validation filter_input php

Question

I’m getting paid for $_POST I would like to know whether it is wrong to validate the if thus.

$romaneio = filter_input(INPUT_POST, 'romaneio');

if($romaneio == ''){
    $romaneio = 'null';
}

Answer 1

There is nothing wrong with your condition, what is interesting to note is the following, the PHP already considers a condition as false in the following cases:

• null
• 0
• array()
• ""

Whereas you do the following assignment:

$romaneio = filter_input(INPUT_POST, 'romaneio');

You can simply put your variable into a condition and use it normally if it is not considered false:

if($romaneio){ //simples assim, sem precisar de comparação
    //e no bloco você a utiliza como quiser
} else { //else opcional caso queira setar algo como no exemplo da pergunta
    $romaneio = 'null';
}

Or even as you put the question, but simply "denying" the condition:

if(!$romaneio){
    $romaneio = 'null'; 
}

Answer 2

The filter_input, in addition to bringing the result, also makes the field validation, bringing FALSE. I recommend doing:

if (!filter_input(INPUT_POST, 'romaneio')) {
    $romaneio = 'null';
}