How to remove only the first occurrence of a line in a file using sed

Question

I have the following file

titi
tata
toto
tata

If I execute

sed -i "/tat/d" file.txt

It will remove all the lines containing tat. The command returns:

titi
toto

but I want to remove only the first line that occurs in the file containing tat:

titi
toto
tata

How can I do that?

Answer 1

You could make use of two-address form:

sed '0,/tat/{/tat/d;}' inputfile

This would delete the first occurrence of the pattern.

Quoting from info sed:

 A line number of `0' can be used in an address specification like
 `0,/REGEXP/' so that `sed' will try to match REGEXP in the first
 input line too.  In other words, `0,/REGEXP/' is similar to
 `1,/REGEXP/', except that if ADDR2 matches the very first line of
 input the `0,/REGEXP/' form will consider it to end the range,
 whereas the `1,/REGEXP/' form will match the beginning of its
 range and hence make the range span up to the _second_ occurrence
 of the regular expression.

Answer 2

If you can use awk, then this makes it:

$ awk '/tata/ && !f{f=1; next} 1' file
titi
toto
tata

To save your result in the current file, do

awk '...' file > tmp_file && mv tmp_file file

Explanation

Let's activate a flag whenever tata is matched for the first time and skip the line. From that moment, keep not-skipping these lines.

• /tata/ matches lines that contain the string tata.
• {f=1; next} sets flag f as 1 and then skips the line.
• !f{} if the flag f is set, skip this block.
• 1, as a True value, performs the default awk action: {print $0}.

Another approach, by Tom Fenech

awk '!/tata/ || f++' file

|| stands for OR, so this condition is true, and hence prints the line, whenever any of these happens:

• tata is not found in the line.
• f++ is true. This is the tricky part: first time f is 0 as default, so first f++ will return False and not print the line. From that moment, it will increment from an integer value and will be True.