JSON数据：从MySQL

<?php 
    require 'dbinfo.php'; 
    try { 
     $db = new PDO($dsn, $username, $password); 
     $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     $sth = $db->query("SELECT * FROM user_tracks"); 
     $loc = $sth->fetchAll();  
     $locations = array_values($loc); 
     echo json_encode(array("user"=>($locations))); 
    } catch (Exception $e) { 
     echo $e->getMessage(); 
    } 
?>

获取的代码应该返回：JSON数据：从MySQL

{"user":[{"id":"1","Latitude":"12.9555033333","Longitude":"80.2461883333","Time":"06:32:57","Date":"2012-03-13","Speed":"0","Course":"183.92"},{...},{....}]}

当它返回：

{"user":[{"id":"1","0":"1","Latitude":"12.9555033333","1":"12.9555033333","Longitude":"80.2461883333","2":"80.2461883333","Time":"06:32:57","3":"06:32:57","Date":"2012-03-13","4":"2012-03-13","Speed":"0","5":"0","Course":"183.92","6":"183.92"},{...},{....}]}

我不确定发生了什么......的问题在哪里？

在此先感谢！

来源

2012-04-03 Karthick

fetchAll（）returns both（注意'fetch_style'参数注释/注释）字符串和数字键值来自查询结果的默认数据。如果你只想要字符串版本，你必须做

$loc = $sth->fetchAll(PDO::FETCH_CLASS);

来源

2012-04-03 03:26:54

它的工作原理！它的另一种方式.. fetchAll（PDO :: FETCH_CLASS）;谢谢您的帮助！ – Karthick 2012-04-03 04:02:38

Woops，对。我的错字。我会纠正答案。 – 2012-04-03 04:52:10

JSON数据：从MySQL

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