PHP form and populate dropdown issue

Question

I am trying to connect to MySQL and populate the data to Dropdown. Here is my code. Some reason dropdown is not getting populated. Please suggest.

Here is the code.

<html>
<body>

<?php
$mysqli_connection = new mysqli($db_host, $db_username, $db_password, $db_database);
if (isset($_POST['Submit_1'])) {
    require 'submit.php';
    require 'validate.php';
    if ($form_errors = validate_form()) {
        show_form($form_errors);
    }
    else
    {
        form_submit_1();
    }
}
else
{
    show_form();
}
function show_form($errors = '')
{
    // were there any errors?  
    if ($errors) {
        //show errors
    }

    ?>
<form name="myForm" id="myForm" method="post">
    <?php
    $sql = "SELECT id, code FROM table1";
    $result11 = $mysqli_connection->query($sql);
    echo " <select name = \"state1\" id=\"state1\">";
    while ($row = $result11->fetch_assoc()) {
        echo "<option value = $row[id]>$row[code]</option>";
    }
    echo "</select>";
    ?>
</form>
    </body>
</html>
<?php
} // End of show_form()
?>

jValdron · Accepted Answer · 2011-12-16 21:04:53Z

4

This part:

echo "<option value = $row[id]>$row[code]</option>";

Is not valid, it should be something like this:

echo "<option value=\"{$row['id']}\">{$row['code']}</option>";

answered Dec 16, 2011 at 21:04

jValdron

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1 Comment

nav100 Over a year ago

It still doesn't work. If I move the dropdown code logic above the function show_form($errors = ''), it works.

Hasteur · Accepted Answer · 2011-12-16 22:10:11Z

0

You'll have to escape from the string back to the PHP layer.

echo "<option value=\"".$row['id']."\">".$row['code']."</option>";

That way you'll concatanate the PHP value of the variables with the string elements you're creating.

answered Dec 16, 2011 at 22:10

Hasteur

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