How to compare array values preserving keys?-CodePudding

I can't figure out how to check if array values are equal or not. Array always have only 2 keys, need to preserve them, keys unknown. E.g.:

$arr = array(
    5 => 180,
    3 => 120
);

if ($arr['key1_value'] != $arr['key2_value']) {
    $variable = $arr['key1'];
} else {
    $variable = ($arr['key1']   $arr['key2']) / 2;
}

CodePudding user response：

Use array_values() or array_keys() to get an array of the value in this array, but with a zero based (predictable) set of keys

$arr = array(
    5 => 180,
    3 => 120
);
#$v = array_values($arr);
$v = array_keys($arr);

if ($v[0] != $v[1]) {
    $variable = $v[0];
} else {
    $variable = ($v[0]   $arr[1]) / 2;
}

CodePudding user response：

This is not so clear... But if I refer to your comments, I should say that this approach could solve your problem:

// Here is your data
$arr = array(
    5 => 180,
    3 => 120
);

// You seem to be interested by keys as a return, so flip the array
$flip = array_flip($arr);

// Return a mean or a key, depending on comparison between flip and original array
$variable = (count($flip) < 2) ? array_sum(array_keys($arr))/2 : array_shift($flip);

To clarify the last line:

If the number of values in $flip is lower, that means your two values are equals (because two same keys can not be accepted after the flip): in this case, we return the mean of keys (here it works because you confirmed that keys are always intergers)
If not, the first key is returned (I think it is what you expected, but not sure)

EDIT: In my proposition, I did a count on the array. As you said it is always 2 compared values, I replaced it directly by "2".