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Home/ Questions/Q 9195385
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T21:37:01+00:00 2026-06-17T21:37:01+00:00

I have been using the code below to validate my user input by $_POST:

  • 0

I have been using the code below to validate my user input by $_POST:

if(isset($_POST['name']) && !empty($_POST['name'])) {
    $n=$_POST['name'];
}
else {
    $errors[] = "Please give a name";
}

This code checks whether ‘name’ was actually set, which is obvious and clear and needed.
Secondly, it checks whether user typed something in textfield to give a value for name.
However, if user gives SPACE ” ” as input it accepts it because it is not empty it has SPACE.

I found one way of doing it right:

if(isset($_POST['name'])) {
    $n = trim($_POST['name']);
    if(empty($n)) {
        $errors[] = "Please give a name";
    }
}

else {
$errors[] = "Please give a name";
}

But here I am repeating same error message twice, so how can it be optimized?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T21:37:02+00:00Added an answer on June 17, 2026 at 9:37 pm

    If you don’t need the trimmed string, you can move the trim itself to the if-clause:

    if(isset($_POST['name']) && (trim($_POST['name']) != '') ) {
        $n=$_POST['name'];
    }
    else {
        $errors[] = "Please give a name";
    }
    

    If you further need it, you could modify the input before checking:

    $_POST['name'] = trim( $_POST['name'] );
    if(isset($_POST['name']) && !empty($_POST['name'])) {
        $n=$_POST['name'];
    }
    else {
        $errors[] = "Please give a name";
    }
    
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