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Home/ Questions/Q 8583787
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T21:37:11+00:00 2026-06-11T21:37:11+00:00

The following code works when i run it in localhost, but when I host

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The following code works when i run it in localhost, but when I host it online, it gives an error.

if(isset($_POST['congressman'])){
    $congressman = $_POST['congressman'];
    $no_of_votes = mysql_result(mysql_query("SELECT no_of_votes FROM `tbcandidates` WHERE `identification_no`=$congressman"),0) + 1;
    $query=mysql_query("UPDATE tbcandidates SET no_of_votes = '$no_of_votes' WHERE identification_no = $congressman");

This line:

no_of_votes = mysql_result(mysql_query("SELECT no_of_votes FROM `tbcandidates` WHERE `identification_no`=$congressman"),0) + 1;

It says mysql_result(): supplied argument is not a valid MySQL result resource in etc..

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T21:37:12+00:00Added an answer on June 11, 2026 at 9:37 pm

    try to add ob_start() at the top of the code and ob_flush() at end

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