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這裏讀POST數據是我在做什麼我有2個數據庫表:無法從列表框中
- 攝像機和字段名等領域
- 用戶電子郵件ID等領域
我已經使用列表框填充它們,我能夠從MySQL數據庫中檢索數據。我使用表單發佈數據並稍後捕獲它。這裏是我的一段代碼:
<form action="share.php" method="post">
Camera Name:
<select>
<?php
while($fetch_options1 = mysql_fetch_array($data1)) { //Loop all the options retrieved from the query
?>
<option id ="<?php echo $fetch_options1['id']; ?>" value="<?php echo $fetch_options1['name']; ?>"><?php echo $fetch_options1['name']; ?></option>
<?php
}
?>
</select>
Share With Email ID:
<select>
<?php
while($fetch_options = mysql_fetch_array($data)) {
?>
<option id ="<?php echo $fetch_options['id']; ?>" value="<?php echo $fetch_options['email']; ?>"><?php echo $fetch_options['email']; ?></option>
<?php } ?>
</select>
<input type="submit" name="submit" value="Share The camera">
</form>
而且在同一頁我用這個代碼來獲取發佈的價值觀,它不給我任何答案上:
<?php
// START FORM PROCESSING
if (isset($_POST['submit'])) { // Form has been submitted.
$errors = array();
$_POST= array('name'=>'','email'=>'');
echo $cname = $_POST['name'];
echo $email1 = $_POST['email'];
echo $user_id=$_SESSION['id'];
} else { // Form has not been submitted.
}
?>
我想獲取選中當用戶點擊提交時,列表框中的電子郵件ID和名稱。我究竟做錯了什麼?
我正在使用$ _POST = array(..因爲我得到錯誤$ _POST ['name']中變量的未定義索引名稱; 謝謝我得到了溶劑 – 2013-02-20 15:32:24
當然......... ... – 2013-02-20 15:56:58