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Asked: 2026-06-03T01:37:44+00:00

I wanted to make a dropdown area by getting the values from a mysql

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I wanted to make a dropdown area by getting the values from a mysql table. This code does not seem to work; all it prints out is the box for the dropdown without the content, how can I get this to work? or is there an alternative procedure in doing this?

<?
$connection = mysql_connect("localhost", "root", "");
mysql_select_db("test", $connection);
$query = "SELECT full_name FROM test";
$names = mysql_query($query);
function dropDown($content, $result)
 {
while($row = mysql_fetch_row($result))
    {
        $name = $row[0];
        $content .= "<option value='$name'>$name</option>";
    }
 }
$content=<<<EOT
           <html>
              <body>
                 <select name="name">
EOT;

dropDown($content, $names)

$content.=<<<EOT
                 </select>
              </body>   
            </html>          


EOT;

echo $content;
?>
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1 Answer

  1. Editorial Team

    return the string. PHP is not C where you use out parameters just because they are sometimes handy.

    function dropDown($result, $fieldName)
{
    $content = '';
    while($row = mysql_fetch_row($result)) {
        $name = $row[0];
        $content .= "<option value='$name'>$name</option>";
    }
    return '<select name="'.$fieldName.'">'.$content.'</select>';
}

$content = '<html><body>';
$content .= dropDown($names, 'name');
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