How to print #!/bin/bash using echo command

Question

I have tried the escape character but it doesn't seems works with !. I need this to auto-configure packages using scripts.

Solution 1

! is magic in a default interactive bash session, but not in scripts (set +H to disable in an interactive prompt). In any case:

echo '#!/bin/bash'

Solution 2

Or with no quotes and escaping the # and the !:

echo \#\!/bin/bash

Solution 3

printf should work too:

printf '#!/bin/bash\n'

Comments

[fedorqui]

That is, using single quotes you avoid variable expansion.

[Adam H. Peterson]

printf is beside the point. The issue is the shell's interpreting of the hashmark, not which command it's passed to. (Also, if you're going to use printf, you'll want to include a \n at the end, or your shebang line will run into the next line printed.

[Adam H. Peterson]

(Correction: I meant bang, rather than hashmark.)

[Avi Das]

Whoops you are right, it does need the \n

[ElmoVanKielmo]

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post.

[Reinstate Monica Please]

@ElmoVanKielmo It does provide an answer to the question, though the issue OP was having wasn't really related to echo/printf so it's basically the same as that other guy's

[ElmoVanKielmo]

@BroSlow this was a review and as you probably know - messages are predefined in that case.