Question

<?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();
    }
?>

代码应返回：

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

返回时：

{"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......这里的问题在哪里？

提前致谢！

Answer 1

fetchAll（）returns both（请注意'fetch_style'参数注释/注释）字符串和默认情况下来自查询结果的数字键控数据。如果只需要字符串版本，则必须执行

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

JSON数据格式不正确：从mysql获取

1 个答案: