How to escape bash parameter expansion (when a "!" followed by letters is a parameter)

Asked Jul 26 '18 at 19:41

Active Jul 26 '18 at 19:41

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I have a bash script I need to call with a string parameter, which string starts with a exclamation mark "!" (this is mandatory in what I'm doing):

myCommand.sh "!HELLO"

Because of parameter expansion, bash reports an error:

-bash: !HELLO: event not found

Is there any way to escape the "!" that produces this? Another way to pass such parameter?

asked Jul 26 '18 at 19:41

Bob

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Use single quotes: `myCommand.sh '!HELLO'` – anubhava Jul 26 '18 at 19:42
Especially [this fantastic answer](https://stackoverflow.com/a/25021905/3266847). – Benjamin W. Jul 26 '18 at 19:51

How to escape bash parameter expansion (when a "!" followed by letters is a parameter)

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