Linux：如何获取名称以大于10的数字结尾的文件

Question

I have some files that have names like this: 'abcdefg_y_zz.jpg' The 'abcdefg' is a sequence of digits, while the 'y' and 'zz' are letters. I need to get all the files that have the sequence of digits ending with a number greater than 10. The files that have 'fg' greater that 10.

Does anyone have an idea on how to do that in a bash script?

Answer 1

好的，从技术上讲，基于你的所有信息......

ls | grep '[0-9]{5}[1-9][0-9]_[[:alpha:]]_[[:alpha:]]{2}.jpg'

Answer 2

How about this? Just exclude ones which have 0 in position f.

ls -1 | grep -v "?????0?_?_??.jpg"

Update Since you want > 10 and not >= 10, you'll need to exclude 10 too. So do this:

ls -1 | grep -v "?????0*_?_??.jpg" | grep -v "??????10_?_??.jpg"

Answer 3

with more scripting

#!/bin/bash
for i in seq FIRST INCREMENT LAST
do
cp abcde$i_y_zz.jpg /your_new_dir //or whatever you want to do with those files
done

so in your example line with seq will be

for i in seq 11 1 100000000

Answer 4

If the filenames are orderly named this awk solution works:

ls | awk 'BEGIN { FIELDWIDTHS = "5 2" } $2 > 10'

Explanation

• FIELDWIDTHS = "5 2" means that $1 will refer to the first 5 characters and $2 the next 2.
• $2 > 10 matches when field 2 is greater than 10 and implicitly invokes the default code block, ie '{ print }'

Answer 5

只需一个过程：

ls ?????+(1[1-9]|[2-9]?)_?_??.jpg

Answer 6

All the solutions provided so far are fine, but anybody who's had some experience with shell programming knows that parsing ls is never a good idea and must be avoided . This actually doesn't apply in this case, where we can assume that the names of the files follow a certain pattern, but it's a rule that should be remembered. More explanation here .

What you want can be achieved much safer with GNU find - assuming that you run the command in the directory where the files are, it would look something like this :

find . -regextype posix-egrep -regex '\./[0-9]{5}[1-9][0-9]_[[:alpha:]]_[[:alpha:]]{2}.jpg$'