I'm looking for a way to only execute replacement when the last character is a newline, using sed.

For instance:


is replaced, but this is not:


Since sed does not work well with newlines, it is not as simple as

$ sed -E "s/[a-zA-Z]*\n/replace/" file.txt

How can this be accomplished?


4 Answers 4


With standard sed, you will never see a newline in the text read from a file. This is because sed reads line by line, and there is therefore no newline at the end of the text of the current line in sed's pattern space. In other words, sed reads newline-delimited data, and the delimiters are not part of what a sed script sees.

Regular expressions can be anchored at the end of the line using $ (or at the beginning, using ^). Anchoring an expression at the start/end of a line forces it to match exactly there, and not just anywhere on the line.

If you want to replace anything matching the pattern [A-Za-z]* at the end of the line with something, then anchor the pattern like this:


...will force it to match at the end of the line and nowhere else.

However, since [A-Za-z]*$ also matches nothing (for example, the empty string present at the end of every line), you need to force the matching of something, e.g. by specifying




So, your sed command line will thus be

$ sed 's/[A-Za-z]\{1,\}$/replace/' file.txt

I did not use the non-standard -E option here because it's not strictly needed. With it, you could have written

$ sed -E 's/[A-Za-z]+$/replace/' file.txt

It's a matter of taste.

sed "s/[a-zA-Z]*$/replace/" input.txt > result.txt

Or, the long complex unnecessary way:

I've found out, this can be done, still using sed, with the help of tr. You can assign another character to represent the end of the line. Another temporary character has to be used, in this case "`". Let's use "~" to represent the end of the line:

tr '\n' '`' <input.txt >output.txt
sed -i "s/`/~`/" output.txt
tr '`' '\n' <output.txt >result.txt

And then to perform the actual search and replace, use "~" rather than "\n":

sed -i -E "s/[a-zA-Z]*~/replace/" result.txt

And then clean up the extra character on the other lines:

sed -i "s/~//" result.txt

Obviously, this can all be piped together resulting in something like:

tr '\n' '`' <input.txt | sed -e "s/`/~`/" | tr '`' '\n' | sed -E -e "s/[a-zA-Z]*~/replace/" | sed "s/~//" > result.txt
  • 4
    Not sure I understand... Why don't you just anchor to end of line with $ ? e.g s/[a-zA-Z]*$/replace/ Jun 1, 2015 at 23:53
  • 1
    2 points: 1) You'd better use \+ instead of * since the latter allows zero letters at end of string; 2) You can use a character class [[:alpha:]]. So: sed 's/[[:alpha:]]\+$/replace/' file Jun 2, 2015 at 0:25
  • @glennjackman What's the backslash for before the plus? Wouldn't that match the addition character? Jun 2, 2015 at 0:39
  • 1
    GNU sed without the -r option uses this regular expression syntax. Jun 2, 2015 at 1:28
  • I had to use -i '', otherwise it was interpreting the regex as the name of a backup file. Nov 20, 2020 at 21:18

From the (broken) code snippet you posted, you seem to want to replace the newline as well. In that case, regex anchoring by itself can't help you. The following is a solution:

sed '/[[:alpha:]]\+$/{N;s/[[:alpha:]]\+\n/replace/}' your_file

Broken down:

  • /[a-zA-Z]\+$/{} means apply whatever comes inside the curlies to lines that match the regex.
  • The regex is the one that uses anchoring as seen in your own answer, modified to take glenn jackman's comments into account.
  • Inside the curlies, N means "append the next line to the active buffer" (what sed calls the 'pattern space')
  • Finally the s/// statement is your required substitution. It now works because the pattern space contains two successive lines and the newline is therefore a part of it.

To find the end of line, just use the $-sign:

Without end of line anchor:

sed -n '/pattern/p' file 

With end of line anchor:

sed -n '/pattern$/p' file

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