Mysqli查询不起作用

我想检索一年中所有最大的疾病使用情况。我有一个名为view_accepted_appoinments的表，它有列疾病和日期。我想这个查询，但它不工作Mysqli查询不起作用

select distinct doc_specilization as disease, 
DATE_FORMAT(date,'%Y') as year, 
count(Distinct doc_specilization) as count 
from view_accepted_appoinments 
group by year 
order by year desc;

查询返回这里面给出错误的病名

+--------------+------+-------+ 
| disease  | YEAR | count | 
+--------------+------+-------+ 
| Primary Care | 2014 |  8 | 
| Primary Care | 2013 |  1 | 
+--------------+------+-------+

来源

2014-12-13 Zeeshan

在这个意义上不工作？？你可以更清楚地了解你所得到的错误信息，或者显示的数据与你所期望的输出不同，并且你也可以向我们显示表格的列 – ManojGeek 2014-12-13 17:00:20

这个查询是否在执行时返回你预期的记录集通过像MySQL Workbench或phpMyAdmin这样的数据库客户端？ – budhajeewa 2014-12-13 17:02:16

你可以分享一些样本数据和你试图达到的结果吗？ – Mureinik 2014-12-13 17:04:59

select distinct d , spec,c from(
     select date_format(date,"%Y") as d, 
     doctor, doc_specilization as spec, 
     count(doc_specilization) as c from 
     view_accepted_appoinments 
     group by doc_specilization, d)f 
     group by d;

来源

2014-12-13 17:57:37

尝试

SELECT doc_specialization AS disease, 
DATE_FORMAT(date, '%Y') AS year, 
COUNT(doc_specialization) AS count, 
FROM view_accepted_appoinments 
GROUP BY doc_specialization, year 
ORDER BY year DESC, doc_specialization ASC;

来源

2014-12-13 17:08:35 budhajeewa

发生此错误您的SQL语法错误;请检查与您的MySQL服务器版本相对应的手册，以便在第4行的'FROM view_accepted_appoinments GROUP BY doc_specialization，year ORDER BY D'附近使用正确的语法。 – Zeeshan 2014-12-13 17:11:55

@Zeeshan：您是否可以提供转储 - 最好带有少量数据库的实际数据？所以我可以导入它并测试？ – budhajeewa 2014-12-13 17:14:36

我为此创建了一个逻辑表（视图）。 – Zeeshan 2014-12-13 17:19:52

Mysqli查询不起作用

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