Question

让我以此开头作为我的序言，我知道在这里（例如here等页面）已经多次问过这个问题，我已经仔细阅读了每个答案并尝试了解决方法，但是一直空着json数组。

我有一个如下的mysql查询：

$myQuery = $wpdb->get_results('SELECT player_name, player_team, 
SUM(ruckContests) AS contests FROM afl_master 
WHERE player_team = "COLL" AND year = 2019 AND ruckContests > 0 
GROUP BY player_id ORDER BY contests DESC LIMIT 2');

此查询的输出如下：

Array ( [0] => stdClass Object ( [player_name] => Brodie Grundy [player_team] => COLL [contests] => 661 ) [1] => stdClass Object ( [player_name] => Mason Cox [player_team] => COLL [contests] => 51 ) )

我想要做的是将其转换为php中的以下json对象：

{
 player_name: "Brodie Grundy", 
 player_team: "COLL", 
 contests: 661
},
{
 player_name: "Mason Cox", 
 player_team: "COLL", 
 contests: 51
}

我已经使用here方法尝试过mysql_fetch_array()和fetch_assoc()，但无济于事。我将不胜感激。

在评论中推荐的这种简单解决方案有效：

json_encode($myQuery)

Answer 1

$JsonData=array();
foreach($myQuery as $val){


            $JsonData[]='
               {
                 "player_name": "'.$val->player_name.'", 
                 "player_team": "'.$val->player_team.'", 
                 "contests": $val->contests 
               }
             ';
}
$expectedJson=implode(',',$JsonData);

Answer 2

只需将结果转换为使用 fetch_array（）或（数组）$ myQuery的数组，然后：

json_encode($myQuery);
json_last_error(); // to see the error

Answer 3

您可以在单个SQL语句中执行此操作（为了更好地理解，我将其格式化为多行）。内部的Select语句将您的查询用作子查询，同时使用JSON_OBJECT生成每一行的JSON对象，最后使用GROUP_CONCAT将每一行与逗号合并，然后添加“ [“和”]“创建完美的JSON数组的最终结果。（换句话说，除了 json_encode（$ myQuery）之外，更简单的版本：）。）

SELECT CONCAT('[', 
    GROUP_CONCAT( 
        JSON_OBJECT( 
            'player_name',   player_name, 
            'player_team' ,  player_team, 
            'contests', contests) 
    ) , ']') as json  
FROM (
        SELECT  player_name,  player_team, SUM(ruckContests) as contests
        FROM afl_master 
        WHERE player_team = "COLL" AND year = 1995 AND ruckContests > 0 
        GROUP BY player_name, player_team 
        ORDER BY SUM(ruckContests) DESC 
        LIMIT 2 
) AS TT

通过PHP将MySQL转换为JSON

3 个答案: