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带有 where 子句和求和值的 Mysql 数据透视表

转载 作者:行者123 更新时间:2023-11-29 08:48:47 26 4
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SELECT * FROM ( 
select student_id, class_id,
sum(number*(1-abs(sign(subject-1)))) as sub1,
sum(number*(1-abs(sign(subject-2)))) as sub2,
sum(number*(1-abs(sign(subject-3)))) as sub3,
sum(number*(1-abs(sign(subject-4)))) as sub4,
sum(number*(1-abs(sign(subject-5)))) as sub5,
sum(number*(1-abs(sign(subject-6)))) as sub6
from result GROUP BY student_id,class_id) m
where class_id = '7'

它的查询工作正常。但如果主题值为空,则该主题中不会显示所有学生记录。我还需要再发烧一次。

最佳答案

我不知道是否有直接使用mysql的方法,但我可以建议你使用php

$sql = mysql_query("SELECT m.* FROM (  
select student_id, class_id,
sum(number*(1-abs(sign(subject-1)))) as sub1,
sum(number*(1-abs(sign(subject-2)))) as sub2,
sum(number*(1-abs(sign(subject-3)))) as sub3,
sum(number*(1-abs(sign(subject-4)))) as sub4,
sum(number*(1-abs(sign(subject-5)))) as sub5,
sum(number*(1-abs(sign(subject-6)))) as sub6
from result GROUP BY student_id,class_id) m
where m.class_id = '7'");

$row = mysql_fetch_array($sql);
$sum = 0;
for($i = 1; $i < count($row); $i++) {
$sum += isset($row['sub' . $i]) ? $row['sub' . $i] : 0;
}

关于带有 where 子句和求和值的 Mysql 数据透视表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11857715/

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