Question

我有一个包含多个Tabletennis匹配项的数据库。要进行排名，我需要计算多列。

我想加起来独特的回合，以了解人们参加的频率。我也想将所有获奖的游戏加起来。有时，我们玩“超级回合”（SR），获胜游戏将翻倍。 SR默认设置为1。

要知道总得分，我想将两个结果（参与和total_games）加起来。

我现在所拥有的：

$sql_user = $conn->query("
SELECT user_id, count(distinct round) as participation, 
sum(IFNULL(GAMES_WIN,0) * SR) AS total_games 
FROM tt_game group by user_id ORDER BY user_id ASC");

是否可以执行以下操作：

$sql_user = $conn->query("
SELECT user_id, count(distinct round) as participation, 
sum(IFNULL(GAMES_WIN,0) * SR) AS total_games 
sum(participation + total_games) AS total_score
FROM tt_game group by user_id ORDER BY total_score DESC");

我该如何回应这些结果？非常感谢！

Answer 1

否，您不能在SELECT子句本身中引用SELECT子句中定义的别名。

您可以重复以下表达式：

SELECT
  user_id,
  count(distinct round) as participation, 
  sum(IFNULL(GAMES_WIN,0) * SR) AS total_games, 
  count(distinct round) + sum(IFNULL(GAMES_WIN,0) * SR) AS total_score
FROM tt_game 
GROUP BY user_id
ORDER BY total_score DESC;

或使用子查询：

SELECT user_id, participation, total_games, participation + total_game AS total_score
FROM
(
  SELECT
    user_id,
    count(distinct round) as participation, 
    sum(IFNULL(GAMES_WIN,0) * SR) AS total_games
  FROM tt_game 
  GROUP BY user_id
) aggregated
ORDER BY total_score DESC;

Answer 2

另一个选择呢？

我认为您需要另一个选择，您可以在其中计算total_score。

尝试这样的事情：

$sql_user = $conn->query("
SELECT t.user_id, t.participation, t.total_games, (participation + total_games) AS total_score
FROM (SELECT user_id, count(distinct round) as participation, sum(IFNULL(GAMES_WIN, 0) * SR) AS total_games
      FROM tt_game
      group by user_id
      ORDER BY total_score DESC) AS t
");

sql将两列加起来，然后将两者加起来

2 个答案: