in_array broke

[Monkuar]

$colorarray = isset($_POST['form']['color']) ? intval($_POST['form']['color']) : 0;

$color = array(
"1",
"2",
"3", 
"4",
"5",
"6",
"7",
"8",
"8",
"9",
"10",
"11",
"12",
"13",
"14",
"15",
"16"
);
if (in_array($color, $colorarray)) {
    echo "search found";
exit;
}else{
echo "hacker";
exit;
}

 

i submit  my $_POST['form']['color'] as 1 through 16 and it always says hacker? 

[dragon_sa]

try this maybe

 

$colorarray = (isset($_POST['form']['color']) ? intval($_POST['form']['color']) : 0);

 

from what I have read you need the brackets to get the expected values, tbh I am not certain if that is the only issue here

[creata.physics]

Please use print_r($colorarray) so we can see what is returned.

[GingerRobot]

1.) You seem to have the arguments to in_array() the wrong way around: http://php.net/manual/en/function.in-array.php

 

2.) Unless this is a simplification, why use in_array() anyway? Surely this would suffice:

 

if($i >= 1 && $i <= 16){
    //valid
}else{
    //invalid
}

[requinix]

1.) You seem to have the arguments to in_array() the wrong way around: http://php.net/manual/en/function.in-array.php

Either that or they're the right way around and the variable names are backwards...