如何使用bash对文件夹中的文件进行排序？

时间：2022-04-17 15:55:23

I have these files in a folder:

我将这些文件放在一个文件夹中:

chap11-solutions.pdf
chap12-solutions.pdf
chap13-solutions.pdf
chap14-solutions.pdf
chap15-solutions.pdf
chap16-solutions.pdf
chap17-solutions.pdf
chap21-solutions.pdf
chap22-solutions.pdf
chap23-solutions.pdf
chap24-solutions.pdf
chap25-solutions.pdf
chap26-solutions.pdf
chap2-solutions.pdf
chap3-solutions.pdf
chap4-solutions.pdf
chap5-solutions.pdf
chap6-solutions.pdf
chap7-solutions.pdf
chap8-solutions.pdf
chap9-solutions.pdf

how do I sort them in this way: chap1..., chap...2, ...., chap11..., chap12,... using Ubuntu bash shell? Thanks.

我如何以这种方式对它们进行排序:chap1 ...,chap ... 2,....,chap11 ...,chap12,...使用Ubuntu bash shell?谢谢。

4 个解决方案

#1

ls|sort -V

The -V parameter ensures that chap10 is considered upper that chap9.

-V参数确保chap10被认为是chap9的上限。

#2

GNU ls has a version sort built-in:

GNU ls内置了一个版本排序:

ls -lv

#3

If you have ruby(1.9.1+)

如果你有红宝石(1.9.1+)

ruby -e 'puts Dir["chap*pdf"].sort_by{|x|x[/\d+/].to_i}'

#4

Assuming that you want to rename the files so you don't have to keep sorting them later:

假设您要重命名文件,以便以后不必继续对它们进行排序:

for f in chap*-solutions.pdf; do num=`echo $f | grep -o "[0123456789]\+"`; two_num=`printf "%02d" $num`; mv $f chap$two_num-solutions.pdf; done

grep -o "[0123456789]+" outputs the chapter number (one or two digits)

grep -o“[0123456789] +”输出章节号(一位或两位数)

printf returns a string that contains the zero-padded number

printf返回包含零填充数字的字符串

#1

ls|sort -V