Question

我已经提到了：

＆安培;

我想对别名字段进行一些计算但是会抛出错误

以下是我的查询

 select j.*,
    (select sum(stars) from ratingstar where jobid=j.id) as stars,
    (select count(*) from ratingstar where jobid=j.id) as count,
    ((stars/(count*5)*5)) as rating     //I have problem here if I remove this it works fine 
from jobs j inner join proposals p on p.jobid=j.id 
            inner join us_signup u on u.id=p.userid 
            inner join hired h on h.proposalid=p.id 
            where h.status="finished"

但它会抛出错误

此选择行中的错误

select j.*,
    (select sum(stars) from ratingstar where jobid=j.id) as stars,
    (select count(*) from ratingstar where jobid=j.id) as count,
    ((stars/(count*5)*5)) as rating     //I have problem here if I remove this it works fine

，错误是

致命错误：未捕获的异常＆＃39;异常＆＃39;消息＆＃39;未知专栏＆＃39;明星＆＃39;在＆＃39;字段列表＆＃39;查询：选择j。，（选择总和（星星）来自ratingstar，其中jobid = j.id）作为星星，（从中选择计数（）） ratingtar，其中jobid = j.id）作为计数，（（星/（计数* 5）* 5））作为评级来自作业j内部联接提议p on p.jobid = j.id内部联接 us_signup u on u.id = p.userid inner join hired h on h.proposalid = p.id 其中h.status =＆＃34;完成＆＃34;＆＃39;在第3637行的E：\ wamp \ www \ sugumar \ mysuite \ includes \ classes \ MysqliDb.php

Answer 1

我认为你最后错过了a）

((stars/(count*5)*5))

编辑*你没有错过它，你刚刚在开头添加了一个无用的

(stars/(count*5)*5)

Answer 2

Count是一个保留的SQL术语，如果你想使用is作为一个字段，你应该这样做：

select j.*,
    (select sum(stars) from ratingstar where jobid=j.id) as stars,
    (select count(*) from ratingstar where jobid=j.id) as count,
    ((stars/(`count`*5)*5)) as rating     //I have problem here if I remove this it works fine 
from jobs j inner join proposals p on p.jobid=j.id 
            inner join us_signup u on u.id=p.userid 
            inner join hired h on h.proposalid=p.id 
            where h.status="finished"

Answer 3

由于count是保留字/函数，因此需要使用反引号引用它：

select j.*,
    (select sum(`stars`) from `ratingstar` where `jobid` = j.`id`) as `stars`,
    (select count(*) from `ratingstar` where `jobid` = j.`id`) as `count`,
    (`stars` / (`count` * 5) * 5) as `rating`
from `jobs` j 
inner join `proposals` p 
    on p.`jobid` = j.`id` 
inner join `us_signup` u 
    on u.`id` = p.`userid` 
inner join `hired` h 
    on h.`proposalid` = p.`id` 
where h.`status` = "finished"

mysql：计算mysql中的别名字段

3 个答案: