sed: delete all lines before matching one, including this one

Question

TARGET

How to delete all lines in a text file before a matching one, including this one?

Input file example:

apple
pear
banana
HI_THERE
lemon
coconut
orange

Desired output:

lemon
coconut
orange

The aim is to do it in sed to use the "-i" option (direct editing).

CLEAN SOLUTION ?

Most answers for similar problems propose something like:

sed -n '/HI_THERE/,$p' input_file

But the matched line is not deleted:

HI_THERE
lemon
coconut
orange

Then, knowing this will delete all from matched line (including it) to end of file:

sed '/HI_THERE/,$d' input_file

I tried something like this:

sed '^,/HI_THERE/d' input_file

But then sed complains:

sed: -e expression #1, char 1: unknown command: `^'

DIRTY SOLUTION

The last (dirty) solution is using pipeline:

sed -n '/HI_THERE/,$p' input_file | tail -n +2

but then, direct edit of the file doesn't work:

sed -n '/HI_THERE/,$p' input_file | tail -n +2 > input_file
cat input_file # returns nothing

and one must use a temporary file like that...

sed -n '/HI_THERE/,$p' input_file | tail -n +2 > tmp_file
mv tmp_file input_file

Answer 1

Similar to your "clean solution":

sed -e '1,/HI_THERE/d' input_file

The first line in the file is line 1 - there's no special ^ address because you always know that, while $ is needed for the end because you don't (necessarily) know which line that is.

This does fall over if the matching line is the first line of the file. With GNU sed you can use 0 instead of 1 to deal with that. For POSIX sed and for portability (which seem to be different in this case) it's more complex (see comments below and this follow-up question).

Answer 2

If you want to stick to Posix sed you could use this:

sed -ne '
  /HI_THERE/!d
  :loop
    n
    p
  bloop
' inp.file

Or, written in a condensed manner:

sed -n '/HI_THERE/!d;:a;n;p;ba' inp.file

Answer 3

$ perl -ne 'print if 1 <(/HI_THERE/...eof)' input_file

Where we make use of the range operator ... to form the proper range and further constrain it to reject the first element within the selected range.

Answer 4

sed

There is no simple way in sed to do what you ask for.
The simplest portable POSIXly solution for sed is something like:

sed -ne '/HI/{:1' -e 'n;p;b1' -e '}'

Other simple solutions are:

sed '0,/HI/d'      ./file             # GNU sed
awk 'f; /HI/{f=1}' ./file

---

ed

The closest POSIXly non-sed solution is with ed. Just remove the range from the first line of the file to the regex /HI/, even if HI is on the same first line.

printf '%s\n' 1,/HI/d ,p Q | ed -s file

Or

ed -Gs imfile2 <<-\edscript
1,/HI/d
,p
Q
edscript

Which means:

• printing no additional information (number of lines read) with option -s
• remove (delete) all lines from the first line of the file to the regex /HI/ (1,/HI/d).
• then print the whole file (,p).
• and exit even if the file has been modified (Q).

If you want to modify the file replace ,p Q with just w (write to file).

Why sed fails on 1,/HI/d while ed works ?

Because sed expects the regex to match on the next line.

In ed, either 3,3 3,/3/ /3/,3 /3/,/3/ will print only one 3:

$ printf '%s\n' 3,3p   3,/3/p   /3/3p   /3/,/3/p   Q | ed -s <(seq 5)
3
3
3
3

While sed will do this:

$ sed -ne 3,3p  <(seq 5)
3

$ sed -ne 3,/3/p  <(seq 5)
3
4
5

$ sed -ne /3/,3p  <(seq 5)
3

$ sed -ne /3/,/3/p <(seq 5)
3
4
5

A regex at the end of a range is expected to match a line following the line matched on the start of the range (the left of the ,). There is no 3 following line numbered 3, so sed prints all following lines (4 and 5).

That is why GNU sed solves the issue with 0,/HI/.

Why :1;n;p;b1 ?

A way to print all lines (except the first) of a range is to use a loop that first asks for the next line and then prints it:

$     sed -n '5{:1;n;p;b1}' <(seq 8)    # GNU syntax
6
7
8

So, simply match the regex you need /HI/ and get inside such loop.

sed -n '/HI/{:1;n;p;b1}' file        # GNU syntax

Which needs to be expanded to a more complex script because some older seds don't allow labels to be finished by the ;.

sed -n -e '/HI/{:1' -e 'n;p;b1' -e '}' file        # portable syntax

Answer 5

grep -nw HI_THERE file.txt |awk -F":" '{print $1}' | xargs -I % sed '1,%d' file.txt

Explanation: I grep using exact word w, get the line number n
Next, I pull the line number using awk, the separator is :
Further I pipe this using xargs and delete until then.

Answer 6

sed is for doing s/old/new on individual strings that is all. For anything else you should use awk, e.g. with any awk in any shell on every UNIX box and handling "HI_THERE" appearing on any line:

$ awk 'f; /HI_THERE/{f=1}' file
lemon
coconut
orange

If you want GNU sed for -i then use GNU awk for -i inplace instead. See https://stackoverflow.com/a/17914105/1745001 for other scripts to select sections of files.