显示从表中数据为下拉菜单

Question

我有2个表

domains_info和TB2

我已经得到形式运作良好，将数据输入数据库

这里是顶部我的网页

<?php 
$action = isset($_POST['action']) ? $_POST['action'] : ""; 

if($action=='create'){ 
//include database connection 
include 'db_connect.php'; 

//write query 
$query = "insert into domains_info 
set 
domain = '".$mysqli->real_escape_string($_POST['domain'])."', 
domain_account = '".$mysqli->real_escape_string($_POST['domain_account'])."', 
renew_date = '".$mysqli->real_escape_string($_POST['renew_date'])."'"; 

if($mysqli->query($query)) { 
//if saving success 
header("Location:domains.php"); 
}else{ 
echo "Database Error: Unable to create record."; 
} 
$mysqli->close(); 
} 

这里是形式

<select id="domain_account" name="domain_account" class="txtBox"> 
<option value="">-select-</option> 
<option value="a">a</option> 
<option value="b">b</option> 
<option value="c">c</option>     
</select> 

我尝试和改变了我的页面的顶部像这样

<?php 
$action = isset($_POST['action']) ? $_POST['action'] : ""; 

if($action=='create'){ 
//include database connection 
include 'db_connect.php'; 

//write query 
$query = "insert into domains_info 
set 
domain = '".$mysqli->real_escape_string($_POST['domain'])."', 
domain_account = '".$mysqli->real_escape_string($_POST['domain_account'])."', 
renew_date = '".$mysqli->real_escape_string($_POST['renew_date'])."'"; 

if($mysqli->query($query)) { 
//if saving success 
header("Location:domains.php"); 
}else{ 
echo "Database Error: Unable to create record."; 
} 
$mysqli->close(); 
} 

$query = "select id, data 
         from tb2 
         where id='".$mysqli->real_escape_string($_REQUEST['id'])."' 
         limit 0,1"; 

$result = $mysqli->query($query); 
$row = $result->fetch_assoc(); 

$id = $row['id']; 
$data = $row['data']; 

，并更新了我的形式，因为这

<select id="domain_account" name="domain_account" class="txtBox"> 
<option value="">-select-</option> 
<option value="<?php echo$data; ?>"><?php echo$data; ?></option>  
</select> 

，你能告诉我是很新的这一点，它不工作。

对不起，我没有解释我想要实现的目标，我试图用数据库中的数据显示下拉式表单。

Answer 1

试试这个用于填充使用数据库下拉：

<?php 
$query = "select id, data from tb2"; 
$result = $mysqli->query($query); 

echo '<select id="domain_account" name="domain_account" class="txtBox">'; 
echo '<option value="">-select-</option>'; 
while ($row = $result->fetch_assoc()){ 
?> 
    <option value="<?php echo $row['data']; ?>"><?php echo $row['data']; ?></option> 
<?php  
} 
echo "</select>"; 
?>