Using the Sort, Cut, and Awk commands

UNIX for Beginners Q & A UNIX for Dummies Questions & Answers
1

Hello,

I am trying, utilizing the few commands I know, to extract all records within my file that were generated in November of 2007. Each record within the file has a "date" field giving the month, day, and year (9-8-88). How do I extract those records to their own file?

Once I extract those records I want to remove any records that are duplicates in another field. Please advise.

Thank you.

update: I'm leaving work in 10 minutes so hopefully someone can assist me. Thank you.

2

I'm leaving, but I will check this message board tomorrow.

Thanks.

3

Please provide an example of the file. You could use awk to separate the month, day and year

echo 9-8-88|awk -F "-" '{print $2}'

That will print the second field separated by '-'. If you wish to print the other fields just use $1 or $3 instead. To print more than one field, "print $2,$3".

4

Thank you for your response.

Here is a (hypothetical) example of what is in our file:

Bill Adams: ABC Contracting: 11-14-07: 6919408:314-463-2219
Bill Adams: ABC Contracting: 2-22-07: 4109663:314-463-2219
Bill Adams: ABC Contracting: 11-6-07: 6918997:314-463-2219
Steve Leitch: ERB Equipment: 11-6-07: 6918967:314-225-6403

There are thousands of records and I only want the ones from 11-07 and I only want the name and company of the contractor to be displayed once in my output.

Thanks again.

5

Assuming the same file formation as the one shown above, and filtering is done based on the contractor's name, you can try:

  awk -F: '$3 ~ /11-[0-3]?[0-9]-07/ && !a[$1]++ { print $1, $2 > "Nov_2007" }' file

$  cat Nov_2007
Bill Adams  ABC Contracting
Steve Leitch  ERB Equipment

For a fancier output use printf.