Question

<?php    
$query = mysql_query("Select id, name From table");
        while($row = mysql_fetch_array($query)) {
            echo $_GET["jsoncallback"] . '(<option value='.$row['id'].'>'.$row['name'].'</option>)';
        }
?>

When I echo result, it is error, how to fix it?

Answer 1

Since you didn’t write exactly what you’re trying to do, i’m guessing you’re trying to return a list of HTML options that a JS callback function will place in your document.

try this:

<?php    
$options = '';
$query = mysql_query("Select id, name From table");
while ($row = mysql_fetch_array($query)) {
  $options .= '<option value="'.$row['id'].'">'.$row['name'].'</option>' . "\n";
}
echo $_GET["jsoncallback"] . "('" . $options . "');";
?>

This will first create all the options as a string, and only then build the callback.