I have a text file, named my_data.txt, with the following contents:

#  define var1 and var2
var1=101
var2=202
// display var1 and var2
echo ${var1}
echo ${var2}

I want search all occurrences of var1 but not those in a line starts with '#' or '//'. I can do this:

grep var1 my_data.txt | grep -v '^#' | grep -v '^//'

output:

var1=101
echo ${var1}

The results is correct. The question: is there any way to pass both values '^#' and '^//' to a single -v option?

CodePudding user response：

I suggest:

grep -v -e '^#' -e '^//' file | grep 'var1'

Or with an extended regular expression (-E):

grep -v -E '^(#|//)' file | grep 'var1'

Output:

var1=101
echo ${var1}

CodePudding user response：

If you can make use of grep -P you can use a Perl-compatible regular expression and a single grep command.

grep -P "^(?!#|//).*\bvar1\b" my_data.txt

The pattern matches

• ^ Start of string
• (?!#|//) Negative lookahead, assert not # or //
• .*\bvar1\b Match the word var1 in the line

Or using awk skipping the line that starts with # or // and print a line that contains var1

awk '/^(#|\/\/)/{next};index($0, "var1")' my_data.txt

The examples will output:

var1=101
echo ${var1}