I'm trying to understand this sed regex.
sed 's/.*\(ADDR=[^|]*\) |.*/\1/'
If I'm not wrong, the above will search for the pattern ADDR=<something> anywhere in a line and
replace it with the first group. I don't get the meaning of [^|] here. Thanks for any help.
\(ADDR=[^|]*\) |.*/\1/
Here
[^|] matches anything other than | and the quantifier * quantifies zero or more occurrences of it. ^ in character class negates the character class.
| matches the character |
NOTE In sed metacharacters like | ( ) etc will lose its meaning so | is not an alternation but matches a | character. If you want to treat the metacharacters as such, then -r (extended regular expression) will do so (with GNU sed; use -E with BSD sed). Or escape \|.
Example:
$ echo "hello ADDR= hello | world " | sed 's/.*\(ADDR=[^|]*\) |.*/\1/'
ADDR= hello
Here (ADDR=[^|]*\) matches from ADDR= hello which contains anything other than |.
[^...] Matches any single character that is not in the class.
| The vertical bar separates two or more alternatives. A match occurs if any of the alternatives is satisfied. For example, gray|grey matches both gray and grey.
[^|] matches anything other than a |. ^ in character class negates the character class while | is loose it's actual meaning when using with sed.
