How to remove part of the middle of a line/string by matching two known patterns in front and behind variable text to be removed - bash

Question

How to remove part of the middle of a line/string by matching two known patterns, one in front of text to be removed and one behind the text to be removed?
I have a Linux text file with thousands of one line, comma delimited records. unfortunately, all records are not the same format. Each line may have as many as four comma delimited fields of which only the first and last are constant, the two middle fields may, or may not, be present.
Examples of existing line (record) formats. Messy data but the first field is always present, as is the last field, starts with word ADDED.
FNAME LNAME, SOME COMMENT, JOINED DATE, ADDED TO DB DATE
FNAME LNAME, ADDED TO DB DATE
FNAME LNAME, SOME COMMENT, ADDED TO DB DATE
FNAME LNAME, JOINED DATE, ADDED TO DB DATE
Objective is to keep field one including the comma, throw away everything following the first comma, keeping the word "ADDED" and everything that follows to the end of line and insert a space between the first comma and the word ADDED.
For each line in parse the file from start of line to the first comma (keep this).
Parse rest of line up to the space before the word “Added” and throw it away.
Keep everything from the space before the word “ADDED” to end of line and concatenate the first part and last part to form one record per line with two fields separated by a comma and a space.
(if record is already in desired format, change nothing)
Final file to look like:
FNAME LNAME, ADDED TO DB DATE
or
Fred Flintstone, ADDED on January 1st 2015 By Barney Rubble
Thanks!

Answer 1

If you don't care about blank lines:
awk '{print $1,$NF}' FS=, OFS=, input
(Blank lines will be output as a single comma)
If you want to just skip blank lines, use:
awk 'NF>1{print $1,$NF}' FS=, OFS=, input
If you want to keep them:
awk '{printf( "%s%s\n", $1, NF>1 ? ","$NF : "")}' FS=, OFS=, input
Note that this will not ensure a single space after the comma, but will retain the spacing as in the final column of the original file. (that is, if there are 3 spaces after the final column in the original, you'll get 3 in the output). It's not clear to me from the description, but that seems like desirable behavior.

Answer 2

A Perl solution
perl -ne 'print join ", ", (split /,\s*/)[0,-1]' myfile
or
perl -pe 's/,.*(?=,)//' myfile
Both of those solutions work fine for me with the data you have given, but you may like to try
perl -pe 's/,.*(?=,\s*ADDED)//' myfile

Answer 3

You can use backreference:
sed 's/\(^[^,]*,\).* ADDED/\1 ADDED/' file

Answer 4

one more approach with awk could help here.
awk -F, '{val=$1;sub(/FNAME.*\,/,",");print val $0}' Input_file
Where I am making field separator as (,) then saving first field to variable named val, now substituting FNAME till comma with (,) in current line, now printing the value of variable val and new edited current line.

Answer 5

Using perl
#!/usr/bin/perl
use strict;
use warnings;
open my $fh, "<", "file.txt" or die "$!: couldn't open file\n";
while(<$fh>) {
my #arr = split(/,/);
my $text = $arr[0] . ", " . $arr[$#arr];
print "$text\n";
}