convert multiple variable output to a table in bash - bash

I have 3 variables, $commonName, $expiryDate and $DaysRemInUnixEpoch. Each variable has 3 lines as below output. I want to display output of all 3 variables in to 3 different columns. I tried looking for solution using printf but no luck. Can anyone please advise if they have done this in the past using printf and how? Any help will be much appreciated.
Below are 3 variables output together in one column. I want to split in to 3 columns having 3 rows in each column.
bash-4.1$ echo -e "$commonName\n$expiryDate\n$daysRemInUnixEpoch"
mycertificate_mycert.mycomp.net
PSIN0P551
ROOTROOTCA
Feb 6 2022 11:57:32 GMT
Jan 9 2023 18:51:25 GMT
Mar 12 2035 18:24:54 GMT
682
1020
5465
bash-4.1$
desired output I am looking for is something like below
mycertificate_mycert.mycomp.net Feb 6 2022 11:57:32 GMT 682
PSIN0P551 Jan 9 2023 18:51:25 GMT 1020
ROOTROOTCA Mar 12 2035 18:24:54 GMT 5465

With bash (Process Substitution), paste and column:
paste -d ';' <(echo "$commonName") <(echo "$expiryDate") <(echo "$daysRemInUnixEpoch") | column -s ';' -t
Output:
mycertificate_mycert.mycomp.net Feb 6 2022 11:57:32 GMT 682
PSIN0P551 Jan 9 2023 18:51:25 GMT 1020
ROOTROOTCA Mar 12 2035 18:24:54 GMT 5465
I assume that your variables do not contain ;.
see: man paste and man column

Related

Sort on 3, 31 and months consequently

I have these lines in a file:
Oct 29 23:14:39
Oct 30 19:45:15
Oct 31 13:15:19
Nov 1 10:34:15
Nov 2 18:39:20
Nov 3 12:34:59
Nov 4 16:34:59
Nov 5 20:34:59
When I run sort -r -k2 it gives me the following:
Nov 5
Nov 4
Nov 3
Oct 31
Oct 30
Oct 29
Nov 2
Nov 1
How do I get it like so:
Nov 5
Nov 4
Nov 3
Nov 2
Nov 1
Oct 31
Oct 30
Oct 29
Would appreciate any pointers, comments, advices at all. Do I also need to sort on months in reverse order? How? -M -r?
Assuming the dates are sorted in the file, just reverse the line order:
tac file
This works well!
sort -k1Mr -r -k2nr file.txt
file.txt content
Oct 6 20:34:59
Oct 29 23:14:39
Oct 30 19:45:15
Oct 31 13:15:19
Nov 1 10:34:15
Nov 2 18:39:20
Nov 3 12:34:59
Nov 4 16:34:59
Nov 5 20:34:59
Output:
Nov 5 20:34:59
Nov 4 16:34:59
Nov 3 12:34:59
Nov 2 18:39:20
Nov 1 10:34:15
Oct 31 13:15:19
Oct 30 19:45:15
Oct 29 23:14:39
Oct 6 20:34:59
NOTE:
The following answers fail to sort the entry Oct 6 20:34:59 if exists.
sort -k1,1 -k2r,2
sort -rM
This works for me:
sort -k1Mr,1 -k2r,2
month-sorted by field #1 reversely, then sorted by field #2 reversely
Nov 5 20:34:59
Nov 4 16:34:59
Nov 3 12:34:59
Nov 2 18:39:20
Nov 1 10:34:15
Oct 31 13:15:19
Oct 30 19:45:15
Oct 29 23:14:39
Edit:
A key specification like -k2 means to take all the fields from 2 to the end of the line into account.
To sort by a single column, use -k2,2 as the key specification. This means to use the fields from #2 to #2, i.e. only the second field.
man sort
-k, --key=KEYDEF
sort via a key; KEYDEF gives location and type
...
KEYDEF is F[.C][OPTS][,F[.C][OPTS]] for start and stop position, where
F is a field number and C a character position in the field; both are
origin 1, and the stop position defaults to the line's end.
The arrangement of the fields is lucky, in that the primary one comes first, and the secondary one second (and the rest of the line doesn't matter because you have at most one entry per day). But your example has only Oct and Nov, and I don't think you ought to rely on that. If you add e.g. Aug 7 you'll need a more robust solution. I suggest:
sort -rM
That is, sort in the default way (by the first field, then by the second), treat any month abbreviation found as a month abbreviation, and reverse the whole thing.
EDIT: This is incorrect. #SaSkY's answer is the correct one.

Find and Echo only the date (with format) in String Output on Bash

I am trying to get the date "+%a %b %d %R:%S %Y" in bash.
here's the sample command and output
$ xscreensaver-command --time
XScreenSaver 5.32: screen non-blanked since Thu Oct 29 12:15:05 2015 (hacks: #184, #60)
I am trying to get the the value Thu Oct 29 12:15:05 2015 on the string.
How can I achieve this?
Try to append with GNU grep:
2>&1 | grep -Po 'since \K.*(?= \()'
Output:
Thu Oct 29 12:15:05 2015

Calculate date time difference [duplicate]

This question already has answers here:
Difference in date time
(3 answers)
Closed 9 years ago.
I have 2 variables.
GMDCOM variable stores the date time in below format
Tue Oct 1 13:37:38 2013
Tue Oct 1 13:32:40 2013
Tue Oct 1 13:37:53 2013
GMDRRS variable stores the date time in the below format
Tue Oct 1 13:35:33 2013
Tue Oct 1 13:34:33 2013
Tue Oct 1 13:32:33 2013
I want to calculate the datetime difference
e.g Tue Oct 1 13:37:38 2013 - Tue Oct 1 13:35:33 2013 in hh:mm:ss format
and store it in another variable. I dont want to use PERL, AWK or SED. Instead i want to use normal BASH Shell commands to achieve it. Please help.
Try following
#!/bin/bash
GMDCOM='Tue Oct 1 13:37:38 2013'
GMDRRS='Tue Oct 1 13:35:33 2013'
d1=$(date -d "$GMDCOM" +%s)
d2=$(date -d "$GMDRRS" +%s)
dd=$(($d1-$d2))
ss=$(($dd%60))
mm=$((($dd/60)%60))
hh=$((($dd/3600)%60))
printf "%02d:%02d:%02d\n" "$hh" "$mm" "$ss"

sed: convert time(3) seconds in a table into printable date (spamdb)

I get the following from spamdb, where the third field represents the time in seconds since the Epoch.
Cns# spamdb | fgrep TRAPPED
TRAPPED|113.163.117.129|1360836903
TRAPPED|113.171.216.201|1360837481
TRAPPED|122.177.159.61|1360844596
TRAPPED|36.231.9.231|1360865649
TRAPPED|37.146.207.209|1360832096
TRAPPED|212.156.98.210|1360837015
TRAPPED|59.99.160.62|1360839785
TRAPPED|86.127.116.162|1360840492
TRAPPED|92.83.139.194|1360843056
TRAPPED|219.71.12.150|1360844704
I want to sort this table by the time, and print the time field with date -r, such that it's presentable and clear when the event has occurred.
How do I do this in tcsh on OpenBSD?
Sorting with sort is easy, and so is editing with sed; but how do I make sed execute date -r or equivalent?
There are indeed a few obstacles here: first, you basically have to separate the data, and then one part of it is presented as-is, whereas another part has to be passed down to date -r for date formatting, prior to being presented to the user.
Another obstacle is making sure the output is aligned: apparently, it's quite difficult to handle the tab character in the shell, possibly only on the BSDs:
sed replace literal TAB
Replacing / with TAB using sed
Also, as we end up piping this to sh for execution, we have to use a different separator for the fields other than the pipe character, |.
So far, this is the best snippet I could come up with, it seems to work great in my tcsh:
Cns# spamdb | fgrep TRAPPED | sort -n -t '|' -k 3 | sed -E -e 's#\|###g' \
-e 's#^([A-Z]+)#([0-9.]+)#([0-9]+)$#"echo -n \2_"; "date -r \3"#g' | \
xargs -n1 sh -c | awk '{gsub("_","\t",$0); print;}'
37.146.207.209 Thu Feb 14 00:54:56 PST 2013
113.163.117.129 Thu Feb 14 02:15:03 PST 2013
212.156.98.210 Thu Feb 14 02:16:55 PST 2013
113.171.216.201 Thu Feb 14 02:24:41 PST 2013
59.99.160.62 Thu Feb 14 03:03:05 PST 2013
86.127.116.162 Thu Feb 14 03:14:52 PST 2013
92.83.139.194 Thu Feb 14 03:57:36 PST 2013
122.177.159.61 Thu Feb 14 04:23:16 PST 2013
219.71.12.150 Thu Feb 14 04:25:04 PST 2013
36.231.9.231 Thu Feb 14 10:14:09 PST 2013

Remove duplicate entries in a Bash script [duplicate]

This question already has answers here:
How to delete duplicate lines in a file without sorting it in Unix
(9 answers)
Closed 7 years ago.
I want to remove duplicate entries from a text file, e.g:
kavitha= Tue Feb 20 14:00 19 IST 2012 (duplicate entry)
sree=Tue Jan 20 14:05 19 IST 2012
divya = Tue Jan 20 14:20 19 IST 2012
anusha=Tue Jan 20 14:45 19 IST 2012
kavitha= Tue Feb 20 14:00 19 IST 2012 (duplicate entry)
Is there any possible way to remove the duplicate entries using a Bash script?
Desired output
kavitha= Tue Feb 20 14:00 19 IST 2012
sree=Tue Jan 20 14:05 19 IST 2012
divya = Tue Jan 20 14:20 19 IST 2012
anusha=Tue Jan 20 14:45 19 IST 2012
You can sort then uniq:
$ sort -u input.txt
Or use awk:
$ awk '!a[$0]++' input.txt
It deletes duplicate, consecutive lines from a file (emulates "uniq").
First line in a set of duplicate lines is kept, rest are deleted.
sed '$!N; /^\(.*\)\n\1$/!P; D'
Perl one-liner similar to #kev's awk solution:
perl -ne 'print if ! $a{$_}++' input
This variation removes trailing whitespace before comparing:
perl -lne 's/\s*$//; print if ! $a{$_}++' input
This variation edits the file in-place:
perl -i -ne 'print if ! $a{$_}++' input
This variation edits the file in-place, and makes a backup input.bak
perl -i.bak -ne 'print if ! $a{$_}++' input
This might work for you:
cat -n file.txt |
sort -u -k2,7 |
sort -n |
sed 's/.*\t/ /;s/\([0-9]\{4\}\).*/\1/'
or this:
awk '{line=substr($0,1,match($0,/[0-9][0-9][0-9][0-9]/)+3);sub(/^/," ",line);if(!dup[line]++)print line}' file.txt

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