MySQL 从两个表中除法计算答案

【问题标题】：MySQL divide calculation from two tablesMySQL 从两个表中除法计算
【发布时间】：2020-06-12 13:23:56
【问题描述】：

对于我的页面，我必须将两个表中的值分开。

我从表一（游戏）中得到计算的第一个值，其中两个值相加。从表二 (game_stats) 我得到发票的第二个值。 Rounds 和 Kills 值现在应该在 game_stats_more 表中划分并保存或更新为一个值。

我目前的问题是，当我计算值时，我得到一个错误，即 AS 值无法识别。我的选择错误是从哪里得到的？

SELECT SUM(game.match_score_team_1 + game.match_score_team_2) AS Rounds, game_stats.match_stats_kills AS Kills, Rounds / Kills AS KPR FROM game INNER JOIN game_stats WHERE game.match_id = 1 AND game_stats.user_id = 1

错误信息：

#1054 - Unknown table field 'Rounds' in field list

示例架构：

CREATE TABLE `game` (
  `match_id` int(3) NOT NULL,
  `team_id` int(3) NOT NULL,
  `match_team_2` varchar(255) COLLATE utf8_german2_ci NOT NULL,
  `match_score_team_1` int(2) NOT NULL,
  `match_score_team_2` int(2) NOT NULL,
  `match_score_role_id` int(3) NOT NULL,
  `match_role_id` int(3) NOT NULL,
  `date` date NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_german2_ci;

INSERT INTO `game` (`match_id`, `team_id`, `match_team_2`, `match_score_team_1`, `match_score_team_2`, `match_score_role_id`, `match_role_id`, `date`) VALUES
(1, 7, 'TestGegner', 7, 2, 1, 1, '2020-06-10'),

CREATE TABLE `game_stats` (
  `match_stats_id` int(3) NOT NULL,
  `match_id` int(3) NOT NULL,
  `user_id` int(3) NOT NULL,
  `match_stats_kills` int(2) NOT NULL,
  `match_stats_deaths` int(2) NOT NULL,
  `match_stats_entry_kill` int(2) NOT NULL,
  `match_stats_entry_death` int(2) NOT NULL,
  `match_stats_clutch` int(2) NOT NULL,
  `match_stats_plants` int(2) NOT NULL,
  `match_stats_hs` int(2) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_german2_ci;

INSERT INTO `game_stats` (`match_stats_id`, `match_id`, `user_id`, `match_stats_kills`, `match_stats_deaths`, `match_stats_entry_kill`, `match_stats_entry_death`, `match_stats_clutch`, `match_stats_plants`, `match_stats_hs`) VALUES
(1, 1, 1, 7, 5, 1, 3, 1, 4, 4),
(2, 1, 2, 6, 6, 2, 2, 0, 1, 3);

CREATE TABLE `game_stats_more` (
  `game_stats_id` int(3) NOT NULL,
  `game_id` int(3) NOT NULL,
  `game_stats_more_kpr` int(3) NOT NULL,
  `game_stats_more_hsp` int(3) NOT NULL,
  `game_stats_more_kd` int(3) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_german2_ci;

INSERT INTO `game_stats_more` (`game_stats_id`, `game_id`, `game_stats_more_kpr`, `game_stats_more_hsp`, `game_stats_more_kd`) VALUES
(1, 1, 0, 0, 0);

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标签： mysql

【解决方案1】：

您不能在同一选择语句中使用为字段定义的别名

    SELECT (game.match_score_team_1 + game.match_score_team_2) AS Rounds, 
           game_stats.match_stats_kills AS Kills, 
           (game.match_score_team_1 + game.match_score_team_2) / 
            game_stats.match_stats_kills AS KPR 
    FROM game 
    INNER JOIN game_stats ON game.match_id=game_stats.match_id
    WHERE game.match_id = 1 AND game_stats.user_id = 1

【讨论】：