Negate while loop condition in bash

Question

In part of the script that I'm working on, I want to validate that the inputted IP address is in the correct format. I want to make a loop while the input format is NOT correct. The following works for a loop while the format IS correct.

while  [[  $range =~ ^[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ]]
      do

How can I turn this around to say DOES NOT fit the format. I was hoping !=~ would work, but I'm getting a syntax error.

Answer 1

The solution is so simple, I'm sure you will all find it funny.

No need to get worked up about negation syntax details, just use until instead of while.

Answer 2

For comparison of string, one should use != instead of !=~.

From man bash

string1 != string2
          True if the strings are not equal.

As you are using a regular expression on the right, you indeed need =~. You can chose to negate the whole expression.

while  [[  !( $range =~ ^[0-9]{1,3]}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ) ]]

Example:

$ bash --version | head -n1
GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu)
$ cat script
range='1'; while [[ !( $range =~ ^[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ) ]]; do echo "$range NOT OK"; range+='.2'; sleep 1; done
echo "Finished"
$ bash script
1 NOT OK
1.2 NOT OK
1.2.2 NOT OK
Finished
$ 

Answer 3

Just put "!" before statement, like:

while ! [[  $range =~ ^[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ]]
do
   run-command
done