交换文件名中的第一个和最后四个字符(Swap first and last four characters in a filename)

交换文件名中的第一个和最后四个字符(Swap first and last four characters in a filename)

堆栈溢出。

我收集了一些工作笔记。 我把它们作为markdown文件保存，并且用日期和年份格式化它们 - 例如，今天的标题是06132017.md

我工作了一年，所以我有很多这样的文件。 我希望将命名约定从第一个月/第一年改为第一年，这样我就可以按字母顺序对它们进行排序，并轻松找到我需要的日期。

因此06132017.md将成为20170613.md - 这将使2016年和2017年不再按照aplha顺序混合。 是否有可以在文件夹上运行的命令来执行此操作？

StackOverflow.

I have a colleciton of notes from work. I keep them as markdown files, and had been formatting them with the date and year - for example, today's is titled 06132017.md

I am coming up on a year at work, so I have quite a few of these files. I wish to change the naming convention from month/day first to year first, so that I can sort them alphabetically and easily find dates I need.

So 06132017.md would become 20170613.md - this would keep 2016 and 2017 from mixing in aplha order. Is there a command I can run on a folder to do this?

最满意答案

如果您有Perl重命名实用程序，则执行起来相当简单：

$ prename 's/^(....)(....)(\.md)$/$2$1$3/' *.md 06132017.md renamed as 20170613.md

点匹配任何字符，括号组，替换侧的$N插入组中捕获的字符。

或者只是在Bash：

$ for x in ????????.md ; do mv -v "$x" "${x:4:4}${x:0:4}.md" ; done '06132017.md' -> '20170613.md'

${var:n:m}取一个长度为m的子字符串，从变量var位置n开始。

If you have the Perl rename utility, it's rather simple to do:

$ prename 's/^(....)(....)(\.md)$/$2$1$3/' *.md 06132017.md renamed as 20170613.md

The dots match any character, the parenthesis group, and $N on the replacement side inserts the characters captured in the groups.

Or just in Bash:

$ for x in ????????.md ; do mv -v "$x" "${x:4:4}${x:0:4}.md" ; done '06132017.md' -> '20170613.md'

${var:n:m} takes a substring of length m, starting at position n from variable var.