日期:2014-05-17  浏览次数:20865 次

急!两个问题,求大神
数据库地址:http://115.com/file/bejzce1g#RDCsetup.sql

1. 编写一个脚本,可以列出每个成员、已付费用、未付费用。将championship和social两类会员分别合计,并给出总和(划分2大类,是否有注册编号,若有,是CM;若没有,是SM。CM,SM分别统计,再给出总和所有的)
championship member和social member的区别是CM有Registrationnum
feespaid是已经付的费用,CM要交185元,SM要交55元
我写的代码是:
SELECT givenname, familyname, feespaid, 185-feespaid as "Fees", sum(feespaid), sum(185-feespaid)
FROM MEMBER
WHERE Registrationnum IS NOT NULL;
GROUP BY givenname, familyname;

总报错

2. FIXTURES表格中列有去年比赛的列表(俱乐部与俱乐部之间有比赛)。核实查证这些fixtures已被正确完成,并编写一个脚本来确认可能存在的问题。你故意去破坏fixtures表格,即数据改错几个地方,来检测这个脚本是否能查到错误。
(一个星期,一个球队比赛一次,比赛6伦,一轮有4个比赛。)
我的算法是:
- 四个球队,每个球队每个礼拜都有比赛(若此条成立,则每支球队必然一共打了6场比赛)
- 3个客场3个主场
所以结果输出两个表: 1. 计算每周比赛球队数量 2. 计算每个球队的主客场数量
 WEEK HOME AWAY  
---------- ---- ----  
  1 GG HH  
  1 RR SS  
  2 HH RR  
  2 SS GG  
  3 HH SS  
  3 RR GG  
  4 HH GG  
  4 SS RR  
  5 RR HH  
  5 GG SS  
  6 GG RR  
  6 SS HH  


第一个表代码是:
SELECT Weeks, Team,
CASE
WHEN Team=4
THEN 'True'
ELSE 'False'
 END "Check"
FROM
(SELECT x as Weeks, y as Team
from
(select week as x, COUNT(DISTINCT home)+COUNT(DISTINCT away) as y 
FROM fixture
WHERE home<>away
)
GROUP BY x
ORDER BY x);

又报错 说not a single-group group function 崩溃啊






------解决方案--------------------
具体内容没怎么看 分组计算要将单个字段都列出来 才能计算其他函数 2个类型差不多
SQL code

--1
SELECT givenname, familyname, feespaid, 185-feespaid as "Fees", sum(feespaid), sum(185-feespaid)
FROM MEMBER
WHERE Registrationnum IS NOT NULL;
GROUP BY givenname, familyname,feespaid;

--2
select week as Weeks, COUNT(DISTINCT home)+COUNT(DISTINCT away)as Team  
FROM fixture
WHERE home<>away
GROUP BY week 
ORDER BY week

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