php-使用以下命令向MySQL添加外键<option><select>

Question

This is a follow up question to my previous question in here. PHP - Dropdown menu(select) not working properly to while loop

I follow the code given to me and I think it is working correctly. But my only concern is that, the query in my code is actually for the shelf table, and not for the book table which have the foreign key for the shelf. Here is my code for the textfield. It is in a form which have a button that redirects to check_book.php to add it

<div class="form-group">

      <label for="inputEmail" class="col-lg-2 control-label">Shelf</label>
      <div class="col-lg-10">
    <?php   
    $sel_admin = "select * from shelf";
    $rs_admin = mysql_query($sel_admin);
    echo '<select class="form-control">';
    while($row = mysql_fetch_array($rs_admin))
                {

                echo"<option value='" .$row['shelf_id']."'>" . $row['shelf_description'] ."</option>";
                }
                echo'</select>';
                ?>
      </div>
    </div>

It's giving me this error Notice: Undefined index: shelf_id in C:\\xampp\\htdocs\\OCatalog\\admin\\check_book.php on line 8 but the other non-foreign key textfield is just fine.

And here is my check_book.php

$book_id = $_POST['book_id'];
    $book_title = $_POST['book_title'];
    $author_id = $_POST['author_id'];
    $book_quantity = $_POST['book_quantity'];
    $shelf_id = $_POST['shelf_id'];

How can I view all the shelf then adding it as a foreign key?

PS. The select is also in the form which when i click the submit button it will go to check_book.php

Answer 1

您忘记在选择中添加名称属性

echo '<select class="form-control" name="shelf_id">';