Unix pattern datetime match - regex

Question

I want to edit this line:
1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350
and i want the output to be :
1987,4,12,31,4,1987-12-31 00:00:00.000,UA,19977,UA,,631,12197,1219701,31703,HPN,White
Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350
I want to find each pattern of: ****-**-** **:**:**.0000000
and erase the last 4 digits ( 0000 ) so I get ****-**-** **:**:**.000.
If its helpful this date format is in the 6th columns and the n-1 columns.

Answer 1

To get the value of the 6th column and erase the last four digits you can use:
awk -F, '{print substr($6, 0, length($6)-4) }'
Similarly, the N-1 column can be reached by:
awk -F, '{print substr( $(NF-1), 0, length($(NF-1))-4) }'
Edit:
To only replace the values in the columns, but still print everything use:
awk 'BEGIN{ FS=","; OFS=","}
{ $6=substr($6, 0, length($6)-4);
$(NF-1)=substr( $(NF-1), 0,length($(NF-1))-4);
print $0}'

Answer 2

Awk based solution
Nicely formatted, portable script:
#!/usr/bin/awk -f
BEGIN {
FS = "," # input: fields are separated by ,
OFS = "," # output: fields are separated by ,
}
{
sub(/[0-9][0-9][0-9][0-9]$/, "", $6) # remove last 4 digits from the 6th column
sub(/[0-9][0-9][0-9][0-9]$/, "", $(NF-1)) # remove last 4 digits from the n-1 column
print
}
One-line, less portable version using gawk:
gawk --re-interval -F , -v OFS=, '{sub("[0-9]{4}$", "", $6); sub("[0-9]{4}$", "", $(NF-1)); print}'
N.B. The regular expression engine of the traditional awk doesn't support the {n} repetition operator, so gawk version 3 or older needs to be run with --re-interval. For other flavors of awk e.g. nawk, you need to explicitly repeat the regular expression as in the portable longer script from above.
sed based solution
sed -r 's/^(([^,]*,){5})([^,]+)[0-9]{4},(([^,]*,)*)([^,]+)[0-9]{4}(,[^,]*)$/\1\3\4\6\7/'
(tested with GNU sed-4.2.2-6)

Answer 3

You could try this GNU sed command also,
$ sed -r 's/^.*,([^,]*)....,.*$/\1/g' file
1987-12-31 08:09:12.000
If you want just replacing then try this,
$ sed -r 's/^(.*,)([^,]*)....(,.*)$/\1\2\3/g' file
1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350
I think you want the the output to be like this,
$ grep -oP '[0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\....' file
1987-12-31 00:00:00.000
1987-12-31 08:09:12.000
Update:
$ echo '1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350' | sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\....)..../\1/g'
1987,4,12,31,4,1987-12-31 00:00:00.000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350

Answer 4

Here's a solution in Perl.
Update - Edited to output the full CSV line with the timestamp replaced with the truncated one
Update 2 - Update both timestamp columns, not just the first one
#!/usr/bin/env perl
use strict;
use warnings;
use feature 'say';
use Text::CSV;
my $CSV = Text::CSV->new();
while (my $line = readline(STDIN)) {
$CSV->parse($line) or die "Unable to parse line '$line'";
my #fields = $CSV->fields();
for my $f (#fields) {
$f =~ s/
^ # start of string
( # start capture to $1
\d{4} - # year
\d{2} - # month
\d{2} \s+ # day
\d{2} : # hour
\d{2} : # minute
\d{2} [.] # second
\d{3} # milisecond
) # end capture to $1
\d{4} # unwanted sub-second precision
$ # end of string
/$1/gmsx;
}
$CSV->combine(#fields);
say $CSV->string();
}
For example:
alex#yuzu:~$ cat input.txt
1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350
alex#yuzu:~$ ./csv.pl < input.txt
1987,4,12,31,4,"1987-12-31 00:00:00.000",UA,19977,UA,,631,12197,1219701,31703,HPN,"White Plains"," NY",NY,36,"New York",22,13930,1393001,30977,ORD,Chicago\," IL",IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,"1987-12-31 08:09:12.000",519494350
On a Debian-like system such as Ubuntu you should already have Perl, and you can install Text::CSV with:
$ sudo apt-get install libtext-csv-perl