Switch from grep to sed

Question

I have a files students.txt, with lines of the form:

Surname, Forename: Day.Month.Year: Degree

For example:

Smith, John: 15.01.1986: MSc IT
Taylor, Susan: 04.05.1987: MSc IT
Thomas, Steve: 19.04.1986: MSc MIT
Sellen, Jo: 03.07.1987: MSc CSE

How can I change the command below into a sed command to return all the surnames of students who were born in 1987?

$ grep 1987 students.txt | grep -o "^\(.*\),"

Answer 1

sed -n '/1987/s/^\([^,]*\),.*$/\1/p' students.txt

or as Glenn pointed out:

sed -n '/1987/s/,.*//p' students.txt

Answer 2

If you want to go with awk I would suggest a different answer to Johnsyweb's. The power of awk includes choosing the field separator(s):

awk -F'[,: ]*' '$3 ~ /1987$/ {print $1}' students.txt

But the question was about sed, so I believe SiegeX should get the better vote.

Answer 3

This kind of problem is well-suited to awk:

% awk '/\.1987:/ {gsub(/,.*/, ""); print}' students.txt 
Taylor
Sellen

Or, specifying a column, rather than the whole line (making it easier for you to customise, should you wish to add other columns in :

% awk '/\.1987:/ {
        gsub(/,$/, "", $1);          
        print $1;
}' students.txt
Taylor
Sellen