bash打印x使用ls找到文件之前的文件数

Question

I have a directory which has a bunch of files that are in yyyymmdd.hhmmss format. I am looking for the x number of files before a certain time is inquired.

For example, I have a directory with files that are named every 2 minutes (eg, 20190205.000200, 20190205.000400, 20190205.000600, ... , 20190205.235800). I want the name of x files before a time is entered. For an example, let's say I choose the time to be '20190205.200000', and I want 15 files. This would produce the files from 20190205.193000 to 20190205.20000 (15 files at 2-min timestep = half hour). Is there a way for the 'ls' command to do this (perhaps even with awk, grep, sed, etc.)? Or would it be easier to convert to a datetime and loop through every 2 minutes?

Answer 1

You could list the content of the directory, grep the name of the file and display X entry before the match

With these files

 🢒 ls -1   
20190205.0000000
20190205.0000200
20190205.0000400
20190205.0000600
20190205.0000800
20190205.0001000
20190205.0001200
20190205.0001400
20190205.0001600
20190205.0001800
20190205.0002000

You could do

 🢒 ls -1 | grep 20190205.0001400 -B 3 
20190205.0000800
20190205.0001000
20190205.0001200
20190205.0001400

Where the -B flag it used to set the number of lines to show before the match