MySQL SELECT查詢結果轉換為JSON

Question

我有運行以下SELECT查詢的模型。

模型

public function get_details($start, $limit) {
    $sql = $this->db->query("SELECT E.`exam-id` AS `examid`, T.`topic-name` AS `topic`,E.`date`, COUNT(E.`exam-id`) AS `total` FROM `exam` E, `exam-topic` T WHERE E.`exam-dbname`=T.`topic-dbname` GROUP BY `exam-id` ORDER BY `exam-id` DESC LIMIT $start, $limit;");
    if($sql->num_rows() > 0){
        return $sql;
    }
    else{
        return FALSE;
    }
}

我的控制器

public function get_details() {
    //pagination
    $config =   array(
        'base_url'          =>  base_url()."exam/create/created/index",
        'total_rows'        =>  $this->exam->get_total(),
        'first_link'        =>  "First",
        'last_link'         =>  "Last",
        'per_page'          =>  2,
        'uri_segment'       =>  5,
        'full_tag_open'     =>  "<div class='paginate'>",
        'full_tag_close'    =>  "</div>",
        'num_tag1_open'     =>  '<a>',
        'num_tag1_close'    =>  '</a>',
        'cur_tag_open'      =>  "<a class='current'>",
        'cur_tag_close'     =>  "</a>",
        'next_tag1_open'    =>  "<a>",
        'next_tag1_close'   =>  "</a>",
        'prev_tag1_open'    =>  "<a>",
        'prev_tag1_close'   =>  "</a>",
        'first_tag1_open'   =>  "<a>",
        'first_tag1_close'  =>  "</a>",
        'last_tag1_open'    =>  "<a>",
        'last_tag1_close'   =>  "</a>"
    );
    $page=($this->uri->segment(5) != '') ? $this->uri->segment(5) : 0;
    $this->pagination->initialize($config);

    $data   =   array(
        'query' =>  $this->exam->get_details($page, $config['per_page']),
        'links' =>  $this->pagination->create_links()
    );
    echo json_encode($data);
}

在視圖頁面中，我使用AJAX捕獲結果。

我的觀點

function get_details() {
            $.ajax({
                url     :   "<?php echo base_url(); ?>exam/create/created/get_details.html",
                dataType:   "JSON",
                success :   function(data) {
                    var obj = $.parseJSON(data.query);
                    $.each(obj, function(key, val) {
                        alert(key+"\n"+val.uid);
                    });
                }
            });         }

現在，我不知道如何在JSON中獲取data[query]的結果，因為data[query]包含MySQL返回的多個結果。

提前致謝。

Answer 1

在您的控制器中應該是代碼。

 $data   =   array(
        'query' =>  json_encode($this->exam->get_details($page, $config['per_page'])),
        'links' =>  $this->pagination->create_links()
    );
echo json_encode($data);

因為您正在使用json parse解析data.query，所以從數據庫獲取的數據應為json。

謝謝阿米特