什么是变量名称（DropDown List）[重复]

i am new to php and i have run into some problems. I am trying to retrieve data from the database and i have managed to do it and display it as dropdown list, however, i am unable to insert it back to the database with the variable i have selected. Please help. May i know what is the variable that i have post to the next page? (which is the processstaffb.php)

<form action="processstaffb.php" method="post">
 <table id="t01">    
  <td width="1%">
  <?php
    $mysqli = new mysqli(spf, dbuser, dbpw, db);
    $sql="Select sbranch_name from branches";
    $result = $mysqli->query($sql);
      if ($result->num_rows > 0) {
       echo "<select name='sbranch_name'>";
       while($row = $result->fetch_assoc()) {
        echo "<option value='" . $row['sbranch_name'] . "'>" .$row['sbranch_name'] . "</option>";
        }
       echo "</select>";
      }  
     $mysqli->close();
    ?>
   </td>

This is the code at processstaffb.php

    if (!empty($_POST['sbranch_name'])) 
{
   $sbranch_name =$_POST["$sbranch_name"];
   echo "if pass";
   echo "$sbranch_name";
}
else 
{ 
   $sbranch_name = null;
   echo '<p><font color="red">You forgot to enter the branch of the officer!</font></p>';
   echo "if fail";
   echo "$sbranch_name";
}

</div>

Because you gave your select the name "sbranch_name" you can access the value of the selected option in processstaffb.php with $_POST['sbranch_name'].

Because you are planning to insert the values in your database, this is a good point to learn about PDO prepared statements or MySQLi prepared statements and SQL injections.