Question

我的脚本显示了我之前选择的areaID的所有城市（在选择框中）。

例如

- 我选择了区域ID 2我将在第二个选择框中看到所有符合区域2的城市

问题是 - 当我加载页面并希望显示之前选择的城市（来自数据库）

城市选择框中的CITY选项不是来自数据库的cityID上的点。

我需要改变什么？

先谢谢。

<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 == $userDetails['area'])
            echo "<option value='$db_area_id' selected>$db_area_name</option>";         
        else    
            echo "<option value='$db_area_id'>$db_area_name</option>";
    }
    ?>
</select>

<select id='cityID' name='cityID'>  </select>


<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>

getCities.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;

Answer 1

试

jquery的：

    $(document).ready(function(){

        $('#areaID').on('change',function(){
          var SelectedAreaID = $(this).val();
          $.ajax({
              type: "POST",
              url: "ajax/getCities.php",
              data: { areaID: SelectedAreaID }
                }).done(function(data) {
                     $('#cityID').html(data);
                });
          return false;
        });

    });

选择框：

<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 == $userDetails['area']){
            echo "<option value='$db_area_id' selected>$db_area_name</option>";         
        }
         else{
            echo "<option value='$db_area_id'>$db_area_name</option>";
        }           
    }
    ?>
</select>

<select id='cityID' name='cityID'>  </select>

getCities.php

$areaID = $_POST['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;

带选择框的ajax脚本

1 个答案: