PHP Notice: Use of undefined constant idq - assumed 'idq'

Question

Im getting this php notice , and my program works perfectly. here is my code :

    $data =  file_get_contents('php://input');
    $P = isset(json_decode($data)->E->EP->P) ? json_decode($data)->E->EP->P : '0';
    for ($i = 1; $i <= $P; $i++) {
        ${idq.$i} = isset(json_decode($data)->E->EP->{idq.$i}) ? json_decode($data)->E->EP->{idq.$i} : '0';
           }
echo ${idq.$i};

should i fix or suppress the error with handling?

Answer 1

Assuming you want your variables to be named $idq1 , $idq2 , $idq3 etc:

They're currently being considered as constants. If you want them in your variables, you need wrap your idq constant in quotes:

for ($i = 1; $i <= $P; $i++) {
    ${'idq'.$i} = isset(json_decode($data)->E->EP->{'idq'.$i}) ? json_decode($data)->E->EP->{'idq'.$i} : '0';
}

echo ${'idq'.$i};

But ternary statements make your code a bit ugly (at least, in this case). But, this looks a lot more cleaner, IMO:

$data =  file_get_contents('php://input');

if(isset(json_decode($data)->E->EP->P)) {
    $P = json_decode($data)->E->EP->P;
}
else {
    $P = '0';
}

for ($i = 1; $i <= $P; $i++) {

    if(isset(json_decode($data)->E->EP->{'idq'.$i}) ) {
        ${'idq'.$i} = json_decode($data)->E->EP->{'idq'.$i};
    }
    else {
        ${'idq'.$i} = '0';
    }

}

Also, to answer your final, question:

should i fix or suppress the error with handling?

Nope. Never. You should always find and fix the error instead of suppressing it.

Answer 2

始终修正您的错误，压制不是解决任何问题的方法，它只是一颗炸弹，随时待命……