mjhjmtu

I want to check the condition like below. Line should not starts with #
symbol:

if[ ! ["$Line" == "#*"] ]; then

But this is not working.

There are many problems in your snippet, but basically the syntax [![...]] is not valid, in the bash (and many other shells) [[ is a single command, which cannot be split by any other character.

You can negate equality operator instead:

if [[ "$LINE" != #* ]]; then echo yes; fi

Take also closer look to spaces surrounding brackets.

Assuming that you have a string in the variable line and you want to output the string yes if it doesn't start with the character #. In the examples below, I will also output no when the condition is not met.

case $line in
 "#"*) echo no
esac

The case statement takes a variable and then any number of (globbing) patterns, and executes the code for the pattern that matches the value of the variable. In this case we try to match the pattern "#"* (the # has to be quoted as to not be taken as introducing a comment), and if it matches, an echo statement is executed.

If there are more patterns, then the code for all but the last pattern has to be terminated by ;;, as in

case $line in
 "#"*) echo no ;;
 *) echo yes
esac

... and the code for a pattern may be empty:

case $line in
 "#"*) ;;
 *) echo yes
esac

The bash shell also does pattern matching with globs in [[ ... ]] with the == operator:

if [[ $line != "#"* ]]; then
 echo yes
else
 echo no
fi

In bash, you may also use regular expression matching with the =~ operator within [[ ... ]]:

if ! [[ $line =~ ^# ]]; then
 echo yes
else
 echo no
fi

or

if [[ $line =~ ^[^#] ]]; then
 echo yes
else
 echo no
fi

If you're paring a file line by line, then the shell is not the right tool for that job. Instead you may want to use something like awk:

awk '/^[^#]/ print yes ' <inputfile

Related:

• Why is using a shell loop to process text considered bad practice?