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假设我们有一些字符串A“swdfsd/sdfsdf/sdfsd/sdfsd/sdf.file”和B“swdfsd/oooo/”,我们想做一些简单的工作,比如获取文件所在的文件夹名称并从位置B开始获取文件A的路径(结果如swdfsd/oooo/../sdfsdf/sdfsd/sdfsd/sdf.file)(不使用 boost::filesystem)?
A
B
swdfsd/oooo/../sdfsdf/sdfsd/sdfsd/sdf.file
编写类似于 Unix dirname 和 basename 命令的简单路径例程非常简单。那些只是处理路径的最后一个路径分隔符。
但正如有人已经评论过的那样,其他任何事情都可能很棘手。考虑使用 . 和..会使事情复杂化。
I am currently trying to play a list of .wav file using the VLC plugin and the html object tag in the following manner :
$.each(rdata['results'],function(key,data) { newRow += '<div class