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Home/ Questions/Q 7706797
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T00:12:03+00:00 2026-06-01T00:12:03+00:00

I have an odd problem. Basicly my page works fine however after a small

  • 0

I have an odd problem. Basicly my page works fine however after a small bit of php, everything after those lines dont load too the page.

<?php


//GET SCHOOLS
$sql = "SELECT `id` FROM `school`";
$query = mysql_query($sql) or die(mysql_error());
$numschools = mysql_num_rows($query);

echo "<select id=\"schoolselect\" class=\"schoolselect\" value=\"Select School\">
<option id='selectschool' value = \"select\" name=\"select\">Select A School</option>

";
while($result = mysql_fetch_array($query) or die(mysql_error()))
{

    $school = $result['id'];
    echo "<option value = \"$school\" name=\"$school\">".$school ."</option>";


}
echo "</select>";

?>

everything before that works, the php part works but anything after that echo “select” doesnt

any help would be amazing

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  1. Editorial Team
    Editorial Team
    2026-06-01T00:12:04+00:00Added an answer on June 1, 2026 at 12:12 am

    Never call die(mysql_error()) inside a fetch loop. mysql_fetch_array() returns FALSE when no more rows are available, so when the last row is reached, die() is called and your script will terminate leaving the </select> unclosed. You won’t get an error message, but you’ll be left with incomplete HTML that won’t render properly in the browser.

    // Don't call die(mysql_error()) in a fetch loop!
    while($result = mysql_fetch_array($query))
    {
        $school = $result['id'];
        echo "<option value = \"$school\" name=\"$school\">".$school ."</option>";
    }
    
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