Delete Last Line Break Using Sed

How to use sed to remove the last n lines of a file

I don't know about sed, but it can be done with head:

head -n -2 myfile.txt

Removing the new line in the last line of a text file using sed

This might work for you (GNU sed):

sed -z 's/\n\+$//' file

This will remove a newline(s) at the end of a file provided there are no null characters.

N.B. Normal use of sed i.e. without the -z option which slurps the file into memory, will remove any newlines before the sed commands can act on them.

Remove the last line from a file in Bash

Using GNU sed:

sed -i '$ d' foo.txt

The -i option does not exist in GNU sed versions older than 3.95, so you have to use it as a filter with a temporary file:

cp foo.txt foo.txt.tmp
sed '$ d' foo.txt.tmp > foo.txt
rm -f foo.txt.tmp

Of course, in that case you could also use head -n -1 instead of sed.

MacOS:

On Mac OS X (as of 10.7.4), the equivalent of the sed -i command above is

sed -i '' -e '$ d' foo.txt

How can I replace each newline (\n) with a space using sed?

Use this solution with GNU sed:

sed ':a;N;$!ba;s/\n/ /g' file

This will read the whole file in a loop (':a;N;$!ba), then replaces the newline(s) with a space (s/\n/ /g). Additional substitutions can be simply appended if needed.

Explanation:

0. sed starts by reading the first line excluding the newline into the pattern space.
1. Create a label via :a.
2. Append a newline and next line to the pattern space via N.
3. If we are before the last line, branch to the created label $!ba ($! means not to do it on the last line. This is necessary to avoid executing N again, which would terminate the script if there is no more input!).
4. Finally the substitution replaces every newline with a space on the pattern space (which is the whole file).

Here is cross-platform compatible syntax which works with BSD and OS X's sed (as per @Benjie comment):

sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/ /g' file

As you can see, using sed for this otherwise simple problem is problematic. For a simpler and adequate solution see this answer.

Delete second-to-last line of file using sed

This might work for you (GNU sed):

sed -i 'N;$!P;D' file

This sets up a window of two lines in the pattern space and then prints the first of those whenever the second is not the last. The last line is always printed because the next line can not be fetched.

Deleting the second-to-last line of a file based on a pattern can be achieved with:

sed -i 'N;$!P;${/pattern.*\n/!P};D'

Delete last character from line that contains specific word

This would do it:

sed '/in_bytes/ s/,$//'

Where /in_bytes/ is a search pattern ensuring only matching lines will execute the following command s/,$//, which is a standard substitution to remove a trailing comma.

Example: https://ideone.com/3hEXBY

sed - remove line break if line does not end on \

give this awk one-liner a try:

awk '{printf "%s%s",$0,(/"$/?"\n":"")}' file

test

kent$  cat f
"foo"
"bar"
"a long
text with
many many
lines"
"lalala"

kent$  awk '{printf "%s%s",$0,(/"$/?"\n":"")}' f
"foo"
"bar"
"a longtext withmany manylines"
"lalala"

Delete third-to-last line of file using sed or awk

With tac + awk solution, could you please try following. Just set line variable of awk to line(from bottom) whichever you want to skip.

tac Input_file | awk -v line="3" 'line==FNR{next} 1' | tac

Explanation: Using tac will read the Input_file reverse(from bottom line to first line), passing its output to awk command and then checking condition if line is equal to line(which we want to skip) then don't print that line, 1 will print other lines.

2nd solution: With awk + wc solution, kindly try following.

awk -v lines="$(wc -l < Input_file)" -v skipLine="3" 'FNR!=(lines-skipLine+1)' Input_file

Explanation: Starting awk program here and creating a variable lines which has total number of lines present in Input_file in it. variable skipLine has that line number which we want to skip from bottom of Input_file. Then in main program checking condition if current line is NOT equal to lines-skipLine+1 then printing the lines.

3rd solution: Adding solution as per Ed sir's comment here.

awk -v line=3 '{a[NR]=$0} END{for (i=1;i<=NR;i++) if (i != (NR-line)) print a[i]}' Input_file

Explanation: Adding detailed explanation for 3rd solution.

awk -v line=3 '             ##Starting awk program from here, setting awk variable line to 3(line which OP wants to skip from bottom)
{
  a[NR]=$0                  ##Creating array a with index of NR and value is current line.
}
END{                        ##Starting END block of this program from here.
  for(i=1;i<=NR;i++){       ##Starting for loop till value of NR here.
    if(i != (NR-line)){     ##Checking condition if i is NOT equal to NR-line then do following.
      print a[i]            ##Printing a with index i here.
    }
  }
}
' Input_file                ##Mentioning Input_file name here.

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