HTML / PHP调查没有正确地从表中传递ID

Question

I'll try to keep it simple... this is the code I am using to populate a dropdown menu from a database. This populates the dropdown menu correctly.

<form action="formsubmit.php?team_id=6" method="post">
    <label> <br />What did they say?: <br />
        <textarea name="quotetext" rows="10" cols="26"></textarea></label>
    <select name='name'>
        <?php
            while ($temp = mysql_fetch_assoc($query)) {
                echo "<option value=" . $temp['id'] . ">".htmlspecialchars($temp['lastname']) . ", " . htmlspecialchars($temp['firstname']) . "</option>";
            }
        ?>
    </select>
    <input type="submit" value="Submit!" />
</form>

What I'm attempting to do is pass the person's ID into along to the file formsubmit.php which is called on the submission of the form. When I use $die($sql) on my database query in formsubmit.php, the ID of the person is blank.... everything else gets passed through just fine. Here is the relevant code in formsubmit.php:

$quotetext = $_POST['quotetext'];
    $id = $_POST['id'];
    $team_id = $_GET['team_id'];
    $sql = "INSERT INTO quotes SET 
            speaker_id='$id',
            quotetext='$quotetext',
            game_id=2";
    die($sql);

EDIT: Fixed, credit to Michael.

<select name='name'>

Should be:

<select name='id'>

Answer 1

In your form, you are using <select name='name'>, but in your PHP script you're calling for $_POST['id']. Change it to:

<select name='id'>