开发写了几个语句，觉得查询结果跟逻辑有点不相符，就拿到这里一起分析了下。
语句如下：
select tp.title, tp.amount, ifnull(sum(case when tu.type = 1 then ti.invest_amount else 0 end),0) as aInvestAmount, ifnull(sum(case when tu.type = 2 then ti.invest_amount else 0 end),0) as bInvestAmount, ifnull(sum(case when tu.type = 3 then ti.invest_amount else 0 end),0) as cInvestAmount from t_invest ti join t_user tu on ti.user_id = tu.id join t_project tp on ti.project_id = tp.id where tp.id = '48346631623950333337353439383060';
其中t_project 中有id:48346631623950333337353439383060,但在t_invest中是没有此project_id的。所以这条语句理论上应该是没有任何输出，但实际上却输出了如下结果：
为了方便说明此问题，我们来建立如下的表格及数据。
mysql> select * from t1; +----+-------+ | id | name | +----+-------+ | 3 | chen | | 1 | zhang | +----+-------+ 2 rows in set (0.00 sec) mysql> select * from t2; +------+--------+ | id | course | +------+--------+ | 2 | math | +------+--------+ 1 row in set (0.00 sec) 然后执行如下语句1： mysql> select t1.id,t1.name,t2.id,t2.course,sum(t1.id) from t1 join t2 on t1.id=t2.id where t1.id=1 ; +----+-------+------+--------+------------+ | id | name | id | course | sum(t1.id) | +----+-------+------+--------+------------+ | 1 | zhang | NULL | NULL | NULL | +----+-------+------+--------+------------+ 1 row in set (0.00 sec)
发现竟然有输出，再执行如下的语句2：
mysql> select t1.id,t1.name,t2.id,t2.course,sum(t1.id) from t1 join t2 on t1.id=t2.id where t1.id in(1,null) ; +----+------+------+--------+------------+ | id | name | id | course | sum(t1.id) | +----+------+------+--------+------------+ | NULL | NULL | NULL | NULL | NULL | +----+------+------+--------+------------+ 1 row in set (0.00 sec)
发现全部为null。
执行如下语句3，返回Empty set。
mysql> select t1.id,t1.name,t2.id,t2.course,sum(t1.id) from t1 join t2 on t1.id=t2.id where t1.id=1 group by t1.id; Empty set (0.00 sec)
mysql中的语法并不是特别的严格，语句1与语句2其实在oracle中是会报语法检查不通过的。会报：ORA-00937: 不是单组分组函数
一般情况下用到聚合函数一般得加上group by会比较严格些，而出来这样的状况只有在有join的时候出来，单表查询还是没问题的，联合表查询聚合函数有使用的话推荐用语句3的写法。