问题描述
我的文件夹结构如下:
mydomain.com
->Folder-A
->Folder-B
我有一个来自数据库的字符串'../Folder-B/image1.jpg',它指向Folder-B中的图像.
I have a string from Database that is '../Folder-B/image1.jpg', which points to an image in Folder-B.
在Folder-A中的脚本中,我使用目录名( FILE )来获取文件名,然后我得到了mydomain.com/Folder-A
.在此脚本中,我需要获取一个表示'mydomain.com/Folder-B/image1.jpg
的字符串.我尝试过
Inside a script in Folder-A, I am using dirname(FILE) to fetch the filename and I getmydomain.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';
这显示为mydomain.com%2FFolder-A%2F..%2FFolder-B%2Fimage1.jpg
这是针对Facebook分享按钮的,无法获取正确的图像.有人知道如何正确获取路径吗?
This is for a facebook share button, and this fails to fetch the correct image. Anyone know how to get the path correctly?
我希望获得一个网址>>> mydomain.com%2FFolder-B%2Fimage1.jpg
I hope to get a url >>>mydomain.com%2FFolder-B%2Fimage1.jpg
推荐答案
对于PHP< 5.3使用:
For PHP < 5.3 use:
$upOne = realpath(dirname(__FILE__) . '/..');
在PHP 5.3至5.6中使用:
In PHP 5.3 to 5.6 use:
$upOne = realpath(__DIR__ . '/..');
在PHP> = 7.0中使用:
In PHP >= 7.0 use:
$upOne = dirname(__DIR__, 1);
这篇关于PHP如何在目录名(__FILE__)上向上一级的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!