使用 PHP &MySQL 填充下拉列表

Using PHP amp; MySQL to populate dropdown(使用 PHP amp;MySQL 填充下拉列表)

问题描述

我正在尝试使用从使用 PHP 和 MySQL 的第一个下拉列表中选择的值填充第二个下拉列表，并且不刷新页面.我认为这很简单，但无法使其正常工作，因此非常感谢您的帮助.

I am trying to populate a second dropdown list using the value selected from a first dropdown list using PHP and MySQL, and without refreshing the page. I thought this would be simple but can't get it to work so any help would be much appreciated.

到目前为止，我有以下几点:

So far, I have the following:

HTML 表单 (form.php)

<select name="list1" id="list1">
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
</select>

<select name="list2" id="list2">

</select>

JavaScript(在 form.php 中)

<script type="text/javascript">
  $("#list1").change(function() {
    $("#list2").load("get_list2.php?id=" + $("#list1").val());
  });
</script>

get_list2.php

require_once("config.php");

$q1 = mysql_query("SELECT * FROM mytable WHERE id = '$_GET[id]'");
while($row1 = mysql_fetch_assoc($q1)){
  echo "<option>".$row1['item']."</option>";
}

谢谢！

推荐答案

就像其他成员所说的，你应该使用 PDO(带有准备好的语句)而不是 mysql_.

Like other members have says, you should use PDO (with prepared statements) instead of mysql_.

一种可能的实现:

HTML (form.php)

HTML (form.php)

<select name="list1" id="list1">
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
</select>

<select name="list2" id="list2"></select>

<script type="text/javascript">
$("#list1").change(function() {
    $.ajax({
        url : "get_list2.php?id=" + $(this).val(),                          
        type: 'GET',                   
        dataType:'json',                   
        success : function(data) {  
            if (data.success) {
                $('#list2').html(data.options);
            }
            else {
                // Handle error
            }
        }
    });
});
</script>

PHP (get_list2.php)

PHP (get_list2.php)

require_once("config.php");

$id = $_GET['id'];

if (!isset($id) || !is_numeric($id))
    $reponse = array('success' => FALSE);
else {
    // Where $db is a instance of PDO

    $query = $db->prepare("SELECT * FROM mytable WHERE id = :id");
    $query->execute(array(':id' => $id));
    $rows = $query->fetchAll(PDO::FETCH_ASSOC);

    $options = "";
    foreach ($rows as $row) {
        $options .= '<option value="'. $row .'">'. $row .'</option>';
    }

    $response = array(
        'success' => TRUE,
        'options' => $options
    );
}

header('Content-Type: application/json');
echo json_encode($response);

PS:没有经过测试，但它应该可以工作......我猜.

PS : not tested but it should works... I guess.

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