sed find and replace from left most character

Question

sed appears to find and replace from right to left.

for example:

echo "a_b_c_d" | sed 's/.*$_.*$/\1/'

outputs

_d

but why doesn't

echo "a_b_c_d" | sed 's/^.*$_.*$/\1/'

or

echo "a_b_c_d" | sed 's/.*$_.*$$/\1/'

output

_b_c_d

since these do not output _b_c_d how should this be done?

How should sed be used to find on first character and not last character when performing a find and replace?

`.*` is a greedy one, rather try something like: `echo "a_b_c_d" | sed 's/[^_]*$_.*$/\1/'` to get `_b_c_d`. Which has everything before first `_` before matched and rest of first `_` saved into back reference to be used later. — RavinderSingh13, Dec 16 '20 at 18:42
Does this answer your question? [Non greedy (reluctant) regex matching in sed?](https://stackoverflow.com/questions/1103149/non-greedy-reluctant-regex-matching-in-sed) — Mark Reed, Dec 16 '20 at 18:42

score 1 · Answer 1 · answered Dec 16 '20 at 18:42

.* is greedy pattern that matches longest possible substring before matching following pattern, _ in this case. So placing .* before _ makes it match longest possible match before matching last _ in your input.

since these do not output _b_c_d how should this be done?

echo "a_b_c_d" | sed 's/^[^_]*\(_.*$\)/\1/'

_b_c_d

Here [^_]* is negated bracket expression (called character class in modern regex flavors) that matches 0 or more of any character that is not _.

sed find and replace from left most character

1 Answers1