Postgresql SQL查询，以确定一个表是否具有比另一个表更多的属性行

我的家庭作业练习有问题，如下所示：

Table "university.prof" Column | Type | Modifiers -----------+-----------------------+----------- pnum | character(9) | not null lastname | character varying(12) | not null firstname | character varying(12) | dept | character(4) | office | character(6) | rank | character(1) | date_emp | date | not null salary | numeric(8,2) | manager | character(9) | Indexes: "prof_pkey" PRIMARY KEY, btree (pnum) Table "university.student" Column | Type | Modifiers -----------+---------------+----------- snum | character(9) | not null lastname | character(12) | firstname | character(12) | dept | character(4) | date_ent | date | cr_comp | smallint | gpa | numeric(4,3) | advisor | character(9) | Indexes: "student_pkey" PRIMARY KEY, btree (snum) SET search_path TO university; SELECT prof.dept FROM prof, student WHERE prof.dept = student.dept GROUP BY prof.dept HAVING count(DISTINCT prof.*) > count(DISTINCT student.*); 找到教授多于学生的所有系

所涉及的表格如下：

Table "university.prof" Column | Type | Modifiers -----------+-----------------------+----------- pnum | character(9) | not null lastname | character varying(12) | not null firstname | character varying(12) | dept | character(4) | office | character(6) | rank | character(1) | date_emp | date | not null salary | numeric(8,2) | manager | character(9) | Indexes: "prof_pkey" PRIMARY KEY, btree (pnum) Table "university.student" Column | Type | Modifiers -----------+---------------+----------- snum | character(9) | not null lastname | character(12) | firstname | character(12) | dept | character(4) | date_ent | date | cr_comp | smallint | gpa | numeric(4,3) | advisor | character(9) | Indexes: "student_pkey" PRIMARY KEY, btree (snum) SET search_path TO university; SELECT prof.dept FROM prof, student WHERE prof.dept = student.dept GROUP BY prof.dept HAVING count(DISTINCT prof.*) > count(DISTINCT student.*); 这个问题本质上要求提供一个值列表，其中左表的值行比右表的多。我该怎么做？此查询不起作用：

SET search_path TO university; SELECT dept FROM prof GROUP BY dept HAVING count(*) > ( SELECT count(*) FROM student GROUP BY dept ); 导致以下错误：

ERROR: more than one row returned by a subquery used as an expression

SELECT dept
FROM prof
GROUP BY dept
HAVING count(*) > (
    SELECT count(*) /* This subquery would yield rows equal to no of different  
    FROM student        Department to which students belong.
    GROUP BY dept       Thus you can't compare it with > symbol.*/      
);

此代码也不起作用，导致没有返回行。正确答案返回两行：

SET search_path TO university; SELECT prof.dept FROM prof, student GROUP BY prof.dept HAVING count(prof.*) > count(student.*); 我该怎么做？我不是在寻找完整的答案，而是关于如何构造查询的提示。我正在Windows 7上使用PostgreSQL 9.1.3

编辑：谢谢你的帮助。正确答案如下：

Table "university.prof" Column | Type | Modifiers -----------+-----------------------+----------- pnum | character(9) | not null lastname | character varying(12) | not null firstname | character varying(12) | dept | character(4) | office | character(6) | rank | character(1) | date_emp | date | not null salary | numeric(8,2) | manager | character(9) | Indexes: "prof_pkey" PRIMARY KEY, btree (pnum) Table "university.student" Column | Type | Modifiers -----------+---------------+----------- snum | character(9) | not null lastname | character(12) | firstname | character(12) | dept | character(4) | date_ent | date | cr_comp | smallint | gpa | numeric(4,3) | advisor | character(9) | Indexes: "student_pkey" PRIMARY KEY, btree (snum) SET search_path TO university; SELECT prof.dept FROM prof, student WHERE prof.dept = student.dept GROUP BY prof.dept HAVING count(DISTINCT prof.*) > count(DISTINCT student.*);

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编辑2：有更好的解决方案。我已接受包含此解决方案的答案。

您第一次查询中导致错误的部分：

ERROR: more than one row returned by a subquery used as an expression

SELECT dept
FROM prof
GROUP BY dept
HAVING count(*) > (
    SELECT count(*) /* This subquery would yield rows equal to no of different  
    FROM student        Department to which students belong.
    GROUP BY dept       Thus you can't compare it with > symbol.*/      
);

相反，您必须遍历选择的教授表，并为每个教授计算该系的教授人数，然后相应地找到与上述教授属于同一系的学生。 大概是这样的：

select t.prof from prof t where
(
select count(*) from prof where dept=t.dept
)
>
(
select count(*) from student where dept=t.dept
)

找到教授多于学生的所有系

考虑一下这一点：

SELECT p.dept, p.prof_ct, s.stud_ct
FROM (
   SELECT dept, count(*) AS prof_ct
   FROM   university.prof
   GROUP  BY dept
   ) p
LEFT JOIN (
   SELECT dept, count(*) AS stud_ct
   FROM   university.student
   GROUP  BY dept
   ) s USING (dept)
WHERE p.prof_ct >= s.stud_ct OR s.stud_ct IS NULL
ORDER BY p.dept;

应该比你现在的解决方案快得多，因为学生和教授在数量级联接之前就被计算出来了

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它还观察到一个系没有学生的情况，而您的解决方案目前忽略了这一点。交叉联接消除了所有没有学生或教授的系。

另一种解决方案使用视图：

根据需要创建或替换临时视图p 选择dept，count*num 来自教授 按部门分组 ; 根据需要创建或替换临时视图 选择dept，count*num 来自学生 按部门分组 ; 选择不同的p.dept 从p，s 其中p.num>s.num；

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此代码正确处理给定部门没有教授或学生的情况。

使用have count*>是最可能的错误原因。第二次尝试将错过一个连接谓词，该谓词按部门连接两个表，其中prof.dept=student.dept和关键字在两个计数中都是不同的，因为这样它会计数教授和学生的重复，总是一样的。