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如何通过选择的下拉列表自动填充 db 中的数据？我的下拉结果也已经出现，代码如下：

<?php
    echo '<tr>
    <td>'.$customer.'</td>
    <td><select name="customer_id">';

    foreach ($customers as $customer) {
        if ($customer['customer_id'] == $customer_id) {
            echo '<option value="'.$customer['customer_id'].'" selected="selected">'.$customer['name'].'</option>';
        } else {
            echo '<option value="'.$customer['customer_id'].'">'.$customer['name'].'</option>';
        }
    }
    echo '</select>
        </td>
    </tr>';
?>

上面列出的下拉列表的结果为

  • 行政
  • 客户1
  • 自由

从以下数据库加载

INSERT INTO `my_customer` (`customer_id`, `name`, `firstname`, `lastname`) VALUES
(8, 'admin', '', ''),
(6, 'customer1', 'ok', ''),
(7, 'FREE', 'marj', 'om');

so whenever dropdown selected i want the all data below:

<tr>
<td><?php echo $firstname; ?></td>
<td><?php echo $lastname; ?></td>
</tr>

也自动填充，似乎需要 javascript/ajax/jquery 来修复它，我想知道是否有人可以帮助我，提前感谢

添加JSON CALL

我已经有如下的json调用：（假设这个放在customer.php的url index.php?p=page/customer）

public function customers() {
    $this->load->model('account/customer');
    if (isset($this->request->get['customer_id'])) {
        $customer_id = $this->request->get['customer_id'];
    } else {
        $customer_id = 0;
    }

    $customer_data = array();
    $results = $this->account_customer->getCustomer($customer_id);
    foreach ($results as $result) {
        $customer_data[] = array(
            'customer_id' => $result['customer_id'],
            'name'       => $result['name'],
            'firstname'       => $result['firstname'],
            'lastname'      => $result['lastname']
        );
    }

    $this->load->library('json');
    $this->response->setOutput(Json::encode($customer_data));
}

和分贝

public function getCustomer($customer_id) {
    $query = $this->db->query("SELECT DISTINCT * FROM " . DB_PREFIX . "customer WHERE customer_id = '" . (int)$customer_id . "'");
    return $query->row;
}
4

1 回答 1

0

假设您正在使用 jQuery，您将侦听选择更改事件，然后对将返回数据的 PHP 函数进行 ajax 调用。然后将数据输出到适当的地方。我建议为下一个标签设置 id 属性：<select>，<td>名称，<td>姓氏，如下所示：

<select name="customer_id" id="customer_id>...</select>

<td id="firstname"> echo firstname </td>
<td id="lastname"> echo lastname </td>

然后是jquery代码：

<script type="text/javascript">//<!--
$(document).ready(function(){
    $('select#customer_id').change(function(){
        $.post(
            "http://www.domain.com/my_php_script.php",
            {customer_id: $(this).val()},
            function(data){
                $('td#firstname').html(data.firstname);
                $('td#lastname').html(data.lastname);
            }
        );
    });
});
//--></script>

假设您my_php_script.php通过给定的 customer_id 从数据库中检索数据$_POST['customer_id']并返回一个 JSON 对象，如echo json_encode(array('firstname' => FIRSTNAME_FROM_QUERY, 'lastname' => LASTNAME_FROM_QUERY));

补充：有两种方法可以解决它 - 在 JS 中而不是

$.post()

你必须使用

$.get(...)

或在您的 PHP 脚本中，而不是

$this->request->get['customer_id']

你必须使用

$this->request->post['customer_id']

在每个地方......这应该这样做......例如：

<script type="text/javascript">//<!--
$(document).ready(function(){
    $('select#customer_id').change(function(){
        $.get(
            "http://www.domain.com/my_php_script.php",
            {customer_id: $(this).val()},
            function(data){
                $('td#firstname').html(data.firstname);
                $('td#lastname').html(data.lastname);
            }
        );
    });
});
//--></script>
于 2011-05-04T12:41:13.467 回答