MySQL重新格式化mysql结果来满足json hansontable | MySQL重新格式化mysql结果来满足json

本文介绍了MySQL重新格式化mysql结果来满足json hansontable的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

根据我以前解决的问题：。
$ b

into

["Superior","Deluxe - City View","Deluxe - Balcony","Junior Suite","Andaman Studio"]

From these codes:

$sql_rName="select title from room_db where hotel='1' order by id asc";
$result_rName=mysql_db_query($dbname,$sql_rName);
while($rec_rName=mysql_fetch_array($result_rName)){
    $_rName=$rec_rName['title'];
    $_array[]=$_rName;
}
echo "{\"data\": ".json_encode($_array)."}";

mysql Table : room_db

Please suggest.

ps. Thanks to Olaf Dietsche for all of these help.

解决方案

To make a correct JSON, Try :`

$result="select title from room_db where hotel='1' order by id asc";
$messages = array();
            while($message_data = mysql_fetch_assoc($result)) {
                $message = array(
                'id' => $message_data['userid'],
                'title' => $message_data['title']
                );
                $messages[] = $message;
                }
                echo json_encode($messages);
            }
`

and on the reseiver side do this :

`
data1=$.parseJSON(data);

            if(data1.length===0){

                $('#table > #table_body').append('<tr><td colspan="4" align="center" style="color:red">NO matching data </td></tr>');
                }
        else{
            for(var i=0;i<data1.length;i++)
            {
                $('#table > #table_body').append('<tr id="' + data1[i]['id'] +'"> <td id="' + data1[i]['id'] +'" align="center" <td>'+data1[i]['title']+'</td> </tr>');
            }
            }
            $('#table').append('</tbody>');

    `

这篇关于MySQL重新格式化mysql结果来满足json hansontable的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！