如何從下拉選擇框列表中將用戶選擇的選項的值存儲到mysql數據庫中？

Question

有一個帶有下拉選擇框列表的表單。 如何從下拉選擇框列表中將用戶選擇的選項的值存儲到mysql數據庫中？ 謝謝。

形成

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">


<select name="extrafield5">
 <option value="NOW" selected="selected">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
<button id="btn1" type="submit" value="Submit">Submit</button>

</select>


</form>

PHP文件

<?php

define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');

$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}

$db_selected = mysqli_select_db($connection, DB_NAME);

if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}

echo 'Connected successfully';


if (isset($_POST['extrafield5'])){
    $extrafield5 = $_POST['extrafield5'];
}

else {$extrafield5 = '';}




$sql = "INSERT INTO seguin_orders (extrafield5) 
        VALUES ('$extrafield5')";

if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}

數據庫

http://oi60.tinypic.com/9ppc0i.jpg

Answer 1

您需要提交表格。

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">
<select name="extrafield5">
 <option value="NOW" selected="selceted">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
</select>
<input type="submit" name="submit" value="Submit"/>
</form>

並在PHP中獲取它：

if (isset($_POST['submit'])){
  if (isset($_POST['extrafield5'])){
    $extrafield5 = $_POST['extrafield5'];
  }
  else {$extrafield5 = '';}
}

Answer 2

提交表格之前，您無法進入下一頁（xxx.php） 。 因此，請在表單中放置“提交”按鈕。

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">
<select name="extrafield5">
 <option value="NOW" selected="selceted">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
</select>
<input type="submit" name="submit" value="Submit"/>
</form>

它將解決問題。