嗨,所有专家都在那里,我的网站问题是我的下拉列表作为POST方法(对于一个国家,州和城市)工作完全正常,除了一个问题,它将将ID插入我的数据库而不是实际名称。
我感到很困惑,因为它在网页上显示一切正常(一个适当的名字显示在一个国家,州和城市。
请帮我确定问题在我的代码中的位置。
提前多多感谢。
这是我的注册页码
<div>
<select name="country" class="country">
<option value="<?php echo $row['country_name'] ?>">--Select Country--</option>
<?php
$stmt = $conn->prepare("SELECT * FROM countries");
$stmt->execute();
while($row=$stmt->fetch(PDO::FETCH_ASSOC))
{
?>
<option value="<?php echo $row['country_id']; ?>"><?php echo $row['country_name']; ?></option>
<?php
}
?>
</select>
<select name="state" class="state">
<option value="<?php echo $row['state_name'] ?>">--Select State--</option>
</select>
<select name="city" class="city">
<option value="<?php echo $row['city_name'] ?>">--Select City--</option>
</select>
</div>
<input type="submit" name="registration" value="REGISTER">
<input type="reset" value=". . . RESET . . .">
以下是从以下字段中获取数据的另一个代码
获取州代码
<?php
include('db_conn.php');
if($_POST['id'])
{
$id=$_POST['id'];
$stmt = $conn->prepare("SELECT * FROM states WHERE country_id=:id");
$stmt->execute(array(':id' => $id));
?><option selected="selected">--Select State--</option><?php
while($row=$stmt->fetch(PDO::FETCH_ASSOC))
{
?>
<option value="<?php echo $row['state_id']; ?>"><?php echo $row['state_name']; ?></option>
<?php
}
}
?>
获取城市
<?php
include('db_conn.php');
if($_POST['id'])
{
$id=$_POST['id'];
$stmt = $conn->prepare("SELECT * FROM cities WHERE state_id=:id");
$stmt->execute(array(':id' => $id));
?><option selected="selected">--Select City--</option>
<?php while($row=$stmt->fetch(PDO::FETCH_ASSOC))
{
?>
<option value="<?php echo $row['city_id']; ?>"><?php echo $row['city_name']; ?></option>
<?php
}
}
?>
和Ajax数据
<?php
//Include database configuration file
include('db_conn.php');
if(isset($_POST["country_id"]) && !empty($_POST["country_id"])){
//Get all state data
$query = $conn->query("SELECT * FROM states WHERE country_id = ".$_POST['country_id']." AND status = 1 ORDER BY state_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
//Display states list
if($rowCount > 0){
echo '<option value="">Select state</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['state_id'].'">'.$row['state_name'].'</option>';
}
}else{
echo '<option value="">State not available</option>';
}
}
if(isset($_POST["state_id"]) && !empty($_POST["state_id"])){
//Get all city data
$query = $conn->query("SELECT * FROM cities WHERE state_id = ".$_POST['state_id']." AND status = 1 ORDER BY city_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
//Display cities list
if($rowCount > 0){
echo '<option value="">Select city</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['city_id'].'">'.$row['city_name'].'</option>';
}
}else{
echo '<option value="">City not available</option>';
}
}
?>
答案 0 :(得分:1)
插入选项值。
echo '<option value="'.$row['city_id'].'">'.$row['city_name'].'</option>';
在这种情况下是city_id
如果您想要名称,请将其设置为值:
echo '<option value="'.$row['city_name'].'">'.$row['city_name'].'</option>';