How to delete top and last non empty lines of the file

1
votes

I want to delete top and last non empty line of the file.

Example:

cat test.txt

//blank_line
abc
def
xyz
//blank_line
qwe
mnp
//blank_line

Then output should be: 

def
xyz
//blank_line
qwe

I have tried with commands

sed "$(awk '/./{line=NR} END{print line}' test.txt)d" test.txt

to remove last non empty line. At here there are two command, (1) sed and (2) awk. But I want to do by single command.

5
awksedgrep
But //blank_lines in the middle of the text should stay, right? - James Brown
Please post your efforts in your post, which you have put to solve this problem. - RavinderSingh13
Too bad the question was badly formed, I was looking forward to seeing solutions to this one. OP, fix that question ASAP and let's get this reopened. - James Brown
Hi James, Blank line in the middle of the test should be stay. Just remove first and last non empty line. - Vishal Patel
Hi Ravinder, I have tried with commands - sed "$(awk '/./{line=NR} END{print line}' test.txt)d" test.txt to remove last non empty line. At here there are two command, (1) sed and (2) awk. But I want to do by single command. Thanks - Vishal Patel

5 Answers

3
votes

Reading the whole file in memory at once with GNU sed for -E and -z:

$ sed -Ez 's/^\s*\S+\n//; s/\n\s*\S+\s*$/\n/' test.txt
def
xyz

qwe

or with GNU awk for multi-char RS:

$ awk -v RS='^$' '{gsub(/^\s*\S+\n|\n\S+\s*$/,"")} 1' test.txt
def
xyz

qwe

Both GNU tools accept \s and \S as shorthand for [[:space:]] and [^[:space:]] respectively and GNU sed accepts the non-POSIX-sed-standard \n as meaning newline.

3
votes

This is a double pass method:

 awk '(NR==FNR) { if(NF) {t=FNR;if(!h) h=FNR}; next}
      (h<FNR && FNR<t)' file file

The integers h and t keep track of the head and the tail. In this case, empty lines can also contain blanks. You could replace if(NF) by if(length($0)==0) to be more strict.

This one reads everything into memory and does a simple replace at the end:

$ awk '{b=b RS $0}
       END{ sub(/^[[:blank:]\n]*[^\n]+\n/,"",b);
            sub(/\n[^\n]+[[:blank:]\n]*$,"",b);
            print b }' file
2
votes

A single-pass, fast and relatively memory-efficient approach utilising a buffer:

awk 'f {
    if(NF) {
        printf "%s",buf
        buf=""
    }
    buf=(buf $0 ORS)
    next
}
NF {
    f=1
}' file
0
votes

here is a golfed version of @kvantour's solution

$ awk 'NR==(n=FNR){e=!NF?e:n;b=!b?e:b}b<n&&n<e' file{,}
0
votes

This might work for you (GNU sed):

sed -E '0,/\S/d;H;$!d;x;s/.(.*)\n.*\S.*/\1/' file

Use a range to delete upto and including the first line containing a non-space character. Then copy the remains of the file into the hold space and at the end of file use substitution to remove the last line containing a non-space character and any empty lines to the end of the file.

Alternative:

sed '0,/\S/d' file | tac | sed '0,/\S/d'| tac