Omit repeating lines - UNIX for Dummies Questions & Answers

TheCrunge · #1 02-22-2005

Can someone help me with the following 2 objectives?

1) The following command is just an example. It gets a list of all print jobs. From there I am trying to extract the printer name. It works with the following command:

Code:

lpstat -W "completed" -o | awk -F- '{ print $1}'

Problem is, I want to be able to omit these multiple lines:

Code:

My_office_printer
My_office_printer
My_office_printer
My_home_printer
My_home_printer

So it looks like this:

Code:

My_office_printer
My_home_printer

Now, each line is actually different (because each one is its own print job), but because I use awk to just get the name of the printer, the end result displays the printer name for however many print jobs there were. I just want one instance of each printer name.

2) In the previous section, I want to reduce the number of lines returned to one per printer. Before I do, I want to count how many there are. For example, how many time was My_office_printer listed.

Any help is greatly appreciated....

vgersh99 · #2 02-22-2005

Code:

lpstat -W "completed" -o | awk -F- '{arr[$1]++}END {for (i in arr) printf("Printer->[%s] Jobs->[%d]\n", i, arr[i])'

TheCrunge · #3 02-22-2005

Holy moly, thanks a heap. That works great!!

I'll be back with more questions

For example, I need to make a graphical representation of that. That's a whole other can of worms though.

Thanks again!

TheCrunge · #4 02-22-2005

OK, my next question on this:

Code:

lpstat -W "completed" -o | grep "testguy" | awk -F- '{arr[$1]++}END {for (i in arr) printf("Printer: %s Jobs: %d\n", i, arr[i])}'

How would I get the user name in the result now that I'm filtering by users?

For example:

Code:

Printer: My_office_printer
Jobs: 21
User: testguy

Again, thanks!

vgersh99 · #5 02-22-2005

Quote:

Originally Posted by TheCrunge

OK, my next question on this:

Code:

lpstat -W "completed" -o | grep "testguy" | awk -F- '{arr[$1]++}END {for (i in arr) printf("Printer: %s Jobs: %d\n", i, arr[i])}'

How would I get the user name in the result now that I'm filtering by users?

For example:

Code:

Printer: My_office_printer
Jobs: 21
User: testguy

Again, thanks!

what is the sample output you're trying to parse? pls post a sample with multiple users as it will need to be parsed.

Single line OR multiple lines per user?

TheCrunge · #6 02-22-2005

OK, here's the deal (my head is starting to hurt here)

I need to get the number of Jobs that are printed on each printer, per date, by user.

What you did earlier helped get the printer name and number of jobs. I tried to change it a little, so I can get some more info (the date) out of it.

So, if I enter "testguy" as my user (variable from other script), then I want the following returned:

Code:

Tue Feb 22: My_office_printer: 3 Jobs - My_home_printer: 5 Jobs"
Mon Feb 21: My_office_printer: 2 Jobs - My_home_printer: 8 Jobs"
Fri Feb 18: Someone_elses_printer: 9 Jobs
etc....

I will then need to be able to use each line as a variable so I can find a way to "graph" this later....

I'll try to explain it better, but I thought I would post this anyway...

In case it helps, here is all the info returned by

Code:

 lpstat -W "completed" -o | grep "testguy"

-W "completed" show all jobs that have been completed, not waiting to complete. The -o will show for all printers. There is a -u flag, but grepping the user seems to work better.

Code:

My_office_printer-494       testguy           23552   Fri Feb 18 09:54:42 2005

This is one job entry. The actual result is many lines, all with this format.
I need to strip the unique job number from the name of the printer (different for each entry, in this case it's -494). The 23552 is the number of bytes the job is, don't need that, and I don't need the time and year. So, really, what I need here is $1,$2,$4,$5,$6, and $1 has to be without the -494 or whatever the unique job number is).

vgersh99 · #7 02-22-2005

how about something like that for starters:

lpstat -W "completed" -o | grep "testguy" | nawk -f crunge.awk

here's crunge.awk:

Code:

BEGIN{
  FLDprt="1"
  FLDusr="2"
}

{
  printer=substr($FLDprt,1,index($0,"-")-1)
  date=$(FLDusr+2) " " $(FLDusr+3) " "$(FLDusr+4)
  dates[date]
  arr[date, printer]++;
}
END {
  for (dateI in dates) {
    printf("%s:", dateI);
    for (dataI in arr) {
      split(dataI, dataIa, SUBSEP)
      if (dataIa[1] == dateI) {
           printf(" %s: %d Jobs - ",
                dataIa[2],  arr[dataI])
           delete arr[dataI];
      }
    }
    print "\"\n"
  }
}

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