Is there a simple bash command that will match the date string in a filename like this... File Name 20th May 2019 descr.txt (File Name and descr are strings of variable length) ...and convert it to a filename like this: File_Name_20190520_descr.txt ?
I know there are some other questions that discuss date-based filename conversions but they are either related to Perl or don't specifically mention the ordinal indicator (1st, 2nd, 3rd, 4th).
With zsh:
zmodload zsh/datetime autoload zmv zmv -n '(* )(<1-31>)??( ??? <1900-2100>)( *.txt)' \ '${1// /_}$(strftime %Y%m%d "$(strftime -r "%d %b %Y" $2$3)")${4// /_}' Remove the -n when happy.
That's if the month is the 3 character abbreviation. If it's the full name:
zmodload zsh/datetime zmodload zsh/langinfo autoload zmv zmv -n '(* )(<1-31>)?? ('${(vj:|:)langinfo[(I)MON_*]}')( <1900-2100>)( *.txt)' \ '${1// /_}$(strftime %Y%m%d "$(strftime -r "%d %B %Y" $2$3$4)")${5// /_}' Those assume the file names are in the language of the current locale. If the month names are always in English, you can set LC_ALL=C. In any case, where te language is not English, you can't necessarily expect the abbreviations to always be 3 characters long or the th, nd, rd to be two characters long, so you'd probably need to adapt.
awk:
{ # split by date split($0, a, /[0-9]{1,2}[a-z]{2} +(January|February|March|April|May|June|July|August|September|October|November|December) +[0-9]{4}/, seps) # remove trailing nth from day sub(/[a-z]{2}/, "", seps[1]) # get desired date format from shell command "date -d '" seps[1] "' +%Y%m%d" | getline d # replace space with _ gsub(/ +/, "_", a[1]) gsub(/ +/, "_", a[2]) print a[1] d a[2] } Read comment as explanation.
I use full name of month, you might need to change it to abbreviation.
