I used
cut -d " " -f 8
and
awk '{print $8}'
But this assumes that the 8th field is the last one, which is not always true.
How can I display the last field in a shell script?
Try doing this :
$ awk '{print $NF}'
or the funny
$ echo "foo bar base" | rev | cut -d ' ' -f1 | rev
base
Edit: this answer is wrong now because the question changed. Move along, nothing to see here.
You can use tail to print a specified number of bytes from the end of the input
tail -c 1
Related
I have a file as below
NAME(BOLIVIA) TYPE(SA)
APPLIC(Java) IP(192.70.xxx.xx)
NAME(BOLIVIA) TYPE(SA)
APPLIC(Java) IP(192.71.xxx.xx)
I am trying to extract the values NAME and IP using sed:
cat file1 |
sed ':a
N
$!ba
s/\n/ /g' | sed -n 's/.*\(NAME(BOLI...)\).*\(IP(.*)\).*/\1 \2/p'
However, I'm only getting the output:
NAME(BOLIVIA) IP(192.71.xxx.xx)
What I would like is:
NAME(BOLIVIA) IP(192.70.xxx.xx)
NAME(BOLIVIA) IP(192.71.xxx.xx)
Would appreciate it if someone could give me a pointer on what I'm missing.
TIA
Your first sed commands reformats the file into one long line. You could have used tr -d "\n" for this, but that is not the problem.
The problem is in the second part, where the .* greedy eats as much as possible until finding the last match.
Your solution could be "fixed" with the ugly
# Do not use this:
sed -zn 's/[^\n]*\(NAME(BOLI...)\)[^\n]*\n[^\n]*\(IP([^)]*)\)[^\n]*/\1 \2/gp' file1
Possible solutions:
cat file1 | paste -d " " - - | sed -n 's/.*\(NAME(BOLI...)\).*\(IP(.*)\).*/\1 \2/p'
# or
grep -Eo "(NAME\(BOLI...\)|IP\(.*\))" file1 | paste -d " " - -
# or
printf "%s %s\n" $(grep -Eo "(NAME\(BOLI...\)|IP\(.*\))" file1)
In case you are ok with awk could you please try following. Written and tested in link
https://ideone.com/bJDzgf with shown samples only.
awk '
match($0,/^NAME\([^)]*/){
name=substr($0,RSTART+5,RLENGTH-5)
next
}
match($0,/IP\([^)]*/){
print name,substr($0,RSTART+3,RLENGTH-3)
name=""
}
' Input_file
This might work for you (GNU sed):
sed -n '/NAME/{N;/IP/s/\s.*\s/ /p}' file
If a line contains NAME and the following line contains IP remove everything between and print the result.
An alternative shorter awk:
awk '$1 ~ /^NAME/ {nm = $1} $2 ~ /^IP/ {print nm, $2}' file
NAME(BOLIVIA) IP(192.70.xxx.xx)
NAME(BOLIVIA) IP(192.71.xxx.xx)
The issue in your script is the use .* which matches in a greedy way
so that you have only the first NAME(BOLI...) and last IP(.*)
If you can use python :
#!/bin/bash
python -c '
import re, sys
for ar in re.findall(r"(NAME\(BOLI.*?\)).*?(IP\(.*?\))", sys.stdin.read(), re.DOTALL):
print(*ar)
' < input-file
I have a grep'ed string from curl, and I want to summarize lines of by its content.
e. g.
input
SomethingA v2.3
SomethingA v2.4
SomethingElse v1.1
SomethingElse v1.2
output
SomethingA 2
SomethingElse 1
The numbers in the output are not a must, but if easy to achieve, would be very nice. The "v" as the leading space is a fix prefix for the numerics which don't have to contain a dot.
I tried echo "$str" | grep -Po '(.*(?<=))v[0-9]' but it still contains the "v1" .. and I don't know how to reduce the leading strings by multiple matches.
You can use
$ awk '{print $1}' filename | sort | uniq -c | awk '{print $2,$1}'
Note : This will also give a count of the blank lines. If you want to get rid of those, use :
$ grep -v '^$' filename | awk '{print $1}' | sort | uniq -c | awk '{print $2,$1}'
In my script, have a possible version number: 15.03.2 set to variable $STRING. These numbers always change. I want to strip it down to: 15.03 (or whatever it will be next time).
How do I remove everything after the second . using sed?
Something like:
$(echo "$STRING" | sed "s/\.^$\.//")
(I don't know what ^, $ and others do, but they look related, so I just guessed.)
I think the better tool here is cut
echo '15.03.2' | cut -d . -f -2
This might work for you (GNU sed):
sed 's/\.[^.]*//2g' file
Remove the second or more occurrence of a period followed by zero or non-period character(s).
$ echo '15.03.2' | sed 's/\([^.]*\.[^.]*\)\..*/\1/'
15.03
More generally to skip N periods:
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){2}[^.]*)\..*/\1/'
15.03.2
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){3}[^.]*)\..*/\1/'
15.03.2.3
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){4}[^.]*)\..*/\1/'
15.03.2.3.4
I’m trying to replace the 3rd column of a file for itself plus the value of column 2 (without any space). I get the proper value for variable c and a but then sed doesn't give any output. Any clue?
#!/bin/tcsh
setenv c `cat lig_mod.pdb | awk '{print $3}'`
echo $c
setenv a `cat lig_mod.pdb | awk '{print $3=$3$2}'`
echo $a
sed -i "" 's/^'"${c}"'$/^'"${a}"'$/g' lig_mod.pdb
Even though awk is usually better for columns parsing this one-liner sed can work for you as well:
sed -i 's/ \(\w*\) \(\w*\) / \1 \2\1 /1' lig_mod.pdb
the '/1' at the end denote the instance number you desire to change which for the 2nd and 3rd columns is the first, but you could use it for any adjacent columns.
I have a file with three columns. I would like to delete the 3rd column(in-place editing). How can I do this with awk or sed?
123 abc 22.3
453 abg 56.7
1236 hjg 2.3
Desired output
123 abc
453 abg
1236 hjg
try this short thing:
awk '!($3="")' file
With GNU awk for inplace editing, \s/\S, and gensub() to delete
1) the FIRST field:
awk -i inplace '{sub(/^\S+\s*/,"")}1' file
or
awk -i inplace '{$0=gensub(/^\S+\s*/,"",1)}1' file
2) the LAST field:
awk -i inplace '{sub(/\s*\S+$/,"")}1' file
or
awk -i inplace '{$0=gensub(/\s*\S+$/,"",1)}1' file
3) the Nth field where N=3:
awk -i inplace '{$0=gensub(/\s*\S+/,"",3)}1' file
Without GNU awk you need a match()+substr() combo or multiple sub()s + vars to remove a middle field. See also Print all but the first three columns.
This might work for you (GNU sed):
sed -i -r 's/\S+//3' file
If you want to delete the white space before the 3rd field:
sed -i -r 's/(\s+)?\S+//3' file
It seems you could simply go with
awk '{print $1 " " $2}' file
This prints the two first fields of each line in your input file, separated with a space.
Try using cut... its fast and easy
First you have repeated spaces, you can squeeze those down to a single space between columns if thats what you want with tr -s ' '
If each column already has just one delimiter between it, you can use cut -d ' ' -f-2 to print fields (columns) <= 2.
for example if your data is in a file input.txt you can do one of the following:
cat input.txt | tr -s ' ' | cut -d ' ' -f-2
Or if you better reason about this problem by removing the 3rd column you can write the following
cat input.txt | tr -s ' ' | cut -d ' ' --complement -f3
cut is pretty powerful, you can also extract ranges of bytes, or characters, in addition to columns
excerpt from the man page on the syntax of how to specify the list range
Each LIST is made up of one range, or many ranges separated by commas.
Selected input is written in the same order that it is read, and is
written exactly once. Each range is one of:
N N'th byte, character or field, counted from 1
N- from N'th byte, character or field, to end of line
N-M from N'th to M'th (included) byte, character or field
-M from first to M'th (included) byte, character or field
so you also could have said you want specific columns 1 and 2 with...
cat input.txt | tr -s ' ' | cut -d ' ' -f1,2
Try this :
awk '$3="";1' file.txt > new_file && mv new_file file.txt
or
awk '{$3="";print}' file.txt > new_file && mv new_file file.txt
Try
awk '{$3=""; print $0}'
If you're open to a Perl solution...
perl -ane 'print "$F[0] $F[1]\n"' file
These command-line options are used:
-n loop around every line of the input file, do not automatically print every line
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace
-e execute the following perl code