在 bash 脚本中获取第一个未注释的行（即不以 # 开头）

Question

I am processing a developer commit message in git hook.

let's say the file has the following content

\n new lines here
# this is a sample commit
# only for the developers
Ticket-ID: we fix old bugs and introduces new ones
we always do this stuff

so cool, not really :P
# company name

My intention is to get only this line Ticket-ID: we fix old bugs and introduces new ones

Answer 1

User123's comment is nice and terse: grep -E "^[[:alnum:]]" file |head -n 1 however is does not catch lines of text that start with non-alphanumeric characters that are not a # such as commit messages that start with an emoji, dashes, parenthesis, etc..

• yeah this line is an exception
• --> This is also an edge case
• (so is this)

To catch all edge cases you can loop through the file and check each $line with a negated ! regexp operator =~ for:

1. Not being a newline ! $line =~ (^[^\n ]*$) ! $line =~ (^[^\n ]*$)
2. Not starting with a pound sign ! $line =~ ^# ! $line =~ ^#
3. Not being a line consisting of all spaces ! $line =~ (^[ ]*$) ! $line =~ (^[ ]*$)

Then just echo the $line and break if those conditions are met:

# file parse.sh
#!/bin/bash
if [[ -f $1 ]]; then
  while IFS= read -r line
  do
    [[ ! $line =~ (^[^\n ]*$) && ! $line =~ ^# && ! $line =~ (^[ ]*$) ]] && echo "$line" && break
  done < "$1"
fi

# file commit .txt


# this is a sample commit
# only for the developers
Ticket-ID: we fix old bugs and introduces new ones
we always do this stuff

so cool, not really :P
# company name

Now you can invoke the parse.sh like this

bash parse.sh commit.txt

Or save the results to a variable using a subshell

result=$(bash parse.sh commit.txt); echo "$result"

Answer 2

Below single line grep should work as per your requirement:

grep -E "^[[:alnum:]]" file |head -n 1

Explanation:

^[[:alnum:]] :: to capture only the line starting with any alphanumeric character[0-9A-Za-z]

head -n 1 ::  to capture the first occurrence