使用sed从一行中删除“*”？

Question

I want to obtain the first field of each line, but it happens one of them will be a '*' and I want it to be skipped.

git branch -vv

returns

  master 34a8e20 [origin/master: behind 14] renamed yml's
* ss_doc 3ebc755 [origin/ss_doc: gone] PRD configuration
  ss_fix d0f4a4c [origin/ss_fix: gone] Merge branch 'ss_fix' into 'master'
  ss_v   c3b4635 [origin/ss_v: gone] remove composes

When I apply the following sed command, the result is the following

git branch -vv |  sed -r  's|\*?(\w+).+|\1|'

the result is

  master
* ss_doc
  ss_fix
  ss_v

I cannot understand why it catches the "*" within the matching group. I've tried other workarounds but this is the closest to the goal. How not to catch the "*"?

Answer 1

Note that sed replacement command replaces only what is matched. What is not matched is not replaced.

Your pattern tries to match a * , but if it does not find * at the current position, it tries to match (\\w+).+ pattern (since \\*? matches one or zero asterisks). Since it is not matched, the sed replacement command keeps the unmatched * in the result.

You need to make sure the asterisk is matched. As there is whitespace between the * and word chars, you may match it with \\s* or [[:space:]]* :

sed -r 's|\*?\s*(\w+).+|\1|'

Another way is to match any whitespace and * before the word chars:

sed -r 's|[*[:space:]]*([[:alnum:]_]+).*|\1|'
          ^^^^^^^^^^^^^

Or, use a PCRE pattern with grep to only match what you need:

grep -oP '^\W*\K\w+'

Or, remove any non-word chars at the start and awk out the first field:

sed 's/^[^[:alnum:]_]*//' | awk '{print $1}'

See the online demo .

Answer 2

Just tell awk to print the first field after the leading blanks and asterisk:

$ awk -F'[ *]+' '{print $2}' file
master
ss_doc
ss_fix
ss_v

That will work using any awk in any shell on any UNIX system. If you prefer sed, this will work with any sed:

$ sed 's/^[ *]*\([^ ]*\).*/\1/' file
master
ss_doc
ss_fix
ss_v