从数据库中填充的下拉列表中获取选定值

Question

I want to get the selected SystemID from my drop down menu which is populated by a database. I have named the drop down"data3"

Drop down from database is all working, it is displaying all the SystemIDs from my table.

Code below:

<select id ="data3" style="position:relative; top:-70px; left: 3.6%;" 
name="data3" class="btn btn-light btn-sm dropdown-toggle">
    <option value="SystemID">SystemID</option>
    <?php
        $sql = "SELECT DISTINCT SystemID FROM MyTable ORDER BY 
        SystemID";
    $result = sqlsrv_query($conn, $sql);
    while ($row = sqlsrv_fetch_array($result, SQLSRV_FETCH_ASSOC)) {
        echo "<option value=\"\">" . $row['SystemID'] . "</option>";
        }
    ?>
  </select>

How do I get the selected value from that drop down? I have tried doing

var data3=$("#data3").val();

But this returns nothing

Answer 1

Use the ternary operator to check if the value coming from the database match the one from the select and mark it as selected. Please note that you must have a list of all the values of the select/option statement to be able to compare.

$row['SystemID'] == $value?'selected':''

echo "<option value='".$row['SystemID'] ."' ". ($row['SystemID'] == $value?'selected':'') ."  >" . 
        $row['SystemID'] . 
     "</option>";