Autosys 文件观察程序文件正则表达式

Question

I have been facing issue in identifying similar file name in my file watcher service. My files looks like this:

1. filename1.YYYYMMDD.txt.YYYYMMDDSSSSSS.END regex I am using filename1.*.txt.*.END 
2. filename1.YYYYMMDD_trl.txt.YYYYMMDDSSSSSS.END regex  I am using filename1.*_trl.*.END
3. filename2.YYYYMMDD.txt.YYYYMMDDSSSSSS.END regex  I am using filename2.*.txt.*.END
4. filename2.YYYYMMDD_trl.txt.YYYYMMDDSSSSSS.END regex  I am using filename2.*_trl.*.END

So you see the problem here, 1 get confused with 2 and 3 gets confused with 4. What should I do, to make it more specific?

Answer 1

Since you are looking for regex i would show this by grep . However, ls command doesn't support regex.

Sample file:

$ cat filenames                                                                                                           
filename1.20210225.txt.20210225123456.END                                                                                                 
filename1.20200225_trl.txt.20210225123456.END

Code:

$ egrep "filename1.([0-9]{8}).txt.([0-9]{14}).END" filenames                                                              
filename1.20210225.txt.20210225123456.END                                                                                                 
$ egrep "filename1.([0-9]{8})_trl.txt.([0-9]{14}).END" filenames                                                          
filename1.20200225_trl.txt.20210225123456.END

Explanation:

in egrep "filename1.([0-9]{8}).txt.([0-9]{14}).END" filenames

[0-9] - Any digits from 0123456789

{8} - Exact count of characters. YYYYMMDD is of 8 characters similarly YYYYMMDDSSSSSS - 14