在mysqli生成的下拉框中的一行上显示多个值

Question

Using mysqli code, I have been able to create a dropdown box using data from a mysql database. The sql statement selects the userId, f_Name, and l_Name from the database. This list is part of a form and when the user selects the name, I want to pass only the userId via the $_Post method. How can that be done? Thanks again for your help.

<select name="names">
    <option value = "">---Select---</option>
    <?php
        $queryusers = 'SELECT userID, f_Name, l_Name FROM users';
        $db = mysqli_query($mysqli, $queryusers);
        while ( $names=mysqli_fetch_assoc($db)) {
           echo "<option value='{".$names['f_Name']."}' . 
                {".$names['l_Name']."}'>".$names['f_Name']. " " .
                $names['l_Name']."</option>";
        }
    ?>
</select>

Answer 1

Just do something like this:

echo '<option value="'.$names['userID'].'" >'.$names['f_Name'].' '. $names['last_nm'] . '</option>';

that becomes the value of the option, which is what gets submitted.

Edit

This is your code using this

<select name="names">
    <option value = "">---Select---</option>
    <?php
        $queryusers = 'SELECT userID, f_Name, l_Name FROM users';
        $db = mysqli_query($mysqli, $queryusers);
        while ( $names=mysqli_fetch_assoc($db)) {
           echo '<option value="'.$names['userID'].
            '" >'.$names['f_Name'].' '. $names['last_nm'] . '</option>';
        }
    ?>
</select>

another edit to show the mysqli prepared statement example

Assuming you are inserting these same values into a table, you'd just do it like this:

Your mysqli connection would be something like this:

$conn = new mysqli($host, $username, $password, $dbname)

Then you'd do your insert like this (assuming userID is an autoincrement key)

$stmt = $dbc->prepare("insert into users (userID, f_name, L_name)
                       values (NULL, ?, ?)");
$stmt->bind_param("ss", $names['f_name'],$names['l_name'])
$stmt->execute();
$stmt->close();

The "ss" just means both variables are strings; if they are integers you'd use i. If you are doing an update, you'd just replace any variables you'd use in the insert with those ? placeholders. It works the same way.

Here's the manual: mysqli prepared statements