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Top Forums Shell Programming and Scripting Using && in if statement with 3 expressions Post 302750835 by Don Cragun on Wednesday 2nd of January 2013 03:08:20 PM
Old 01-02-2013
Quote:
Originally Posted by angilulu
how do you write an if statement for something like
Code:
if ((expr 1 >= expr 2 && expr 3 >= expr 4) && expr 5 <= expr 6)

if ((TRUE && TRUE) && TRUE) then
condition...

i've done it this way but it doesn't seem to work.
Code:
if ([[ "$ex_year" -ge "$curr_year"  && "$ex_month" -ge "$curr_month" ]]  && "$ex_day" -le "$curr_day" ); then
       condition...

You don't say what shell you're using and it makes a big difference. With a shell conforming to the standards, the following will work:
Code:
if [ "$ex_year" -ge "$curr_year" ] && [ "$ex_month" -ge "$curr_month" ] && [ "$ex_day" -le "$curr_day" ]
then    echo true
else    echo false
fi

If you're using a recent bash or ksh, the following will also do what you want:
Code:
if [[ "$ex_year" -ge "$curr_year" && "$ex_month" -ge "$curr_month" && "$ex_day" -le "$curr_day" ]]
then    echo true
else    echo false
fi

This User Gave Thanks to Don Cragun For This Post:
 

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EXPR(1) 						      General Commands Manual							   EXPR(1)

NAME
expr - evaluate arguments as an expression SYNOPSIS
expr arg ... DESCRIPTION
The arguments are taken as an expression. After evaluation, the result is written on the standard output. Each token of the expression is a separate argument. The operators and keywords are listed below. The list is in order of increasing precedence, with equal precedence operators grouped. expr | expr yields the first expr if it is neither null nor `0', otherwise yields the second expr. expr & expr yields the first expr if neither expr is null or `0', otherwise yields `0'. expr relop expr where relop is one of < <= = != >= >, yields `1' if the indicated comparison is true, `0' if false. The comparison is numeric if both expr are integers, otherwise lexicographic. expr + expr expr - expr addition or subtraction of the arguments. expr * expr expr / expr expr % expr multiplication, division, or remainder of the arguments. expr : expr The matching operator compares the string first argument with the regular expression second argument; regular expression syntax is the same as that of ed(1). The (...) pattern symbols can be used to select a portion of the first argument. Otherwise, the matching operator yields the number of characters matched (`0' on failure). ( expr ) parentheses for grouping. Examples: To add 1 to the Shell variable a: a=`expr $a + 1` To find the filename part (least significant part) of the pathname stored in variable a, which may or may not contain `/': expr $a : '.*/(.*)' '|' $a Note the quoted Shell metacharacters. SEE ALSO
ed(1), sh(1), test(1) DIAGNOSTICS
Expr returns the following exit codes: 0 if the expression is neither null nor `0', 1 if the expression is null or `0', 2 for invalid expressions. EXPR(1)
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