Question

我有2张桌子 - 汽车制造商和汽车模型。我从数据库中提取数据。当我点击第一个表格中的制作人时，我希望第二个表格只显示来自该制作人的汽车。我尝试使用ajax，但ajax请求似乎没有覆盖$ DBcontent变量，因此第二个表不会被过滤。但是当我使用console.log（数据）时;在成功的Ajax上它显示正确的记录。我误会了什么吗？我省略了SQL查询，因为它们简单且不相关。

控制器CarListController.php：

<?php

class CarListController extends CI_Controller 
{   
    public function index()
    {
                $this->load->model('OutputFromDatabaseModel');
                $this->load->helper(array('form', 'url'));

                $query = $this->OutputFromDatabaseModel->getProducers();
                $DBcontent['PROD'] =  $query;
                $query = $this->OutputFromDatabaseModel->getCars(); // fill with all Cars first
                $DBcontent['CARS'] =  $query; 

                $this->load->view('CarListView', $DBcontent);
    }  
}
?>

查看CarListView.php：

<script>
var URL = "(myURL)/index.php/Process/FilterCars";
$(document).ready(function() 
{
    $(".test").click(function(e) //row from Producer table
    {
        $ID = this.id;
        $.ajax({
                type: "POST",
                url: URL,
                data: {id:$ID},
                dataType    : "json",
            });

        showModels = true;
        if (showModels) // only show the car models table when clicked on a producer
        {
            document.getElementById("modelTable").style.display="";
        }         
    });
});
</script>
<body>
   <div class="col-xs-2">
    <table id="producerTable" class="table table-striped table-bordered table-hover table-responsive">
     <tr><td><strong>Producer</strong></td></tr> 
     <?php foreach($PROD as $producer){?>
     <tr class="test" id="<?=$producer->ProducerID?>"><td><?=$producer->ProducerName;?></td></tr>     
        <?php }?>  
    </table>
   </div> 
   <div class="col-xs-3">
    <table id="modelTable" style="display:none" class="table table-striped table-bordered table-hover table-responsive">
     <tr><td><strong>Car Model</strong></td></tr> 
     <?php foreach($CARS as $car){?>
     <tr id="<?=$car->CarId?>"><td><?=$car->Model;?></td></tr>     
        <?php }?>  
    </table>
   </div> 
</body>

AjaxProcess Process.php：

<?php
class Process extends CI_Controller 
{   
    public function index(){
        $this->load->view('CarListView', $DBcontent); //tried to put it everywhere 
    }  
    function FilterCars(){
        $this->load->model('OutputFromDatabaseModel');
        $this->load->helper(array('form', 'url'));
        $ProducerID = $this->input->post('id');
        $query = $this->OutputFromDatabaseModel->getCarsOfProducer($ProducerID); //filtered by Producer
        $DBcontent['CARS'] =  $query;
    }
}
?>

如何使用更新的$ DBcontent变量启动View？如果采用错误的方法，有哪些替代方案。

Answer 1

您的第二张桌子可能仍然显示所有车辆，因为您的ajax请求没有做任何事情，它只是将数据发布到服务器，您没有使用它返回的数据。你只需设置＆＃34; modelTable＆＃34;在ajax调用后可见。请参阅Jquery Ajax中的success函数http://api.jquery.com/jquery.ajax/

$(".test").click(function(e) //row from Producer table
   {
    $ID = this.id;
    $.ajax({
            type: "POST",
            url: URL,
            data: {id:$ID},
            dataType    : "json",
            success:function(data){ // function to call when the ajax call sucessful
            // i simply replace the content of the table with the response with ajax, 
            //you can use the data to create html here and append it into the table.
            $('#modelTable').html(data); 

            }
        });
  }

AJAX请求不更新第二个表（codeigniter）

1 个答案: