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Home/ Questions/Q 6793127
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T18:03:49+00:00 2026-05-26T18:03:49+00:00

The problem is when I use $fname = isset($_POST[‘fname’]) ? $_POST[‘fname’] : ‘sample’; die($fname);

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The problem is when I use

$fname = isset($_POST['fname']) ? $_POST['fname'] : 'sample';
die($fname);

in PHP side after form submission with filled fname field getting output exactly what I filled, after submission with unfilled fname input field, getting output ABSOLUTELY NOTHING. Used code !empty instead

$fname = !empty($_POST['fname']) ? $_POST['fname'] : 'sample';
die($fname);

after submission with unfilled fname input field, getting output sample.

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  1. Editorial Team
    Editorial Team
    2026-05-26T18:03:50+00:00Added an answer on May 26, 2026 at 6:03 pm

    That’s because the ‘fname’ variable is sent anyway, it’s just empty but sent!

    Try changing your form method from POST to GET and you’ll see by yourself.

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