How to use IF loop in unix for multiple conditions

Question

I need a logic to implement the following logic in unix

if ( $a !="xyz" || $d !="abc" ) && ( $b= $c))
then
echo "YES WORKING"
fi

I tried below code not working

if [ [ [ $a != "xyz" ] -o [ $d != "abc" ] ] -a [ "$b" = "$c" ] ]
then
echo "YES WORKING"
fi

getting error as

:[ :] unexpected operator/operand

@fedorqui just now changed the logic Thanks for pointing that — Pranee
– Pranee, Commented Jun 4, 2014 at 12:09
@TimCastelijns getting error as :[ :] unexpected operator/operand — Pranee
– Pranee, Commented Jun 4, 2014 at 12:11

jmlemetayer · Accepted Answer · 2014-06-04 12:29:28Z

3

You can do something like this:

[ $a != "xyz" -o $d != "abc" ] && [ "$b" = "$c" ] && echo "YES WORKING"

answered Jun 4, 2014 at 12:29

jmlemetayer

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1 Comment

konsolebox Over a year ago

I forgot about -o. Your answer's better than mine with simpler shells. +1

konsolebox · Accepted Answer · 2014-06-04 12:23:58Z

2

Your logic should work easy in shells supporting [[ ]]:

if [[ ($a != "xyz" || $d != "abc") && $b = "$c" ]]; then
    echo "YES WORKING"
fi

Although there's a way for those that doesn't:

if ([ ! "$a" = "xyz" ] || [ ! "$d" = "abc" ]) && [ "$b" = "$c" ]; then
    echo "YES WORKING"
fi

But that's still inefficient since you'd be summoning subshells, so use { } but the syntax is a little ugly:

if { [ ! "$a" = "xyz" ] || [ ! "$d" = "abc" ]; } && [ "$b" = "$c" ]; then
    echo "YES WORKING"
fi

answered Jun 4, 2014 at 12:23

konsolebox

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3 Comments

konsolebox Over a year ago

@Pranee What shell are you using? System? Perhaps you should consider the other methods as [[ ]] is only supported in modern shells like Zsh and Bash.

konsolebox Over a year ago

Also, are you running the lines alone, i.e. it's not included with other lines? If they work alone then perhaps it's a problem with other parts of the script.

Pranee Over a year ago

your code worked but "$b" = "$c" not working may be because $b= "05/31/2014 " and $c="05/31/2014 "

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