Search Results

Using the filenames as the ONLY time reference:
newest=$( ls abcd*| sort -n -k2 -t '.' | tail -1 )

Using file mtimes (last time written to, the default for ls -l )
newest=$( ls -rt '.' | tail...

That depends on what you call "latest". If you can make ABSOLUTELY sure the name's time stamp is correct, can't be tampered with, and coincides with the file creation, why not use it? If you don't,...

if we only assume that the fifteenth field is the last at the same time

awk 'BEGIN {FS="|"; OFS="|"} {gsub("[^a-zA-Z0-9 ]", "", $FN)} 1' file


Please use CODE tags as required by forum rules!...

I'm disappointed that you ignored my request for clarity in the description of your problem again and still won't show us a sample of what you're trying to do.

If we assume that when you say you...

You weren't too far off in your own attempt. Try, after correcting the already commented on syntax errors and anchoring the regex at the line end,



sed 's/[^a-zA-Z|0-9]$/ /; s/ */ /g' file

Hi, rakeshp
Try this. Will only delete the last one
sed 's/[^\w]$//'
or. Will delete range of last characters
sed 's/[^\w]\+$//'

Bad idea to use sed and awk together. Try this
awk 'BEGIN {FS="|"; OFS="|"} {sub("[^[:alnum:]]$", ""); gsub("[^[:alnum:]]", "", $15)} 1'

If the input file has a lot more than 89 records, you might also want to try:
awk 'NR==23;NR==89{print;exit}'
to avoid reading the rest of the file after line/record 89.

Yout can do this with two commands
awk 'NR==23; NR==89'
or with a logical or
awk 'NR==23 || NR==89'
This is the short condition{action} form where the default action is print.