更改第一个框的选定值后,填充第二个和第三个选择框


Populate second and third select box after change the selected value of the first box

有三个选择框,首先禁用"device_sel"answers"prob_sel"。当我从第一个选择框"customer_sel"中选择值时,我使用$.post来填充device_sel。我的问题是,当我在"device_sel"中选择值,然后填充"prob_sel"时,该怎么做。当我尝试填充prob_sel时,我认为在$.post过程中没有看到一些错误。当我更改(或第一次选择)device_sel中的值时,prob_sel会保持begin处于禁用状态,并且不会发生任何事情。

有人能发现错误或打字,花了一个小时想弄清楚却什么都找不到吗。

我的javascript:

<script type="text/javascript">
    $(document).ready(function()
    {
        $('#device_sel').attr('disabled', true);
        $('#prob_sel').attr('disabled', true);

        $('#customer_sel').change(function()
        {
            var cid = $('#customer_sel').attr("value");
            $.post("get_customers_devices.php", {cid:cid}, function(data)
            {
                $('#device_sel').attr('disabled', false);
                $('#device_sel').html(data);
            });
        });
        $('#device_sel').change(function()
        {
            var did = $('#device_sel').attr("value");
            jQuery.post("get_device_problems.php", {did:did}, function(data)
            {
                $('#prob_sel').attr('disabled', false);
                $('#prob_sel').html(data);
            });
        });
    });
</script>

get_customers_devices.php:

    <?php
include "FillSelect.class.php";
echo $fill->FillDeviceSelect($getdata->GetCustomersDevice($_POST['cid']));
?>

get_device_problems.hp:

 <?php
include "FillSelect.class.php";
echo   $fill- >FillProblemSelect($getdata->GetDeviceProblems($_POST['did'])) ;
?>

GetCustomerDevice和GetDeviceProblems:

function GetCustomersDevice($cid)
    {
        try
         {
            $db = new PDO('sqlite:base.db') ;
            $db->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING );
            $pquery = $db->prepare("SELECT * FROM Devices WHERE cid=:cid");
            $data = array("cid"=>$cid);
            $pquery -> execute($data) ;
            $rows = $pquery->fetchAll();
            return $rows;
        }
        catch(PDOException $e)
        {
            echo "Failure: " . $e->getMessage();            
        }
        $db = NULL;
    }
    function GetDeviceProblems($did)
    {
        try
         {
            $db = new PDO('sqlite:base.db') ;
            $db->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING );
            $pquery = $db->prepare("SELECT * FROM Problems p, Devices d WHERE d.did=:did");
            $data = array("did"=>$did);
            $pquery -> execute($data) ;
            $rows = $pquery->fetchAll();
            return $rows;
        }
        catch(PDOException $e)
        {
            echo "Failure: " . $e->getMessage();            
        }
        $db = NULL;
    }

最后FillDeviceSelect和FillProbemSelect:

function FillDeviceSelect($data)
    {
        $device = '<option value="0">Select option</option>';
        foreach($data as $row)
        {
            $device.= '<option value="'. $row['did'] .'">'. $row['device_name'] .'</option>';
        }
        return $device;
    }
function FillProblemSelect($data)
    {
        $problem = '<option value="0">Select option</option>';
        foreach($data as $row)
        {
            $problem .= '<option value="' . $row['pid']  .'">' .  $row['problem'] .'</option>';
        }
        return $problem;
    }

解决了它。非常非常愚蠢的错误,get_device_problems与我的页面不在同一个文件夹中,但我知道文件保存在另一个文件夹,这就是为什么我的postcall不起作用。。。

我不是肯定的,但我想这是因为你对device_sel有onChange效果,然后当你用来自customer_sel的onChange调用填充它时,你会更改它的默认值。试着从$('#device_sel').change(function()中解除第二次更改的绑定,看看你是否有同样的问题。(如果我完全误解了您的问题,我深表歉意)