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	public function upload(){	$path = "/uploads/news/" . date("Ymd");		if (!is_dir($path)){		$res=mkdir($path,0777,true);		if ($res){			echo "目录 $path 创建成功";exit;		}else{			echo "目录 $path 创建失败";exit;		}	}	echo "here";exit;


我尝试创建一个目录,可是不知道为什么老是创建不了,总是会直接跳到 here 这里。


回复讨论(解决方案)

成功或失败总会提示一个吧。

成功或失败总会提示一个吧。



就是if里面根本没有跳进去,直接跳here 不知道怎么回事

那可能你的upload方法根本就没有执行到,自己检查一下。

那可能你的upload方法根本就没有执行到,自己检查一下。



有,upload进去了,if前面那行业执行了,但是就是没有跳进去if里面 纳闷!

$path = "/uploads/news/" . date("Ymd");  去掉最前面的/ 

$path = "/uploads/news/" . date("Ymd");
你把这个目录建在服务器根目录下啦
目录已经存在所以没执行if的语句块啊

$path = "/uploads/news/" . date("Ymd");  去掉最前面的/ 




去掉前面/ 整一个是表示什么意思? 如果我要表示在服务器上的这个项目的根目录下的uploads文件夹的话我要怎么表示?

$path = "/uploads/news/" . date("Ymd");
你把这个目录建在服务器根目录下啦
目录已经存在所以没执行if的语句块啊



看了一下,的确这个文件夹存在了,但是是在直接D盘的uploads文件夹下的,但是我要让它在项目根目录下的uploads文件夹下得怎么写法呢?

$path = $_SERVER['DOCUMENT_ROOT']."/uploads/news/" . date("Ymd")

$path = $_SERVER['DOCUMENT_ROOT']."/uploads/news/" . date("Ymd")



服务器上也可以这样写吗 ? 不知道为什么目录echo出来是正确 但是没办法把图片显示出来 
显示这样 Not allowed to load local resource: file:///D:/wamp/apps/p1weixin/uploads/news/20160129/1454032603836.jpg


$path = $_SERVER['DOCUMENT_ROOT']."/uploads/news/" . date("Ymd")



服务器上也可以这样写吗 ? 不知道为什么目录echo出来是正确 但是没办法把图片显示出来 
显示这样 Not allowed to load local resource: file:///D:/wamp/apps/p1weixin/uploads/news/20160129/1454032603836.jpg
你代码是这样的格式的话  出现这情况就再正常不过了
<img  src="/static/imghwm/default1.png"  data-src="file:///D:/wamp/apps/p1weixin/uploads/news/20160129/1454032603836.jpg"  class="lazy"  / alt="PHP无法创建目录" >

写文件,出问题要么是权限问题,要么是路径问题。把错误信息放出来,基本上就能发现是什么问题了。

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