get the second last line from shell pipeline - shell

I want to get the second last line from the ls -l output.
I know that
ls -l|tail -n 2| head -n 1
can do this, just wondering if sed can do this in just one command?

ls -l|sed -n 'x;$p'
It can't do third to last though, because sed only has 1 hold space, so can only remember one older line. And since it processes the lines one at a time, it does not know the line will be next to last when processing it. awk could return thrid to last, because you can have arbitrary number of variables there, but the script would be much longer than the tail -n X|head -n 1.

In a awk one-liner :
echo -e "aaa\nbbb\nccc\nddd" | awk '{v[c++]=$0}END{print v[c-2]}'
ccc

Try this to delete second-last line in file
sed -e '$!{h;d;}' -e x filename

tac filename | sed -n 2p
-- but involves a pipe, too

Related

Delete the 3 last line of my txt with bash? [duplicate]

I want to remove some n lines from the end of a file. Can this be done using sed?
For example, to remove lines from 2 to 4, I can use
$ sed '2,4d' file
But I don't know the line numbers. I can delete the last line using
$sed $d file
but I want to know the way to remove n lines from the end. Please let me know how to do that using sed or some other method.
I don't know about sed, but it can be done with head:
head -n -2 myfile.txt
If hardcoding n is an option, you can use sequential calls to sed. For instance, to delete the last three lines, delete the last one line thrice:
sed '$d' file | sed '$d' | sed '$d'
From the sed one-liners:
# delete the last 10 lines of a file
sed -e :a -e '$d;N;2,10ba' -e 'P;D' # method 1
sed -n -e :a -e '1,10!{P;N;D;};N;ba' # method 2
Seems to be what you are looking for.
A funny & simple sed and tac solution :
n=4
tac file.txt | sed "1,$n{d}" | tac
NOTE
double quotes " are needed for the shell to evaluate the $n variable in sed command. In single quotes, no interpolate will be performed.
tac is a cat reversed, see man 1 tac
the {} in sed are there to separate $n & d (if not, the shell try to interpolate non existent $nd variable)
Use sed, but let the shell do the math, with the goal being to use the d command by giving a range (to remove the last 23 lines):
sed -i "$(($(wc -l < file)-22)),\$d" file
To remove the last 3 lines, from inside out:
$(wc -l < file)
Gives the number of lines of the file: say 2196
We want to remove the last 23 lines, so for left side or range:
$((2196-22))
Gives: 2174
Thus the original sed after shell interpretation is:
sed -i '2174,$d' file
With -i doing inplace edit, file is now 2173 lines!
If you want to save it into a new file, the code is:
sed -i '2174,$d' file > outputfile
You could use head for this.
Use
$ head --lines=-N file > new_file
where N is the number of lines you want to remove from the file.
The contents of the original file minus the last N lines are now in new_file
Just for completeness I would like to add my solution.
I ended up doing this with the standard ed:
ed -s sometextfile <<< $'-2,$d\nwq'
This deletes the last 2 lines using in-place editing (although it does use a temporary file in /tmp !!)
To truncate very large files truly in-place we have truncate command.
It doesn't know about lines, but tail + wc can convert lines to bytes:
file=bigone.log
lines=3
truncate -s -$(tail -$lines $file | wc -c) $file
There is an obvious race condition if the file is written at the same time.
In this case it may be better to use head - it counts bytes from the beginning of file (mind disk IO), so we will always truncate on line boundary (possibly more lines than expected if file is actively written):
truncate -s $(head -n -$lines $file | wc -c) $file
Handy one-liner if you fail login attempt putting password in place of username:
truncate -s $(head -n -5 /var/log/secure | wc -c) /var/log/secure
This might work for you (GNU sed):
sed ':a;$!N;1,4ba;P;$d;D' file
Most of the above answers seem to require GNU commands/extensions:
$ head -n -2 myfile.txt
-2: Badly formed number
For a slightly more portible solution:
perl -ne 'push(#fifo,$_);print shift(#fifo) if #fifo > 10;'
OR
perl -ne 'push(#buf,$_);END{print #buf[0 ... $#buf-10]}'
OR
awk '{buf[NR-1]=$0;}END{ for ( i=0; i < (NR-10); i++){ print buf[i];} }'
Where "10" is "n".
With the answers here you'd have already learnt that sed is not the best tool for this application.
However I do think there is a way to do this in using sed; the idea is to append N lines to hold space untill you are able read without hitting EOF. When EOF is hit, print the contents of hold space and quit.
sed -e '$!{N;N;N;N;N;N;H;}' -e x
The sed command above will omit last 5 lines.
It can be done in 3 steps:
a) Count the number of lines in the file you want to edit:
n=`cat myfile |wc -l`
b) Subtract from that number the number of lines to delete:
x=$((n-3))
c) Tell sed to delete from that line number ($x) to the end:
sed "$x,\$d" myfile
You can get the total count of lines with wc -l <file> and use
head -n <total lines - lines to remove> <file>
Try the following command:
n = line number
tail -r file_name | sed '1,nd' | tail -r
This will remove the last 3 lines from file:
for i in $(seq 1 3); do sed -i '$d' file; done;
I prefer this solution;
head -$(gcalctool -s $(cat file | wc -l)-N) file
where N is the number of lines to remove.
sed -n ':pre
1,4 {N;b pre
}
:cycle
$!{P;N;D;b cycle
}' YourFile
posix version
To delete last 4 lines:
$ nl -b a file | sort -k1,1nr | sed '1, 4 d' | sort -k1,1n | sed 's/^ *[0-9]*\t//'
I came up with this, where n is the number of lines you want to delete:
count=`wc -l file`
lines=`expr "$count" - n`
head -n "$lines" file > temp.txt
mv temp.txt file
rm -f temp.txt
It's a little roundabout, but I think it's easy to follow.
Count up the number of lines in the main file
Subtract the number of lines you want to remove from the count
Print out the number of lines you want to keep and store in a temp file
Replace the main file with the temp file
Remove the temp file
For deleting the last N lines of a file, you can use the same concept of
$ sed '2,4d' file
You can use a combo with tail command to reverse the file: if N is 5
$ tail -r file | sed '1,5d' file | tail -r > file
And this way runs also where head -n -5 file command doesn't run (like on a mac!).
#!/bin/sh
echo 'Enter the file name : '
read filename
echo 'Enter the number of lines from the end that needs to be deleted :'
read n
#Subtracting from the line number to get the nth line
m=`expr $n - 1`
# Calculate length of the file
len=`cat $filename|wc -l`
#Calculate the lines that must remain
lennew=`expr $len - $m`
sed "$lennew,$ d" $filename
A solution similar to https://stackoverflow.com/a/24298204/1221137 but with editing in place and not hardcoded number of lines:
n=4
seq $n | xargs -i sed -i -e '$d' my_file
In docker, this worked for me:
head --lines=-N file_path > file_path
Say you have several lines:
$ cat <<EOF > 20lines.txt
> 1
> 2
> 3
[snip]
> 18
> 19
> 20
> EOF
Then you can grab:
# leave last 15 out
$ head -n5 20lines.txt
1
2
3
4
5
# skip first 14
$ tail -n +15 20lines.txt
15
16
17
18
19
20
POSIX compliant solution using ex / vi, in the vein of #Michel's solution above.
#Michel's ed example uses "not-POSIX" Here-Strings.
Increment the $-1 to remove n lines to the EOF ($), or just feed the lines you want to (d)elete. You could use ex to count line numbers or do any other Unix stuff.
Given the file:
cat > sometextfile <<EOF
one
two
three
four
five
EOF
Executing:
ex -s sometextfile <<'EOF'
$-1,$d
%p
wq!
EOF
Returns:
one
two
three
This uses POSIX Here-Docs so it is really easy to modify - especially using set -o vi with a POSIX /bin/sh.
While on the subject, the "ex personality" of "vim" should be fine, but YMMV.
This will remove the last 12 lines
sed -n -e :a -e '1,10!{P;N;D;};N;ba'

Getting head to display all but the last line of a file: command substitution and standard I/O redirection

I have been trying to get the head utility to display all but the last line of standard input. The actual code that I needed is something along the lines of cat myfile.txt | head -n $(($(wc -l)-1)). But that didn't work. I'm doing this on Darwin/OS X which doesn't have the nice semantics of head -n -1 that would have gotten me similar output.
None of these variations work either.
cat myfile.txt | head -n $(wc -l | sed -E -e 's/\s//g')
echo "hello" | head -n $(wc -l | sed -E -e 's/\s//g')
I tested out more variations and in particular found this to work:
cat <<EOF | echo $(($(wc -l)-1))
>Hola
>Raul
>Como Esta
>Bueno?
>EOF
3
Here's something simpler that also works.
echo "hello world" | echo $(($(wc -w)+10))
This one understandably gives me an illegal line count error. But it at least tells me that the head program is not consuming the standard input before passing stuff on to the subshell/command substitution, a remote possibility, but one that I wanted to rule out anyway.
echo "hello" | head -n $(cat && echo 1)
What explains the behavior of head and wc and their interaction through subshells here? Thanks for your help.
head -n -1 will give you all except the last line of its input.
head is the wrong tool. If you want to see all but the last line, use:
sed \$d
The reason that
# Sample of incorrect code:
echo "hello" | head -n $(wc -l | sed -E -e 's/\s//g')
fails is that wc consumes all of the input and there is nothing left for head to see. wc inherits its stdin from the subshell in which it is running, which is reading from the output of the echo. Once it consumes the input, it returns and then head tries to read the data...but it is all gone. If you want to read the input twice, the data will have to be saved somewhere.
Using sed:
sed '$d' filename
will delete the last line of the file.
$ seq 1 10 | sed '$d'
1
2
3
4
5
6
7
8
9
For Mac OS X specifically, I found an answer from a comment to this Q&A.
Assuming you are using Homebrew, run brew install coreutils then use the ghead command:
cat myfile.txt | ghead -n -1
Or, equivalently:
ghead -n -1 myfile.txt
Lastly, see brew info coreutils if you'd like to use the commands without the g prefix (e.g., head instead of ghead).
cat myfile.txt | echo $(($(wc -l)-1))
This works. It's overly complicated: you could just write echo $(($(wc -l)-1)) <myfile.txt or echo $(($(wc -l <myfile.txt)-1)). The problem is the way you're using it.
cat myfile.txt | head -n $(wc -l | sed -E -e 's/\s//g')
wc consumes all the input as it's counting the lines. So there is no data left to read in the pipe by the time head is started.
If your input comes from a file, you can redirect both wc and head from that file.
head -n $(($(wc -l <myfile.txt) - 1)) <myfile.txt
If your data may come from a pipe, you need to duplicate it. The usual tool to duplicate a stream is tee, but that isn't enough here, because the two outputs from tee are produced at the same rate, whereas here wc needs to fully consume its output before head can start. So instead, you'll need to use a single tool that can detect the last line, which is a more efficient approach anyway.
Conveniently, sed offers a way of matching the last line. Either printing all lines but the last, or suppressing the last output line, will work:
sed -n '$! p'
sed '$ d'
Here is a one-liner that can get you the desired output, and it can be used more generally for getting all lines from a file except the last n lines.
grep -n "" myfile.txt \ # output the line number for each line
| sort -nr \ # reverse the file by using those line numbers
| sed '1,4d' \ # delete first 4 lines (last 4 of the original file)
| sort -n \ # reverse the reversed file (correct the line order)
| sed 's/^[0-9]*://' # remove the added line numbers
Here is the above command in an actual single line and runnable (can't execute the above due to the added comments):
grep -n "" myfile.txt | sort -nr | sed '1,4d' | sort -n | sed 's/^[0-9]*://'
It's a little cumbersome, and this problem can be solved with more comprehensive commands like ghead, but when you can't or don't want to download such tools, it's nice to be able to do this with the more basic options. I've been in situations where it's simply not an option to get better tools.
awk 'NR>1{print p}{p=$0}'
For this job, an awk one-liner is a bit longer than a sed one.

Printing a line of a file given line number

Is it possible, in UNIX, to print a particular line of a file? For example I would like to print line 10 of file example.c. I tried with cat, ls, awk but apparently either these don't have the feature or I'm not able to properly read the man :-).
Using awk:
awk 'NR==10' file
Using sed:
sed '10!d' file
sed -n '10{p;q;}' example.c
will print the tenth line of example.c for you.
Try head and tail, you can specify the amount of lines and where to start.
To get the third line:
head -n 3 yourfile.c | tail -n 1
head -n 10 /tmp/asdf | tail -n 1
Unfortunately, all other solutions which use head/tail will NOT work incorrectly if line number provided is larger than total number of lines in our file.
This will print line number N or nothing if N is beyond total number of lines:
grep "" file | grep "^20:"
If you want to cut line number from output, pipe it through sed:
grep "" file | grep "^20:" | sed 's/^20://'
Try this:
cat -n <yourfile> | grep ^[[:space:]]*<NUMBER>[[:space:]].*$
cat -n numbers the file
the regex of grep searches the line numbered ;-)
The original mismatched as mentioned in the comments.
Te current one looks for the exact match.
- i.e. in the particular cas we need a line starting with an arbitrary amount () of spaces the followed by a space followed by whatever (.)
In case anyone thumbles over this regex and doesn't get it at all - here is a good tutorial to get you started: http://regex.learncodethehardway.org/book/ (it uses python regex as an example tough).
This might work for you:
sed '10q;d' file

remove n lines from STDOUT on bash

Do you have any bash solution to remove N lines from stdout?
like a 'head' command, print all lines, only except last N
Simple solition on bash:
find ./test_dir/ | sed '$d' | sed '$d' | sed '$d' | ...
but i need to copy sed command N times
Any better solution?
except awk, python etc...
Use head with a negative number. In my example it will print all lines but last 3:
head -n -3 infile
if head -n -3 filename doesn't work on your system (like mine), you could also try the following approach (and maybe alias it or create a function in your .bashrc)
head -`echo "$(wc -l filename)" | awk '{ print $1 - 3; }'` filename
Where filename and 3 above are your file and number of lines respectively.
The tail command can skip from the end of a file on Mac OS / BSD. tail accepts +/- prefix, which facilitates expression below, which will show 3 lines from the start
tail -n +3 filename.ext
Or, to skip lines from the end of file, use - prefixed, instead.
tail -n -3 filenme.ext
Typically, the default for tail is the - prefix, thus counting from the end of the file. See a similar answer to a different question here: Print a file skipping first X lines in Bash

Output specific line huge text file

I have a sql dump with 300mb that gives me an error on specific line.
But that line is in the middle of the file. What is the best approach?
head -n middleLine dump.sql > output?
Or can i output only the line i need?
You could use sed -n -e 123456p your.dump to print line 123456
If the file is long, consider using
sed -n 'X{p;q}' file
Where X is the line number. It will stop reading the file after reaching that line.
If sed is too slow for your taste you may also use
cat $THE_FILE | head -n $DESIRED_LINE | tail -n 1
You can use sed:
sed -n "x p" dump.sql
where x is the line number.
This might work for you:
sed 'X!d;q' file
where X is the line number.
This can also be done with Perl:
perl -wnl -e '$. == 4444444 and print and exit;' FILENAME.sql
4444444 being the line number you wish to print.
You can also try awk like:
awk 'NR==YOUR_LINE_NO{print}' file_name
If you know a phrase on that line I would use grep. If the phrase is "errortext" use:
$ cat dump.sql | grep "errortext"

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