Question

我试图为每个学习者排一行，格式如下：

<div id="left">
        <ul id="menu">
            <li><img src="http://www.w3schools.com/bootstrap/img_chania.jpg"/></li>
            <li><img src="http://www.w3schools.com/bootstrap/img_chania2.jpg"/></li>
            <li><img src="http://www.w3schools.com/bootstrap/img_flower2.jpg"/></li>
            <li><img src="http://www.w3schools.com/bootstrap/img_flower.jpg"/></li>
            <li><img src="http://www.w3schools.com/bootstrap/img_flower2.jpg"/></li>

        </ul>
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    <img id ="right" src="http://www.w3schools.com/bootstrap/img_chania.jpg"/>

学习者有很多提供者，提供者有很多学习者在每个提供者，学习者有很多结果，结果属于一个学习者结果有一个标准，一个标准属于无/多个结果
标准有很多职业

nsn|last_name|first_name|middle_name|gender|ethnicities|providers|qualifications|vocation_PI|vocation_SI|.....

我无法得到每个职业的计数。

我在每个计数字段中得到相同的值（不确定这是否是所有计数的总数或者是什么）

预期产出：

select l.nsn, l.last_name,l.first_name,l.middle_name, l.gender, GROUP_CONCAT(distinct e.name SEPARATOR '|') as ethnicities, GROUP_CONCAT(distinct p.name SEPARATOR '|') as providers, count(case v.code when 'PI' then 1 else 0 end) as v_PI,count(case v.code when 'SI' then 1 else 0 end) as v_SI,count(case v.code when 'CR' then 1 else 0 end) as v_CR, GROUP_CONCAT(distinct q.name SEPARATOR '|') as qualifications
from learners l 
join learner_ethnicity le on l.id = le.learner_id
join ethnicities e on le.ethnicity_id = e.id
join learner_provider lp on l.id = lp.learner_id
join providers p on lp.provider_id = p.id
join results r on lp.id = r.learner_provider_id
join standards s on r.standard_id = s.id
join standard_vocation sv on s.id = sv.standard_id
join vocations v on sv.vocation_id = v.id
join learner_qualification lq on l.id = lq.learner_id
join qualifications q on lq.qualification_id = q.id
group by l.id

实际输出：

123451234 | Doe | John | James | M | European|Maori | SecondaryProvider|TertiaryProvider | NCEA Level 1|NCEA Level 2|Farming PHD | 185 | 430 | .....

有什么建议吗？

感谢。

Answer 1

用SUM替换你的计数你会得到你想要的东西

select l.nsn, l.last_name,l.first_name,l.middle_name, l.gender, GROUP_CONCAT(distinct e.name SEPARATOR '|') as ethnicities, GROUP_CONCAT(distinct p.name SEPARATOR '|') as providers, sum(case v.code when 'PI' then 1 else 0 end) as v_PI, sum(case v.code when 'SI' then 1 else 0 end) as v_SI,sum(case v.code when 'CR' then 1 else 0 end) as v_CR, GROUP_CONCAT(distinct q.name SEPARATOR '|') as qualifications
from learners l 
join learner_ethnicity le on l.id = le.learner_id
join ethnicities e on le.ethnicity_id = e.id
join learner_provider lp on l.id = lp.learner_id
join providers p on lp.provider_id = p.id
join results r on lp.id = r.learner_provider_id
join standards s on r.standard_id = s.id
join standard_vocation sv on s.id = sv.standard_id
join vocations v on sv.vocation_id = v.id
join learner_qualification lq on l.id = lq.learner_id
join qualifications q on lq.qualification_id = q.id
group by l.id

从连接表中分组和计数

1 个答案: