How can I compare a number against a range in bash

问题:

How to script a comparison of a number against a range?

1 is not within 2-5

or

3 is within 2-5

回答1:

It's even better in Perl6.

Chained comparison operators:

if( 2 <= $x <= 5 ){
}

Smart-match operator:

if( $x ~~ 2..5 ){
}

Junctions:

if( $x ~~ any 2..5 ){
}

Given / When operators:

given( $x ){
  when 2..5 {
  }
  when 6..10 {
  }
  default{
  }
}

回答2:

In Perl:

if( $x >= lower_limit && $x <= upper_limit ) {
   # $x is in the range
}
else {
   # $x is not in the range
}

回答3:

In bash:

$ if [[ 1 -gt 2 && 1 -lt 5 ]]; then echo "true"; fi
$ if [[ 3 -gt 2 && 1 -lt 5 ]]; then echo "true"; fi
true

回答4:

The smart match operator is available in Perl 5.10, too:

if ( $x ~~ [2..5] ) {
    # do something
}

回答5:

In Bash:

x=9; p="\<$x\>"; if [[ $(echo {10..20}) =~ $p ]]; then echo true; else echo false; fi

Edited to correctly handle conditions as noted in the comment below.

rangecheck () { local p="\<$1\>"; if [[ $(echo {10..20}) =~ $p ]]; then echo true; else echo false; fi; }
for x in {9..21}; do rangecheck "$x"; done
false
true
.
.
.
true
false

回答6:

The [[ version of test has supported regular expressions since Bash 3.0.

[[ 3 =~ ^[2-5]$ ]]; echo $? # 0

The numeric comparison operators in test sometimes return an error if the input isn't numeric:

[[ 1a -ge 1 ]]; echo $? # value too great for base (error token is "1a")
[[ '$0' -le 24 ]] # syntax error: operand expected (error token is "$o")

You can test if the input is an integer with =~:

x=a23; [[ "$x" =~ ^[0-9]+$ && "$x" -ge 1 && "$x" -le 24 ]]; echo $? # 1
x=-23; [[ "$x" =~ ^-?[0-9]+$ && "$x" -ge -100 && "$x" -le -20 ]]; echo $? # 0

回答7:

In perl

grep {/^$number$/} (1..25);

will give you a true value if the number is in the range and a false value otherwise.

For example:

[dsm@localhost:~]$ perl -le 'print "has `$ARGV[0]`" if grep {/^$ARGV[0]$/} (1..25)' 4
has `4`
[dsm@localhost:~]$ perl -le 'print "has `$ARGV[0]`" if grep {/^$ARGV[0]$/} (1..25)' 456
[dsm@localhost:~]$