mixed arrays (text and strings) in a php function

Question

Im trying to get to grips with php (rather unsuccessfully). I keep getting tripped up on the syntax for defining and calling functions

the code ive written is

$alpha = array(1, "alpha");
$beta = array(2, "beta");

function find_best_provider($provider1, $provider2){
    if($provider1 > $provider2){
        return array($provider1[0], $provider1[1]);
    }
    else {
        return array($provider2[0], $provider2[1]);
    }
}

$winner = find_best_provider($alpha, $beta);
echo $winner;

But i keep getting this notice - Notice: Array to string conversion in /Applications/MAMP/htdocs/find_best_provier_func.php on line 17 Array

Im aware what the problem is but im not quite sure how to solve it, any helps much appreciated !

$provider1 > $provider2 that's not how you should compare two arrays — Nir Alfasi
– Nir Alfasi, Commented Feb 14, 2013 at 22:51
Show us the whole code of the find_best_provier_func.php file. Looks like the problem not in the code you provided — Vadim Ashikhman
– Vadim Ashikhman, Commented Feb 14, 2013 at 22:52
@VadimAshikhman of course the problem is here. $provider1 > $provider2 is not valid checking. — BenM
– BenM, Commented Feb 14, 2013 at 22:52
@BenM you can check whether an array is greater of another array — Vadim Ashikhman
– Vadim Ashikhman, Commented Feb 14, 2013 at 22:54
It doesn't have the desired output here, hence the code here will always return the else... — BenM
– BenM, Commented Feb 14, 2013 at 22:56

BenM · Accepted Answer · 2013-02-14 22:51:11Z

2

If you're trying to evaluate the first element in the array, try this:

function find_best_provider($provider1, $provider2)
{
    if($provider1[0] > $provider2[0])
    {
        return array($provider1[0], $provider1[1]);
    }
    else 
    {
        return array($provider2[0], $provider2[1]);
    }
}

answered Feb 14, 2013 at 22:51

BenM

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