Question

我在尝试生成一个显示客户名称，联系电话，电子邮件地址和任何未付工作总计（总和）至少500美元的客户的未付工作总成本的查询时遇到了很多麻烦。

然后我必须订购结果，以便最大的金额在顶部。

到目前为止，我已经提出了这个问题

SELECT 
  j.job_id, 
  c.name + c.surname, 
  c.phone, 
  c.email_address, 
  SUM(jt.cost * (DATEDIFF(mi, timestart, timecomplete)))
FROM client as c 
LEFT OUTER JOIN job AS j ON c.tax_file_number = c.tax_file_number
LEFT OUTER JOIN job_type AS jt ON j.jobtype_id = jt.jobtype_id
WHERE SUM(jt.cost * (DATEDIFF(mi, timecomplete, timestart) > 500
ORDER BY SUM(jt.cost * (DATEDIFF(mi, timecomplete, timestart))) DESC;

任何帮助将不胜感激

Answer 1

在这里猜测，因为你的要求不明确，但我想你想要这个：

WITH jobinfo as
(
   SELECT j.client_id, SUM(jt.cost * (DATEDIFF(mi, timecomplete, timestart))) as cost
   FROM job as j
   JOIN job_type as jt ON j.jobtype_id = jt.jobtype_id
   GROUP BY j.client_id
)
SELECT 
  c.client_id,
  c.name + c.surname, 
  c.phone, 
  c.email_address, 
  j.cost
FROM client as c 
JOIN jobinfo j ON c.client_id= j.client_id
WHERE j.cost > 500
ORDER BY j.cost DESC

Answer 2

declare _sum int,

设置_sum =（从job_type中选择SUM（jt.cost *（DATEDIFF（mi，timecomplete，timestart）））SELECT j.job_id，c.name + c.surname，c.phone，c.email_address，SUM （jt.cost *（DATEDIFF（mi，timestart，timecomplete）））FROM client as c LEFT OUTER JOIN job AS j ON c.tax_file_number = c.tax_file_number LEFT OUTER JOIN job_type AS jt ON j.jobtype_id = jt.jobtype_id WHERE _sum ＆GT; 500

Answer 3

请尝试以下方法。 SUM子句不允许使用聚合函数（COUNT，WHERE等），您应该使用HAVING。

SELECT 
  j.job_id, 
  c.name + c.surname, 
  c.phone, 
  c.email_address, 
  SUM(jt.cost * (DATEDIFF(mi, timestart, timecomplete)))
FROM client as c 
LEFT OUTER JOIN job AS j ON a.tax_file_number = c.tax_file_number
LEFT OUTER JOIN job_type AS jt ON j.jobtype_id = jt.jobtype_id
HAVING SUM(jt.cost * (DATEDIFF(mi, timecomplete, timestart) > 500
ORDER BY SUM(jt.cost * (DATEDIFF(mi, timecomplete, timestart))) DESC;

修改已修复ON a.tax_file_number = c.tax_file_number

创建SQL查询以显示客户欠款

3 个答案: