带有 AJAX 代码的选择框 - selectbox with ajax code

selectbox with ajax code

本文关键字：选择代码 AJAX 带有 | 更新日期: 2023-09-27

>我有 2 个选择框

1.区域编号

2.城市编号

当用户在第一个选择框中选择区域时，第二个选择框将自动更改，仅显示所选区域的城市。

这很好用，除非我从数据库加载数据或用户提交表单。在这些情况下，不会选择正确的城市 - 指针始终位于第一个选项（第一个城市）上。

我需要更改什么？

<p><label>AREA</label> 
    <select name='areaID' id='areaID'>
        <?PHP
        $query = mysql_query("SELECT * FROM `areas` ORDER BY id ASC "); 
        while($index = mysql_fetch_array($query)) 
        {
            $db_area_id = $index['id'];
            $db_area_name = $index['name'];
            if ($db_area_id == $userDetails['areaID'])
                echo "<option value='$db_area_id' selected>$db_area_name</option>";         
            else    
                echo "<option value='$db_area_id'>$db_area_name</option>";
        }
        ?>
    </select><span>*</span>
</p>
<p><label>CITY</label>
    <select id='cityID' name='cityID'>  </select>
</p>

<script>
<?PHP if ($_POST) { ?>
    $(document).ready(function(){
        $('#areaID').filter(function(){
            var areaID=$('#areaID').val();
            var cityID=<?PHP echo $userDetails['cityID'] ?>;
            $('#cityID').load('ajax/getCities.php?areaID=' + areaID+'&cityID=' + cityID);
            return false;
        });
    }); 
<?PHP }else { ?>
$(function () {
    function updateCitySelectBox() {
        var areaID = $('#areaID').val();
        $('#cityID').load('ajax/getCities.php?areaID=' + areaID);
        return false;
    }
    updateCitySelectBox();
    $('#areaID').change(updateCitySelectBox);
});
<?PHP } ?>
</script>

获取城市.php ：

<?PHP
    $areaID = (int) $_GET['areaID'];

    $second_option = "";
    $query2 = mysql_query("SELECT * FROM `cities` WHERE area_id = $areaID ORDER BY id ASC");
    while($index = mysql_fetch_array($query2)) 
    {
        $id = $index['id'];
        $name  = $index['name'];
        $name = iconv('windows-1255', 'UTF-8', $name);
        $second_option .= "<option value='$id'>$name</option>";
    }
    echo $second_option;
//  exit; 
?>

更改以下内容：

<p><label>CITY</label>
    <select id='cityID' name='cityID'>  </select>
</p>

对此：

<p><label>CITY</label>
    <select id='cityID' name='cityID'>
    <?php
    if (isset($userDetails['areaID']))
    {
        $query = mysql_query("SELECT * FROM `cities` WHERE area_id = {$userDetails['areaID']} ORDER BY id ASC "); 
        while($index = mysql_fetch_array($query)) 
        {
            $db_city_id = $index['id'];
            $db_city_name = $index['name'];
            if (isset($userDetails['cityID']) && $db_city_id == $userDetails['cityID'])
                echo "<option value='$db_city_id' selected>$db_city_name</option>";         
            else    
                echo "<option value='$db_city_id'>$db_city_name</option>";
        }
    }
    else
    {
        echo '<option value="0">Select Area...</option>';
    }
    ?>
    </select>
</p>