如何在ajax的结果函数中使用php json_encode($var)答案

【问题标题】：how to use php json_encode($var) in result function in ajax如何在ajax的结果函数中使用php json_encode($var)
【发布时间】：2013-12-27 05:17:09
【问题描述】：

//这是ajax代码

var survey_id = $(this).val();
var user_id = $('#SurveyFilterUserId').val();
            $.ajax({
                type        : 'POST',
                dataTyle    : 'Json',
                url         : '<?php echo BASEURL; ?>/admin/Users/survey_filters_edit/',
                data        : {
                    'data[find][survey_id]'     :survey_id,
                    'data[find][user_id]'       :user_id
                },
                success :   function(result){
                    // Here is the problem
                }
            });

// 这是我的 CakePHP 查找查询

$surveyFilter = $this->SurveyFilter->find('all', 
                                        array( 'conditions'=>
                                          array('SurveyFilter.survey_id' => $survey_id,
                                                'SurveyFilter.user_id' => $user_id)));

            echo json_encode($surveyFilter);
            die();

// 这是我的输出 // 这个输出是查找查询的 json_encode 输出

[{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"1","object_type":"devices","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}},{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"2","object_type":"devices","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}},{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"3","object_type":"devices","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}},{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"1","object_type":"alerts","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}},{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"10","object_type":"alerts","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}},{"SurveyFilter":{"survey_id":"1","user_id":"8","object_id":"11","object_type":"alerts","created":"2013-12-27 09:34:04","modified":"2013-12-27 09:34:04"}}]

【问题讨论】：

标签： php jquery ajax json cakephp

【解决方案1】：

我假设您正在尝试在您的 javascript 函数中访问 JSON 数据。试试这个

          success :   function(result){
                // Here is the problem
                var returned_data = $.parseJSON(result);

                //Now manipulate parsed JSON data.
                //For test purposes log returned data

                console.log(returned_data);

                //access any field like this, for example access survey_id.
                returned_data.SurveyFilter.survey_id  
            }

【讨论】：

【解决方案2】：

您可以通过“.”访问这些 json 数据。像这样：

 var survey_id = $(this).val();
    var user_id = $('#SurveyFilterUserId').val();
                $.ajax({
                    type        : 'POST',
                    dataTyle    : 'Json',
                    url         : '<?php echo BASEURL; ?>/admin/Users/survey_filters_edit/',
                    data        : {
                        'data[find][survey_id]'     :survey_id,
                        'data[find][user_id]'       :user_id
                    },
                    success :   function(result){
                       console.log(result.survey_id);
                    }
                });

【讨论】：

【解决方案3】：

success :   function(data) {
               $.each(data, function (key, val) {
                        alert(val.SurveyFilter);
               })

}

您必须处理嵌套对象，请记住这一点。

【讨论】：

【解决方案4】：

试试这样，你有错别字

更改数据类型 =>数据类型

$.ajax({
    type        : 'POST',
    dataType    : 'Json',
    url         : '<?php echo BASEURL; ?>/admin/Users/survey_filters_edit/',
    data        : {
        'data[find][survey_id]'     :survey_id,
        'data[find][user_id]'       :user_id
    },
    success :   function(result){
        $.each(result,function(k,obj){
                console.log(obj.SurveyFilter.survey_id);
                console.log(obj.SurveyFilter.user_id);
        })
    }
});

【讨论】：

【解决方案5】：

你可以使用Firebug -> network 查看json类型的结果，然后你就可以知道结果的结构并进行下一步

【讨论】：