Help script find file most recent

Shell Programming and Scripting
1

Hi,

I need to find the most recent files by their name from an X repertoire.
The problem is that the name of the files is of type:

POWERL10_20151203000.xml
POWERL10_20151203001.xml
POWERL10_20151202000.xml

FIXED VALUE_DATENNN.xml

NNN = Sequential number

I would need to recover the first two files in the example.

He tried to use find:

find / home / Rec12 -name $ 1 -type f | queue -1 >> test.txt

I don't get resultat.

Could anyone help me, please? Thank you

2 
ls -1t $(find /somedir -type f -name "$1") | head -2
1 Like
3

Thanks!

But if I write

head -2

, only I will have the last two values. In the repertoire there are thousands of files every day.
I wanted to give an example of the files.

4

Then, specify your request correctly and detailedly, and post representative samples.

5

You don't mention what OS you are using this solution uses GNU find so that we can match files with version info within their name:

look=${1:-[A-Z]*}
find /home/rec12 -type f -regextype posix-extended -regex ".*/${look}_[0-9]{9,}$" -print | awk -F / '
{ fn=$NF
  split(fn, a, "_")
  fval=a[2]
  fname[a[1]]

  if (fv[a[1], 1] < fval) {
      fv[a[1], 2] = fv[a[1], 1]
      fp[a[1], 2] = fp[a[1], 1]
      fv[a[1], 1] = fval
      fp[a[1], 1] = $0
  } else if (fv[a[1], 2] < fval) {
      fv[a[1], 2] = fval
      fp[a[1], 2] = $0
  }
}
END {
    for (fd in fname) {
       print fp[fd,1] 
       if(fd SUBSEP 2 in fp) print fp[fd,2]
    }
}'

It picks the most recent (via date and revision #) version of the passed file regardless of the folder it is contained in.

Example:

$ mkdir -p {a,b,c}/{d,e,f}
$ touch /home/rec12/{a,b,c}/{d,e,f}/{TRIAL,DUMMY}_20180329{0,1,2}
$ touch /home/rec12/c/d/TRIAL_201803299
$ touch /home/rec12/a/e/TRIAL_201803297

$ ./verita TRIAL
/home/rec12/c/d/TRIAL_201803299
/home/rec12/a/e/TRIAL_201803297