PHP：基於下拉菜單的MySQL值調用

Question

所以我試圖做的是從我的服務器調用一個值，並根據我從下拉框中選擇的變量輸入它，然後我想要這些變量並通過JavaScript計算出一個新的輸出（我不認爲我想爲最後一部分提供幫助，但我喜歡在前面）。總之，這裏是代碼

$db_name = 'prod_chrotex_db'; 
$conn = new mysqli($host, $user, $pass, $db_name); 
$query = mysqli_query($conn, "SELECT * FROM table1 INNER JOIN table2 ON id "); 

echo '<select name="TITLE">'; 
while ($row = mysqli_fetch_array($query)) { 
    echo '<option value="' . $row['title'] . '">' . $row['title'] . '</option>'; 
    $rows = mysqli_fetch_assoc($query); 
    echo 'ID: ' . $rows['ID'] ; 
    $value = $rows['values'] 
}; 

echo "</select>"; 

Answer 1

$db_name = 'prod_chrotex_db'; 
$conn = new mysqli($host, $user, $pass, $db_name); 
$query = mysqli_query($conn, "SELECT * FROM table1 INNER JOIN table2 ON id "); 

echo '<select name="TITLE" id="selectb" onchange="callfunction();>'; 
while ($row = mysqli_fetch_array($query)) { 
    echo '<option value="' . $rows['ID'] . '" ">' . $row['title'] . '</option>'; 


}; 

echo "</select>"; 

?> 
<script> 

    function callfunction(){ 
     var val = document.getElementById('selectb').value(); 
     // You get the selected value in val variable. Here you can calculate what you want 

     } 

</script>