using compound conditions in bash shell script

Question

I want to do something like below in Bash script. How do I implement it in Bash syntax?

if !((a==b) && (a==c))
then
    # Do something
end if

Answer 1

For numeric comparison, you can do:

if ! (( (a == b) && (a == c) ))

For string comparison:

if ! [[ "$a" == "$b" && "$a" == "$c" ]]

In Bash, the double parentheses set up an arithmetic context (in which dollar signs are mostly optional, by the way) for a comparison (also used in for ((i=0; i<=10; i++)) and $(()) arithmetic expansion) and is used to distinguish the sequence from a set of single parentheses which creates a subshell.

This, for example, executes the command true and, since it's always true it does the action:

if (true); then echo hi; fi 

This is the same as

if true; then echo hi; fi

except that a subshell is created. However, if ((true)) tests the value of a variable named "true".

If you were to include a dollar sign, then "$true" would unambiguously be a variable, but the if behavior with single parentheses (or without parentheses) would change.

if ($true)

or

if $true

would execute the contents of the variable as a command and execute the conditional action based on the command's exit value (or give a "command not found" message if the contents aren't a valid command).

if (($true)) 

does the same thing as if ((true)) as described above.