Given the following ...
$ less -N file.txt
1 first line
2 second line
3 third line
4 fourth line
file.txt (END)
... I'd like to do something like this:
$ less -N --STARTING-NUMBER=0 file.txt
0 first line
1 second line
2 third line
3 fourth line
file.txt (END)
In other words, I'd like to be able to specify which value the line numbering starts from.
Note that this is possible using nl:
$ nl -v 0 file.txt
0 first line
1 second line
2 third line
3 fourth line
But -N in less can be toggled on and off without leaving less, whereas if I pipe the above into less, the line numbers could not be toggled off.
If less has something like nl's -v option - or there were any other way to achieve the same - that would be awesome. But I don't see it in the less(1) man pages.
Here comes a quick patch
--- line.c.bak 2022-12-07 22:30:28
+++ line.c 2022-12-28 01:40:08
## -367,7 +367,7 ##
len = 0;
else
{
- linenumtoa(linenum, buf);
+ linenumtoa(linenum - 1, buf);
len = (int) strlen(buf);
}
for (i = 0; i < linenum_width - len; i++)
or you make a feature request at https://github.com/gwsw/less/issues.
Related
Consider the following data in a file file.txt:
$
$
$
FORCE 10 30 40
* 1 5 4
FORCE 11 20 22
* 2 3 0
FORCE 19 25 10
* 16 12 8
.
.
.
I want to print every 2nd element of the third column, starting from line 4, resulting in:
30
20
25
I have tried:
cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'
However, this is not resulting in anything being printed and no errors generated either.
You might use awk checking that the row number > 3 and then check for an even row number with NR%2==0.
Note that you don't have to use cat
awk 'NR > 3 && NR%2==0 {
print $3
}' file.txt
Output
30
20
25
Using sed
$ sed -En '4~2s/([^ \t]*[ \t]+){2}([^ \t]*).*/\2/p' input_file
30
20
25
I have tried:
cat file.txt | sed 's/\|/ /' | awk 'NR%2==4 {print $3}'
However, this is not resulting in anything being printed and no errors
generated either.
You do not need cat whilst using GNU sed as it can read file on its' own, in this case it would be sed 's/\|/ /' file.txt.
You should consider if you need that part at all, your sample input does not have pipe character at all, so it would do nothing to it. You might also drop that part if lines holding values you want to print do not have that character.
Output is empty as NR%2==4 does never hold, remainder of division by x is always smaller than x (in particular case of %2 only 2 values are possible: 0 and 1)
This might work for you (GNU sed):
sed -nE '4~2s/^((\S+)\s*){3}.*/\2/p' file
Turn off implicit printing by setting the -n option and reduce back slashes in regexps by turning on -E.
From the fourth line and then every second line thereafter, capture the third column and print it.
N.B. The \2 represents the last inhabitant of that back reference which in conjunction with the {3} means the above.
Alternative:
sed -n '4,${s/^\(\(\S\+\)\s*\)\{3\}.*/\2/p;n}' file
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Using the command line, how can I transform something like:
1 first line
2 second line
3 third line
4 fourth line
extra bit
5 fifth line
6 sixth line
into, say:
1 first line
2 second line
3 third line
4 fourth line; extra bit
5 fifth line
6 sixth line
The condition on which I would like to merge, is to remove any newline creating a line which does not start with a number.
I have seen answers to similar questions using the command-line tools awk, sed, and tr.
awk '/^[0-9]/{ printf "%s%s", (NR == 1 ? "" : "\n"), $0; next}
{printf "; %s", $0} END { printf "\n"}' input
I'm not really sure what you want to do when the first line does not begin with a digit, and I'm making the assumption that starting with a digit is the characteristic you are looking for to combine lines. Modify as needed.
With GNU sed:
sed "4{N;s/\n/; /}" file
With GNU awk:
awk -v line=4 'NR==line{x=$0; getline; $0=x "; " $0}1' file
Output:
1 first line
2 second line
3 third line
4 fourth line; extra bit
5 fifth line
6 sixth line
Could you please try following.
Written and tested it in
https://ideone.com/xqk4si
awk -v line_num="5" '
FNR==(line_num-1){
val=$0
next
}
val{
$0=val";"$0
val=""
}
1
' Input_file
Explanation: mentioning awk variable named line_num which has line number which OP wants to merge with its previous line. In main program checking condition if current line is just one lesser than mentioned line number of yes then create variable val and save that line. Then next condition checking if Val is SET then print previous line value semi colon and current line value and next will skip all further statements from there. 1 is way to print the current lines in awk
On second thought, it might be better to merge all lines that do not start with a number, rather than specifying by number each line to be merged.
Easy to do with ed:
printf "%s\n" '2,$g/^[^0-9]/-1s/$/; /\' '.,+1j' w | ed -s input.txt
Translated from ed's rather cryptic commands: For each line that does not start with a digit (Skipping the first line because it has no previous one to merge with), add ; to the end of the previous line, and then join those two lines. Finally save the changed file.
Example:
$ cat input.txt
1 first line
2 second line
extra stuff
3 third line
4 fourth line
extra bit
5 fifth line
6 sixth line
$ printf "%s\n" '2,$g/^[^0-9]/-1s/$/; /\' '.,+1j' w | ed -s input.txt
$ cat input.txt
1 first line
2 second line; extra stuff
3 third line
4 fourth line; extra bit
5 fifth line
6 sixth line
With GNU sed, to join any number of lines not starting with a digit:
sed -E ':a;N;s/\n([^0-9])/; \1/;ta;P;D;' file
Is it possible using awk or sed to get the line number of a line such that it is the first line matching a regex after another line matching another regex?
In other words:
Find line l1 matching regex r1. l1 is the first line matching r1.
Find line l2 below l1. l2 matches regex r2. l2 is the first line matching r2, ignoring lines l1 and above.
Clarification: By match I mean partial match, for most general solution.
A partial match can of course be turned into a full-word match with \<...\> or a full-line match with ^...$.
Example input:
- - '787928'
- stuff
- - '810790'
- more stuff
- - '787927'
- yet more stuff
- - '828055'
- some more stuff
- - '828472'
- some other stuff
If r1 is ^-.*787927.* and r2 is ^- I'd expect the output to be 7, i.e. the number of the line that says - - '828055'.
Input example :
world
zekfzlefkzl
fezekzevnkzjnz
hello
zeniznejkglz
world
eznkflznfkel
hello
zenilzligeegz
world
Command :
pat1="hello"; pat2="world";
awk -v pat1=$pat1 -v pat2=$pat2 '$0 ~ pat1{pat1_match = 1}($0 ~ pat2)&&pat1_match{print NR; exit}' <input>
Output :
6
For an input file that looks like this:
1 pat2
2 x
3 pat1
4 x
5 pat2
6 x
7 pat1
8 x
9 pat2
you could use sed as follows:
$ sed -n '/pat1/,${/pat2/{=;q;};}' infile
5
which works like this:
sed -n ' # suppress output with -n
/pat1/,$ { # for all lines from the first occurrence of "pat1" on...
/pat2/ { # if the line matches "pat2"
= # print line number
q # quit
}
}' infile
The above fails if the first occurrence of pat1 is on the same line as pat2:
1 pat2
2 x
3 pat1 pat2
4 x
5 pat2
6 x
7 pat1
8 x
9 pat2
would print 3. With GNU sed, we can use this instead:
$ sed -n '0,/pat1/!{/pat2/{=;q;};}' infile
5
sed -n ' # suppress output
0,/pat1/! { # for all lines after the first occurrence of "pat1"
/pat2/ { # if the line matches "pat2"
= # print line number
q # quit
}
}' infile
The 0 address is a GNU extension; using 1 instead would break if pat1 was on the first line.
This might work for you (GNU sed):
sed -n '/^-.*787927.*/{:a;n;/^-/!ba;=;q}' file
On encountering a line that begins -.*787927.*, start a loop that replaces the current line with the next, until a line begins - where on print the line number and quit.
How to print all lines if certain condition matches.
Example:
echo "$ip"
this is a sample line
another line
one more
last one
If this file has more than 3 lines then print the whole variable.
I am tried:
echo $ip| awk 'NR==4'
last one
echo $ip|awk 'NR>3{print}'
last one
echo $ip|awk 'NR==12{} {print}'
this is a sample line
another line
one more
last one
echo $ip| awk 'END{x=NR} x>4{print}'
Need to achieve this:
If this file has more than 3 lines then print the whole file. I can do this using wc and bash but need a one liner.
The right way to do this (no echo, no pipe, no loops, etc.):
$ awk -v ip="$ip" 'BEGIN{if (gsub(RS,"&",ip)>2) print ip}'
this is a sample line
another line
one more
last one
You can use Awk as follows,
echo "$ip" | awk '{a[$0]; next}END{ if (NR>3) { for(i in a) print i }}'
one more
another line
this is a sample line
last one
you can also make the value 3 configurable from an awk variable,
echo "$ip" | awk -v count=3 '{a[$0]; next}END{ if (NR>count) { for(i in a) print i }}'
The idea is to store the contents of the each line in {a[$0]; next} as each line is processed, by the time the END clause is reached, the NR variable will have the line count of the string/file you have. Print the lines if the condition matches i.e. number of lines greater than 3 or whatever configurable value using.
And always remember to double-quote the variables in bash to avoid undergoing word-splitting done by the shell.
Using James Brown's useful comment below to preserve the order of lines, do
echo "$ip" | awk -v count=3 '{a[NR]=$0; next}END{if(NR>3)for(i=1;i<=NR;i++)print a[i]}'
this is a sample line
another line
one more
last one
Another in awk. First test files:
$ cat 3
1
2
3
$ cat 4
1
2
3
4
Code:
$ awk 'NR<4{b=b (NR==1?"":ORS)$0;next} b{print b;b=""}1' 3 # look ma, no lines
[this line left intentionally blank. no wait!]
$ awk 'NR<4{b=b (NR==1?"":ORS)$0;next} b{print b;b=""}1' 4
1
2
3
4
Explained:
NR<4 { # for tghe first 3 records
b=b (NR==1?"":ORS) $0 # buffer them to b with ORS delimiter
next # proceed to next record
}
b { # if buffer has records, ie. NR>=4
print b # output buffer
b="" # and reset it
}1 # print all records after that
I need to delete the nth matching line in a file from the match up to the next blank line (i.e. one chunk of blank line delimited text starting with the nth match).
This will delete a chunk of text that starts and ends with a blank line starting with the fourth blank line. It also deletes those delimiting lines.
sed -n '/^$/!{p;b};H;x;/^\(\n[^\n]*\)\{4\}/{:a;n;/^$/!ba;d};x;p' inputfile
Change the first /^$/ to change the start match. Change the second one to change the end match.
Given this input:
aaa
---
bbb
---
ccc
---
ddd delete me
eee delete me
===
fff
---
ggg
This version of the command:
sed -n '/^---$/!{p;b};H;x;/^\(\n[^\n]*\)\{3\}/{:a;n;/^===$/!ba;d};x;p' inputfile
would give this as the result:
aaa
---
bbb
---
ccc
fff
---
ggg
Edit:
I removed an extraneous b instruction from the sed commands above.
Here's a commented version:
sed -n ' # don't print by default
/^---$/!{ # if the input line doesn't match the begin block marker
p; # print it
b}; # branch to end of script and start processing next input line
H; # line matches begin mark, append to hold space
x; # swap pattern space and hold space
/^\(\n[^\n]*\)\{3\}/{ # if what was in hold consists of 3 lines
# in other words, 3 copies of the begin marker
:a; # label a
n; # read the next line
/^===$/!ba; # if it's not the end of block marker, branch to :a
d}; # otherwise, delete it, d branches to the end automatically
x; # swap pattern space and hold space
p; # print the line (it's outside the block we're looking for)
' inputfile # end of script, name of input file
Any unambiguous pattern should work for the begin and end markers. They can be the same or different.
perl -00 -pe 'if (/pattern/) {++$count == $n and $_ = "$`\n";}' file
-00 is to read the file in "paragraph" mode (record separator is one or more blank lines)
$` is Perl's special variable for the "prematch" (text in front of the matching pattern)
In AWK
/m1/ {i++};
(i==3) {while (getline temp > 0 && temp != "" ){}; if (temp == "") {i++;next}};
{print}
Transforms this:
m1 1
first
m1 2
second
m1 3
third delete me!
m1 4
fourth
m1 5
last
into this:
m1 1
first
m1 2
second
m1 4
fourth
m1 5
last
deleting the third block of "m1" ...
Running on ideone here
HTH!
Obligatory awk script. Just change n=2 to whatever your nth match should be.
n=2; awk -v n=$n '/^HEADER$/{++i==n && ++flag} !flag; /^$/&&flag{flag=0}' ./file
Input
$ cat ./file
HEADER
line1a
line2a
line3a
HEADER
line1b
line2b
line3b
HEADER
line1c
line2c
line3c
HEADER
line1d
line2d
line3d
Output
$ n=2; awk -v n=$n '/^HEADER$/{++i==n&&++flag} !flag; /^$/&&flag{flag=0}' ./file
HEADER
line1a
line2a
line3a
HEADER
line1c
line2c
line3c
HEADER
line1d
line2d
line3d