I'm writing a script to grab the last update/patch date on a few hundred servers. Lacking another tool due to various reasons, I've decided to write a script. At the moment I'm using the following command:
sudo yum history | grep [0-9] | grep -E 'Update|Install|U|I' | awk 'NR==1'
Which gives me the first line with an action on it. But it only gives me the first line, I toyed with the idea of grabbing the first 5 rows but that may not be applicable to every situation.
sudo yum history | grep [0-9] | grep -E 'Update|Install|U|I' | awk 'NR>=1&&NR<=5'
So I would like to check the last column or two and if more than x packages have been updated or installed then to grab that row.
Generically speaking the output of yum history is:
18 | first last <username> | 2018-08-30 19:41 | E, I, U | 43 ss
17 | first last <username> | 2018-07-10 15:28 | E, I, U | 230 EE
16 | first last <username> | 2018-04-25 20:08 | E, I, U | 44 ss
15 | first last <username> | 2018-01-30 20:57 | E, I, O, U | 108 EE
14 | first last <username> | 2018-01-30 20:39 | Install | 4
The issue I'm running into is the last two columns can differ in their column position and the last column may just be numeric or it may contain letters or special characters. I want to ignore any last column that has any character that is not numeric, and to evaluate whether the last column has more than 20 packages installed or updated. If the last column is more than 20 packages to then grab that row and only that row.
Use a regular expression, matching for the number in the last column. To print all history records with >=20 alterations:
sudo yum history | \
perl -ne '/\| *(\d+)[^\|]*$/ and $1>=20 and print($_)'
Of - if you only want the time stamp from matching history records:
sudo yum history | \
perl -ne '
#col=split(/\|/);
$col[4]=~/^ *(\d+)/ and $1>=20 and print($col[2],"\n")'
use awk:
sudo yum history | awk -F ' *\\| *' '$4 ~ /\<(Install|Update|U|I)\>/ && $5 > 20'
Related
We have to find the difference(d) Between last 2 nos and display rows with the highest value of d in ascending order
INPUT
1 | Latha | Third | Vikas | 90 | 91
2 | Neethu | Second | Meridian | 92 | 94
3 | Sethu | First | DAV | 86 | 98
4 | Theekshana | Second | DAV | 97 | 100
5 | Teju | First | Sangamithra | 89 | 100
6 | Theekshitha | Second | Sangamithra | 99 |100
Required OUTPUT
4$Theekshana$Second$DAV$97$100$3
5$Teju$First$Sangamithra$89$100$11
3$Sethu$First$DAV$86$98$12
awk 'BEGIN{FS="|";OFS="$";}{
avg=sqrt(($5-$6)^2)
print $1,$2,$3,$4,$5,$6,avg
}'|sort -nk7 -t "$"| tail -3
Output:
4 $ Theekshana $ Second $ DAV $ 97 $ 100$3
5 $ Teju $ First $ Sangamithra $ 89 $ 100$11
3 $ Sethu $ First $ DAV $ 86 $ 98$12
As you can see there is space before and after $ sign but for the last column (avg) there is no space, please explain why its happening
2)
awk 'BEGIN{FS=" | ";OFS="$";}{
avg=sqrt(($5-$6)^2)
print $1,$2,$3,$4,$5,$6,avg
}'|sort -nk7 -t "$"| tail -3
OUTPUT
4$|$Theekshana$|$Second$|$0
5$|$Teju$|$First$|$0
6$|$Theekshitha$|$Second$|$0
I have not mentiond | as the output field separator but still it appears, why is this happening and the difference is zero too
I am just 6 days old in unix,please answer even if its easy
your field separator is only the pipe symbol, so surrounding whitespace is part of the field definitions and that's what you see in the output. In combined uses pipe has the regex special meaning and need to be escaped. In your second case it means space or space is the field separator.
$ awk 'BEGIN {FS=" *\\| *"; OFS="$"}
{d=sqrt(($NF-$(NF-1))^2); $1=$1;
print d "\t" $0,d}' file | sort -n | tail -3 | cut -f2-
4$Theekshana$Second$DAV$97$100$3
5$Teju$First$Sangamithra$89$100$11
3$Sethu$First$DAV$86$98$12
a slight rewrite will eliminate the number of fields dependency and fixes the format.
At the moment
I want to trim .fmbi1a5nn9sp5o4qy3eyazeq5.eddvrl9sa8t448pb38vibj8ef: and .ilwio0k43fgqt4jqzyfadx19v: so the output take less space :)
First step:
docker ps --format "{{.Names}}: {{.Status}}" | sort -k1 | column -t
mon_node-exporter.fmbi1a5nn9sp5o4qy3eyazeq5.eddvrl9sa8t448pb38vibj8ef: Up 7 days
mon_prometheus.1.ilwio0k43fgqt4jqzyfadx19v: Up 7 days
I know
I can do something like:
docker ps --format "{{.Names}}: {{.Status}}" | sort -k1 | rev | cut -d"." -f2- | rev
mon_node-exporter.fmbi1a5nn9sp5o4qy3eyazeq5
mon_prometheus.1
The issue
is that I'm losing the other columns :-/
Idea
It would sound logical to do something like this (with awk) but it does not work. Any ideas?
docker ps --format "{{.Names}} : {{.Status}}" | sort -k1 | awk '{(print $1 | rev | cut -d"." -f2- | rev),$2,$3,$4,$5,$6}' | column -t
Thank you in advance!
P
to cut the last dot extension
$ docker ... | sort | awk '{sub(/\.[^.]*$/,"",$1)}1' file | column -t
mon_node-exporter.fmbi1a5nn9sp5o4qy3eyazeq5 Up 7 days
mon_prometheus.1 Up 7 days
or, delete anything longer than 20 chars after a dot.
$ ... | sed -e 's/\(\.[a-z0-9:]\{20,\}\)* / /' | column -t
mon_node-exporter Up 7 days
mon_prometheus.1 Up 7 days
Works! This trick will make my life so much easier.
(I removed file)
docker ps --format "{{.Names}}: {{.Status}}" | sort -k1 | awk '{sub(/\.[^.]*$/,"",$1)}1' | column -t;
mon_grafana.1 Up 24 hours
mon_node-exporter.fmbi1a5nn9sp5o4qy3eyazeq5 Up 23 hours
Question #2:
Now how would you proceed to cut the characters after the first dot?
Cheers!
How to count number of integers in a file using egrep?
I tried to solve it as a pattern finding problem. Actually, I am facing problem of how to represent range of characters [0-9] continuously which include "space" before the beginning and "space or dot" after the end. I think the latter can be solved by using \< and \> respectively. Also, It should not include dot in between otherwise it will not be an integer. I am unable to convert this logic into regular expression using available tools and techniques.
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
The right answer should be 5 i.e. for 2322, 33, 55, 69 and 33.
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' | wc -w
^^ ^ ^ ^ ^ ^ ^
|| | | | | | |
E = extended regex--------+| | | | | | |
o = extract what found-----+ | | | | | |
starts with new line or space---+ | | | | |
digits--------------------------------+ | | | |
optional dot, question mark, etc.-------------+ | | |
ends with end line or space----------------------------+ | |
repeat 1 time or more (to detect integers like "123 456")--+ |
count words------------------------------------------------------+
Note: 123. 123? 123: are also counted as integer
Test:
#!/bin/bash
exec 3<<EOF
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
two integers 123 456.
how many tables in room 400? 50.
50? oh I thought it was 40.
23: It's late, 23:00 already
EOF
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' <&3 | \
tee >(sleep 0.5; echo -n "integer counted: "; wc -w; )
Outputs:
2322.
33
55.
69.
33.
123 456.
400? 50.
50?
40.
23:
integer counted: 12
Based on the observation that you want 66.88 excluded, I'm guessing
grep -Ec '[0-9]\.?( |$)' file
which finds a digit, optionally followed by a dot, followed by either a space or end of line.
The -c option says to report the number of lines which contain a match (so not strictly the number of matches, if there are lines which contain multiple matches) and the -E option enables extended regular expression syntax, i.e. what was traditionally calned egrep (though the command name is now obsolescent).
If you need to count matches, the -o option prints each match on a separate line, which you can then pass to wc -l (or in lucky cases combine with grep -c, but check first; this doesn't work e.g. with GNU grep currently).
On my ubuntu this code working fine
grep -P '((^)|(\s+))[-+]?\d+\.?((\s+)|($))' test
I have the following command that I use to rewrite some maxscale output to be able to use it in other software:
maxadmin list servers | sed -r 's/[^a-z 0-9]//gi;/^\s*$/d;1,3d;' | awk '$1=$1' | cut -d ' ' -f 1,5 | sed -e 's/ /":"/g' | sed -e 's/\(.*\)/"\1"/' | tr '\n' ',' | sed 's/.$/}\n/' | sed 's/^/{/'
I am thinking this is way to complex for what I want to do, but I am not able to see a simpler version of this myself. What I want is to rewrite this (output of maxadmin list servers):
Servers.
-------------------+-----------------+-------+-------------+--------------------
Server | Address | Port | Connections | Status
-------------------+-----------------+-------+-------------+--------------------
svr_node1 | 192.168.178.1 | 3306 | 0 | Master, Synced, Running
svr_node2 | 192.168.178.1 | 3306 | 0 | Slave, Synced, Running
svr_node3 | 192.168.178.1 | 3306 | 0 | Slave, Synced, Running
-------------------+-----------------+-------+-------------+--------------------
Into this:
{"svrnode1":"Master","svrnode2":"Slave","svrnode3":"Slave"}
My command does a good job but as I said, there should be a simpler way with less sed commands being run hopefully.
You can use awk, like this:
json.awk
BEGIN {
printf "{"
}
# Everything after line for and before the last ------ line
# plus the last empty line (if any).
NR>4&&!/^([-]|$)/{
sub(/,/,"",$9) # Remove trailing comma
printf "%s\"%s\":\"%s\"",s,$1,$9
s="," # Set comma separator after first iteration
}
END {
print "}"
}
Run it like this:
maxadmin list servers | awk -f json.awk
Output:
{"svr_node1":"Master","svr_node2":"Slave","svr_node3":"Slave"}
In comments there came up the question how to achieve that without an extra json.awk file:
maxadmin list servers | awk 'BEGIN{printf"{"}NR>4&&!/^([-]|$)/{sub(/,/,"",$9);printf"%s\"%s\":\"%s\"",s,$1,$9;s=","}END{print"}"}'
Ugly, but works. ;)
If you want to put this into a shell script, consider a multiline version like this:
maxadmin list servers | awk '
BEGIN{printf"{"}
NR>4&&!/^([-]|$)/{
sub(/,/,"",$9)
printf"%s\"%s\":\"%s\"",s,$1,$9
s=","
}
END{print"}"}'
I'm trying to write a shell script that prints the full names of users logged on to a machine. The finger command gives me a list of users, but there are many duplicates. How can I loop through and print out only the unique ones?
Edit:
This is the format of what finger gives me:
xxxx XX of group XXX pts/59 1:00 Feb 13 16:38
xxxx XX of group XXX pts/71 1:11 Feb 13 16:27
xxxx XX of group XXX pts/105 1d Feb 12 15:22
xxxx YY of group YYY pts/102 2:19 Feb 13 14:13
xxxx ZZ of group ZZZ pts/42 2d Feb 7 12:11
I'm trying to extract the full name (i.e. whatever comes before 'of group' in column 2), so I would be using awk together with finger.
What you want is actually fairly difficult in a shell script, here is, for example, my full output of finger(1):
Login Name TTY Idle Login Time Office Phone
martin Martin Tournoij *v0 1d Wed 14:11
martin Martin Tournoij pts/2 22 Wed 15:37
martin Martin Tournoij pts/5 41 Thu 23:16
martin Martin Tournoij pts/7 31 Thu 23:24
martin Martin Tournoij pts/8 Thu 23:29
You want the full name, but this may contain 1 space (as per my example), or it may just be 'Teller' (no space), or it may be 'Captain James T. Kirk' (3 spaces). So you can't just use the space as delimiter. You could use the character position of 'TTY' in the header as an indicator, but that's not very elegant IMHO (especially with shell scripting).
My solution is therefore slightly different, we get only the username from finger(1), then we get the full name from /etc/passwd
#!/bin/sh
prev=""
for u in $(finger | tail +2 | cut -w -f1 | sort); do
[ "$u" = "$prev" ] && continue
echo "$u $(grep "^$u" /etc/passwd | cut -d: -f5)"
prev="$u"
done
Which gives me both the username & login name:
martin Martin Tournoij
Obviously, you can also print just the real name (without the $u).
The sort and uniq BinUtils commands can be used to removed duplicates.
finger | sort -u
This will remove all duplicate lines, but you will still see similar lines due to how verbose the finger command is. If you just want a list of usernames, you can filter it out further to be very specific.
finger | cut -d ' ' -f1 | sort -u
Now, you can take this one step further, and remove the "header/label" line printed out by the finger command.
finger | cut -d ' ' -f1 | sort -u | grep -iv login
Hope this helps.
Other possible solution:
finger | tail -n +2 | awk '{ print $1 }' | sort | uniq
tail -n +2 to omit the first line.
awk '{ print $1 }' to extract the first column.
sort to prepare input for uniq.
uniq remove duplicates.
If you want to iterate use:
for user in $(finger | tail -n +2 | awk '{ print $1 }' | sort | uniq)
do
echo "$user"
done
Could this be simpler?
No spaces or any other special characters to worry about!
finger -l | awk '/^Login/'
Edit: To remove the content after of group
finger -l | awk '/^Login/' | sed 's/of group.*//g'
Output:
Login: xx Name: XX
Login: yy Name: YY
Login: zz Name: ZZ