realpath(__DIR__ . '/..');
In PHP >= 7.0 use:
$upOne = dirname(__DIR__, 1);
回答2:
If you happen to have php 7.0+ you could use levels.
dirname( __FILE__, 2 )
with the second parameter you can define the amount of levels you want to go back.
http://php.net/manual/en/function.dirname.php
回答3:
Try this
dirname(dirname( __ FILE__))
Edit: removed "./" because it isn't correct syntax. Without it, it works perfectly.
回答4:
You could use PHP's dirname function.
<?php echo dirname(__DIR__); ?>
.
That will give you the name of the parent directory of __DIR__
, which stores the current directory.
回答5:
You can use realpath
to remove unnessesary part:
// One level up
echo str_replace(realpath(dirname(__FILE__) . '/..'), '', realpath(dirname(__FILE__)));
// Two levels etc.
echo str_replace(realpath(dirname(__FILE__) . '/../..'), '', realpath(dirname(__FILE__)));
On windows also replace \
with /
if need that in URL.
回答6:
One level up, I have used:
str_replace(basename(__DIR__) . '/' . basename(__FILE__), '', realpath(__FILE__)) . '/required.php';
or for php < 5.3:
str_replace(basename(dirname(__FILE__)) . '/' . basename(__FILE__), '', realpath(__FILE__)) . '/required.php';
回答7:
To Whom, deailing with share hosting environment and still chance to have Current PHP less than 7.0
Who does not have dirname( __FILE__, 2 );
it is possible to use following.
function dirname_safe($path, $level = 0){
$dir = explode(DIRECTORY_SEPARATOR, $path);
$level = $level * -1;
if($level == 0) $level = count($dir);
array_splice($dir, $level);
return implode($dir, DIRECTORY_SEPARATOR).DIRECTORY_SEPARATOR;
}
print_r(dirname_safe(__DIR__, 2));
回答8:
dirname(__DIR__,level);
dirname(__DIR__,1);
level is how many times will you go back to the folder
回答9:
I use this, if there is an absolute path (this is an example):
$img = imagecreatefromjpeg($_SERVER['DOCUMENT_ROOT']."/Folder-B/image1.jpg");
if there is a picture to show, this is enough:
echo("<img src='/Folder-B/image1.jpg'>");