讓我前言本帶我知道這個問題已經被問很多次，就在這里（即這里的其他頁面中）和我已經通過每一個回答的拖網，并試圖解決方案，但一直收到一個空的JSON數組。我有一個 mysql 查詢，如下所示：$myQuery = $wpdb->get_results('SELECT player_name, player_team, SUM(ruckContests) AS contests FROM afl_master WHERE player_team = "COLL" AND year = 2019 AND ruckContests > 0 GROUP BY player_id ORDER BY contests DESC LIMIT 2');此查詢的輸出如下：Array ( [0] => stdClass Object ( [player_name] => Brodie Grundy [player_team] => COLL [contests] => 661 ) [1] => stdClass Object ( [player_name] => Mason Cox [player_team] => COLL [contests] => 51 ) )我想要做的是在 php 中將其轉換為以下 json 對象：{ player_name: "Brodie Grundy", player_team: "COLL", contests: 661},{ player_name: "Mason Cox", player_team: "COLL", contests: 51}我已經嘗試mysql_fetch_array()過以及fetch_assoc()使用這里的方法，但無濟于事。我很感激任何可用的幫助。評論中推薦的這個簡單解決方案有效：json_encode($myQuery)
3 回答
縹緲止盈
TA貢獻2041條經驗 獲得超4個贊
只需使用fetch_array()或 (array) $myQuery將結果轉換為數組，然后：
json_encode($myQuery);
json_last_error(); // to see the error
茅侃侃
TA貢獻1842條經驗 獲得超21個贊
$JsonData=array();
foreach($myQuery as $val){
$JsonData[]='
{
"player_name": "'.$val->player_name.'",
"player_team": "'.$val->player_team.'",
"contests": $val->contests
}
';
}
$expectedJson=implode(',',$JsonData);
蠱毒傳說
TA貢獻1895條經驗 獲得超3個贊
您可以在單個 SQL 語句中執行此操作（為了更好地理解，我剛剛將其格式化為多行）。內部Select語句使用你的查詢作為子查詢，同時使用 JSON_OBJECT生成每一行的JSON對象，最后使用GROUP_CONCAT將每一行用逗號組合起來，然后將“[”和“]”添加到最終結果中用于創建完美的 JSON 數組。（另一種方式，除了json_encode($myQuery)，更簡單的版本:) ）。
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
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