Bash get first N non white space characters

Question

I am trying to check if a certain line in a file is commented out (has // as the first non white space characters of the line). How can I do this in bash? I'm ok with using sed, awk, whatever as long as it comes standard on OS X.

I have tried:

grep textOnlyOnMyLine | cut -c 1-2 myFile.java

but this will pick up white spaces if white spaces are the first 2 characters of the line. Also this command seems to hang after outputting what it finds.

Answer 1

You can use grep:

grep '^[[:blank:]]*//' *.java

To search for a particular line # you can use sed:

sed -n '3s|^[[:blank:]]*//|&|p' file

Answer 2

With awk you can get the lines that are comments with:

awk '/\s*\/\//' file

So it looks for [spaces] plus // in the line and then prints them.

In case you want to check a specifid word you can do:

awk '/\s*\/\// && NR==line_number' file

Test

$ cat a
hello
// this is a coment
 // this is a coment

$ awk '/\s*\/\//' a
// this is a coment
 // this is a coment

$ awk '/\s*\/\// && NR==1' a
$

$ awk '/\s*\/\// && NR==2' a
// this is a coment