如何确定ajax的URL

时间:2014-05-12 04:04:08

标签: ajax codeigniter

如标题中所述......我无法确定我是否正确执行了后期操作,或者操作是否正确执行,那么为什么值为空,因为我在传入之前检查了该值后期操作......这是我的代码:

脚本:

$.ajax({
    url: "<?php echo base_url();?>/Index/viewDayDocuments",
    type: 'post',
    data: {currentDay: 'currentDay', currentMonth: 'currentMonth', currentYear: 'currentYear'},
    success: function(result){
        $('.list').text('');
        $('.list').remove();
        $(".listIncoming").html("<p class = 'list'>This is the: "+ result +"</p>");
        $("#myform").show(500);
    }
 });

控制器代码,它返回一个返回值:

    $data['day'] = $_POST['currentDay'];
        $data['month'] =  $_POST['currentMonth'];
        $data['year'] =  $_POST['currentYear'];

        $date = $data['year']."-".$data['month']."-".$data['day'];

        $this->load->model('search_form');
        $output['details'] = $this->search_form->searchDateRetrievedIncoming($date);

        return $data;

1 个答案:

答案 0 :(得分:0)

您的ajax请求需要一个以字符串形式回显的字符串。

  $data['day'] = $_POST['currentDay'];
    $data['month'] =  $_POST['currentMonth'];
    $data['year'] =  $_POST['currentYear'];

    $date = $data['year']."-".$data['month']."-".$data['day'];

    $this->load->model('search_form');
    $output['details'] = $this->search_form->searchDateRetrievedIncoming($date);

    echo json_encode($data);

如果数组不是数据格式(例如JSON),则无法正确回显数组。

现在我们,在你的javascript中获取数据

$.ajax({
url: "<?php echo base_url();?>/Index/viewDayDocuments",
type: 'post',
data: {currentDay: 'currentDay', currentMonth: 'currentMonth', currentYear: 'currentYear'},
success: function(result){
    $('.list').text('');
    $('.list').remove();
    date = JSON.parse(result);
    $(".listIncoming").html("<p class = 'list'>This is the: "+ date.month +"/"+date.day+ "/"+date.year +"</p>");
    $("#myform").show(500);
  } 
 });

在这个模块中,我将STRING从PHP文件转换为JSON对象,以便正确读取。