grep a string from a specific column My goal is to display a list of log names, from the start time column, grepping for `20161221`. I use this command: $ ls -m1 /var/log/audit.raw.* | grep 20161221 /var/log/audit.raw.20161220173001EST.20161221000004EST.gz /var/log/audit.raw.20161221000004EST.20161221083001EST.gz /var/log/audit.raw.20161221083001EST.20161221163000EST.gz /var/log/audit.raw.20161221163000EST.20161222000004EST.gz But the first entry is from `20161220`. I know I can trim it successfully with this command instead: $ ls -m1 /var/log/audit.raw.* | grep 20161221 | tail -n +2 /var/log/audit.raw.20161221000004EST.20161221083001EST.gz /var/log/audit.raw.20161221083001EST.20161221163000EST.gz /var/log/audit.raw.20161221163000EST.20161222000004EST.gz I wanted to see if there is a more intelligent use of grep to avoid trimming the output with `tail -n +2`

What you want to do may be done using just `ls`:


$ ls /var/log/audit.raw.20161221*