在网页源代码中隐藏真实路径,但是点击链接还是要能在新页面打开。
如果这样写:
$_f = readfile($_p); header("Content-type: image/* "); echo $_f;
然后点击源代码中的 src 路径,结果 php 文件被下载下来了
回复讨论(解决方案)
url rewrite
url rewrite
服务器我动不了的不过倒是可以用 RewriteRule
我觉得你这个需求,又要不改动服务器端,又要源码隐藏img的路径,似乎办不到啊。。。。
我觉得你这个需求,又要不改动服务器端,又要源码隐藏img的路径,似乎办不到啊。。。。
不是。。我是说我改不了服务器配置。后面改了下可以用了,也不会出现下载php的现象。
后来发现在输出图片之前,不能有任何的 echo,哪怕是 echo '' 都会出错。所以全部用重定向到404页面代替了
url rewrite
那你把你码中header("Content-type: image/* ");的image/*改成具体的图片类型,并且该方法在所有输出之前调用

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