php - Display multiple values on one line in mysqli generated dropdown box -

using mysqli code, have been able create dropdown box using data mysql database. sql statement selects userid, f_name, , l_name database. list part of form , when user selects name, want pass userid via $_post method. how can done? again help.

<select name="names">     <option value = "">---select---</option>     <?php         $queryusers = 'select userid, f_name, l_name 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> 

just this:

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

that becomes value of option, gets submitted.

edit

this code using this

<select name="names">     <option value = "">---select---</option>     <?php         $queryusers = 'select userid, f_name, l_name 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 show mysqli prepared statement example

assuming inserting these same values table, you'd this:

your mysqli connection this:

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

then you'd insert (assuming userid autoincrement key)

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

the "ss" means both variables strings; if integers you'd use i. if doing update, you'd replace variables you'd use in insert ? placeholders. works same way.

here's manual: mysqli prepared statements