在mysql的选项列表中存储行的id

Question

I'm echoing a list of attributes of a row from a table into an option list for the user to pick from which later would be used to be inserted into another table. What I'm trying to do is, to store the 'id' of that row -not using the foreign key- separately, but the 'id' I keep getting is the 'id' of the last row on the list. Here is a snippet to my code:

        <?php
        $conn = mysqli_connect('localhost', 'root', '', 'database');

          $result = mysqli_query($conn, "SELECT * FROM myTable WHERE region = '{$thisRegion}' ");
          //$result .= " ");
          while ($row = mysqli_fetch_assoc($result))
          {
            $selected = (isset($_POST['list']) && $_POST['list'] ==  $row['id']) ? 'selected' : '';
              echo "<option value='$row[streetAddress] $row[apt] $row[city] $row[zip] $row[_state] $row[country]' $selected >$row[city] $row[zip] $row[_state] $row[country] $row[streetAddress] $row[apt]</option>";
              $thisId= $row['id'];
              $_SESSION["thisId"] = $thisId;
          }
          ?>
        </select>

Answer 1

Your problem it's on the code

while ($row = mysqli_fetch_assoc($result))
{
    ...
    $thisId= $row['id'];
    $_SESSION["thisId"] = $thisId;
}

That code means that on each iteration, $_SESSION["thisId"] will be replaced with the id of the current row, making it so that at the end of the loop only the last row's ID will be there.

The way I think to do this would be to add the id to the option, and print the row just as display.

echo "<option value='{$row['id']}'> {$row['apt']} {$row['city']}... </option>";

Then, the receiving page could read

$selectedRowId = $_GET['optionKey']

I hope that helps, you should read more about php and forms moving forward as it will make your coding easier. Here is one basic tutorial http://www.w3schools.com/php/php_forms.asp