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multiple if condition in bash

Hi,

how can I write this multiple condition in a bash script?

Code:

if ((a > 30) || ((a < 0) && (b > 30))) then ...

Thanks for your help.

tikit

You have to nest a double parenthesis construct like this:

Code:

if (( a > 30 || (( a < 0 && b > 30 )) )) ; then

aargh...colucix got to it before I did.

Quote:

Originally Posted by w3bd3vil (Post 3338150)

aargh...colucix got to it before I did.

he he he... ;)

You appear to be trying to do numeric comparisons, in which case you're using the wrong operators. See http://www.tldp.org/LDP/abs/html/comparison-ops.html
:)

Moin,

Quote:

Originally Posted by chrism01 (Post 3338834)

You appear to be trying to do numeric comparisons, in which case you're using the wrong operators. See http://www.tldp.org/LDP/abs/html/comparison-ops.html
:)

not in the bash. See "man bash":

Code:

      ((expression))

              The  expression is evaluated according to the rules described below under ARITH-

              METIC EVALUATION.  If the value of the expression is non-zero, the return status

              is  0;  otherwise  the  return  status  is 1.  This is exactly equivalent to let

              "expression".

...

ARITHMETIC EVALUATION

      The  shell  allows  arithmetic expressions to be evaluated, under certain circumstances

      (see the let builtin command and Arithmetic Expansion).  Evaluation is done  in  fixed-

      width  integers with no check for overflow, though division by 0 is trapped and flagged

      as an error.  The operators and their precedence and associativity are the same  as  in

      the C language.

...

      <= >= < >

              comparison

...

      &&    logical AND

      ||    logical OR

Jan