[英]How to add parent dir variable to scandir (PHP)
Hi sorry this is probably a really newbie question, but I'm searching for the answer for 2 days now and no luck.嗨,抱歉,这可能是一个真正的新手问题,但我现在正在寻找答案 2 天,但没有运气。 I'm using a script what generates an image gallery from an uploaded folder.
我正在使用从上传的文件夹生成图片库的脚本。 Script info: Tim-thumb script created by Tim McDaniels and Darren Hoyt with tweaks by Ben Gillbanks.
脚本信息:Tim-thumb 脚本由 Tim McDaniels 和 Darren Hoyt 创建,并由 Ben Gillbanks 进行调整。
You need to add the directory name what contains the images to the script like:您需要将包含图像的目录名称添加到脚本中,例如:
$dirname = "images/" ;
$images = scandir($dirname);
shuffle($images);
What I'm trying to do is make the script work automatically if i upload it in a new folder with new images so I don't need to add the $dirname every time I upload a gallery.如果我将脚本上传到包含新图像的新文件夹中,我想要做的是让脚本自动工作,这样我就不需要每次上传图库时都添加 $dirname。 for example:
例如:
$dirname = $mydir ; $dirname = $mydir ;
where mydir returns the path of the current directory like:其中 mydir 返回当前目录的路径,如:
$mydir = basename(getcwd()) . $mydir = basename(getcwd()) 。 DIRECTORY_SEPARATOR ;
DIRECTORY_SEPARATOR ;
but it's not working.但它不起作用。
also tried to make it work from a function:还试图让它从一个函数工作:
function current_dir()
{$path = dirname($_SERVER[PHP_SELF]);$position = strrpos($path,'/') + 1;print substr($path,$position);}
than -->比-->
$dirname = current_dir() ; $dirname = current_dir() ;
but no luck.但没有运气。 I think I'm missing something here, I'm a totally noob and maybe it's just a syntax issue but can't make it work.
我想我在这里遗漏了一些东西,我完全是个菜鸟,也许这只是一个语法问题,但无法使其工作。 I always get [function.scandir]: failed to open dir ... or creates the gallery but the images not working (I see only alt tags) thank you for any help.
我总是得到 [function.scandir]: failed to open dir ... 或创建图库但图像不起作用(我只看到 alt 标签)感谢您的帮助。
>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>
EDIT : !!!编辑 : !!!
Just realized... :o !!!刚刚意识到... :o !!!
The answer is:答案是:
$dirname = "./" ;
omg我的天啊
<<<<<<<<<<<<<<<<<<< <<<<<<<<<<<<<<<<<<<<<
"I think I'm missing something here..." :D “我想我在这里遗漏了一些东西......”:D
Try尝试
$parent = dirname(__DIR__);
http://php.net/manual/en/language.constants.predefined.php http://php.net/manual/en/language.constants.predefined.php
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