Question

我有一个像这样的mysql表：

CREATE TABLE IF NOT EXISTS `entries` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `domain_name` varchar(255) NOT NULL,
  `presentation_name` varchar(255) NOT NULL,
  `total_score` mediumint(9) NOT NULL,
  `times_played` mediumint(9) NOT NULL,
  `avg_score` float(4,2) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=206 ;

我正在用PHP更新数据但查询计算错误的＆＃39; avg_score＆＃39;在更新时。可以说，整行看起来像这样：

id   domain_name   presentation_name  total_score  times_played   avg_score
1    test.com      test               30           3              10.00

但是当我使用新数据运行此更新查询时：

$score = 6;
$query = "UPDATE `entries` 
                        SET 
                            total_score = (total_score + $score), 
                            times_played = (times_played + 1), 
                            avg_score = ( (total_score + $score) / (times_played + 1) ) 
                    WHERE id = '1'";

它变成这样：

id   domain_name   presentation_name  total_score  times_played   avg_score
1    test.com      test               36           4              8.40

你可以看到＆＃39; avg_score＆＃39;是错的（应该是9.00）。我在phpmyadmin中尝试了相同的查询并得到了相同的错误计算。我真的无法在这里找到我做错的事。

Answer 1

见

total_score = (total_score + $score),

36 = 30 + 6

times_played = (times_played + 1),

4 = 3 + 1

然后你做

avg_score = ( (total_score + $score) / (times_played + 1) )

8.4 =（36 + 6）/ 5

是对的！

Answer 2

来自MySQL manual：

以下语句中的第二个赋值将col2设置为当前（更新）col1值，而不是原始col1值。结果是col1和col2具有相同的值。 此行为不同于标准SQL 。
UPDATE t1 SET col1 = col1 + 1, col2 = col1;

因此，您可以在更新平均值时省略+分数和+ 1：

UPDATE `entries` SET 
    total_score  = total_score  + $score, 
    times_played = times_played + 1, 
    avg_score    = total_score / times_played
WHERE id = '1'

Answer 3

看起来您的前两列已更新，更新第三列时会使用更新的值，请注意8.4 = (36 + 6) / (4 + 1)

因此，您不需要第+1和+ $score作为第三列。

虽然你真的不应该在数据库中存储重复的数据，因为这只会导致这样的问题。

只需在php或mysql中计算您需要的平均值。

Answer 4

尝试：改变顺序（首选方法）

                    SET 
                        avg_score = ( (total_score + $score) / (times_played + 1) ) ,
                        total_score = (total_score + $score), 
                        times_played = (times_played + 1)

或：

                    SET 
                        total_score = (total_score + $score), 
                        times_played = (times_played + 1), 
                        avg_score = ( (total_score) / (times_played) )

您正在引用已在后续计算中调整的字段，但假设它们保持不变。

Answer 5

看起来这些更新正在逐步完成。

即，总得分为36，次数达到4，然后avg_score的计算完成（36 + 6,4 + 1）= 42/5 = 8.4

尝试用另一句话更新avg_score。

不知道这是做什么的。

Answer 6

更改：

$score = 6; $query = "UPDATE `entries` SET total_score = (total_score + $score), times_played = (times_played + 1), avg_score = ( (total_score + $score) / (times_played + 1) ) WHERE id = '1'";

要

$score = 6; $query = "UPDATE `entries` SET total_score = (total_score + $score), times_played = (times_played + 1), avg_score = (total_score /times_played ) WHERE id = '1'";

SQL Fiddle

MySQL更新计算错误的划分

6 个答案: