Use awk to print and edit a date

Question

I'm searching for a one-line command to get my desired output. Normal text looks like: "test_list_20160915_bla.log" Desired output: "2016/09/15"

I could do this with two awk commands (I know the command only print the year, it's just for the purpose):

echo "test_list_20160915_bla.log" |awk -F_ '$3 ~ /[0-9]/ {print $3}' |awk 'BEGIN {OFS="/"} {print substr($1,1,4)}'

But how do I use this within 1 command? Is awk even the right tool for that? Maybe sed could do the same, but I'm more familiar with awk.

I have some trouble with the provided solution. Sometimes we have files like: "test_20161205145213.log". With the sed command my output will look like "2051/45/21", which is pretty bad. Tried several things, but I can't figure it out.

Switched this

 sed -r 's!^.*_([0-9]{4})([0-9]{2})([0-9]{2})_.*$!\1/\2/\3!'

to

 sed -r 's!^.*(20[0-9]{2})([0-9]{2})([0-9]{2}).*$!\1/\2/\3!'

This limits the error outputs to timestamps with hour 2. I would prefer to completely ignore the filename and only go for the timestamp in it.

Timestamp could be yyymmdd_hhmmss or yyymmddhhmmss. I just need the yyyy/mm/dd part. Problem is that there is no fixed length or field seperator.

Answer 1

You can slice and dice in the same awk script:

echo "test_list_20160915_bla.log" |
    awk -F_ '$3 ~ /^[1-9][0-9]*$/ { print substr($3,1,4) "/" substr($3,5,2) "/" substr($3,7,2) }'

Here, we just extract the year, month and day digit groups from the third "_"-separated field and print them out.

Or with sed, which also handles your additional requirement:

(
    echo "test_list_20160915_bla.log"
    echo "test_20161205145213.log"
) |
    sed -r 's!^.*_([0-9]{4})([0-9]{2})([0-9]{2}).*$!\1/\2/\3!'

Here, we use a Regular Expression to capture three digit groups for year, month and day, discarding everything else, and then print the three groups out joined by /.