For example, given:
$ cat test001.txt this is a test bucket bucket line not this one. bucket lein endy bucket and others endy and a ttest How can I change the lines that start with "bucket" with "(bucket", ignoring spaces?
So far I have:
sed -i 's/^\sbucket/bucket(/g' test001.txt but that doesn't work (no change and no error). Then I would like to only do it for lines that also end with 'endy' (so only 2 of the 3 bucket lines).
You can use back references (e.g., \1) in your replacement-term which will replace subexpressions (things enclosed within escaped parentheses — \(…\)) in your search term. E.g.,
$ cat test001.txt | sed -e 's/^\([[:space:]]*\)\(bucket.*endy\)$/\1(\2/' this is a test bucket bucket line not this one. (bucket lein endy (bucket and others endy and a ttest will create two subexpressions:
^\([[:space:]]*\)\(bucket.*endy\)$It will then replace any line that matches with \1(\2, which is the whitespace, followed by an opening parenthesis, followed by the bucket..endy string.
Here's another approach:
sed -e '/^[[:space:]]*bucket.*endy$/s/bucket/(bucket/' file.txt The initial /pattern/ tells sed to only work on lines containing that pattern, and the rest of it is a simple, standard substitution.
The \s is not part of BRE, which sed uses. In BRE \s is the same as a literal s. The equivalent of \s would be [[:space:]]. Try the following:
sed -i 's/^[[:space:]]*bucket/(bucket/g' test001.txt 
sed -i 's/^[[:space:]]*bucket.*endy/(bucket/g' test001.txt..for any character and*for zero or more (greedy).