Php MySQL更新除法计算错误_Php_Mysql_Sql_Sql Update

Php MySQL更新除法计算错误

php mysql sql

Php MySQL更新除法计算错误,php,mysql,sql,sql-update,Php,Mysql,Sql,Sql Update,我有一个mysql表，如下所示： 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) N

我有一个mysql表，如下所示：

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 ;

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

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

我正在用PHP更新数据
但在更新时，查询计算“平均分数”时出错。比如说，整行看起来像这样：

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 ;

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

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

但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：

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 ;

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

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

你可以看到“平均分数”是错误的（应该是9.00）。我在phpmyadmin中尝试了相同的查询，但得到了相同的错误计算。真的找不到我做错了什么。

看起来更新是以增量方式进行的

也就是说，总分变为36，播放次数变为4，然后计算平均分数（36+6，4+1）=42/5=8.4

尝试更新另一句话中的平均分数

但不知道是什么在这样做。

看起来更新是以增量方式进行的

也就是说，总分变为36，播放次数变为4，然后计算平均分数（36+6，4+1）=42/5=8.4

尝试更新另一句话中的平均分数

但是，不知道是什么原因。

看起来前两列得到了更新，更新后的值在更新第三列时使用，请注意

8.4=（36+6）/（4+1）

因此，第三列不需要

+1

和

+score

尽管您确实不应该在数据库中存储重复数据，因为这只会导致类似这样的问题

只要在php或mysql中计算所需的平均值即可。

看起来前两列已更新，更新后的值将在更新第三列时使用，请注意

8.4=（36+6）/（4+1）

因此，第三列不需要

+1

和

+score

尽管您确实不应该在数据库中存储重复数据，因为这只会导致类似这样的问题

在php或mysql中，只要在需要时计算平均值即可。

请参阅

total_score = (total_score + $score),

36=30+6

times_played = (times_played + 1),

4=3+1

那你呢

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

8.4=（36+6）/5

这是正确的

见

total_score = (total_score + $score),

36=30+6

times_played = (times_played + 1),

4=3+1

那你呢

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

8.4=（36+6）/5

这是正确的

试试看：更改顺序（首选方法）

或：

您引用的字段已在后续计算中进行了调整，但假设这些字段保持不变。

请尝试：更改顺序（首选方法）

或：

您引用的字段已在后续计算中进行了调整，但假设它们保持不变。

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'";

  $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'";

至：

$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'";

  $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'";

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'";

  $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'";

至：

$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'";

  $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'";

来自：

下面语句中的第二个赋值将col2设置为当前（更新的）col1值，而不是原始col1值。结果 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忽略+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'

从：

下面语句中的第二个赋值将col2设置为当前（更新的）col1值，而不是原始col1值。结果 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忽略+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'

尝试将

avg_分数

字段转换为

十进制（4,2）

字段。在进行此类计算时，浮动可能很棘手。没有浮动不是问题。请尝试将

avg_分数

字段转换为

十进制（4,2）

字段。在进行这样的计算时，浮动可能会很棘手。没有浮动不是问题。