SQL查詢，以確定是否一個表具有比其他

Question

我有如下一個課外練習麻煩的屬性更多的行：

Find all departments with more professors than students.

涉及的表如下：

 
      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) 

這個問題實質上是要求一個值列表，其中左表具有的值多於右表的值。我該怎麼做呢？該查詢不起作用：

 
SET search_path TO university; 

SELECT dept 
FROM prof 
GROUP BY dept 
HAVING count(*) > (
    SELECT count(*) 
    FROM student 
    GROUP BY dept 
); 

導致以下錯誤：

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

此代碼不能正常工作，或者說，導致沒有行返回（正確答案返回兩行）：

 
SET search_path TO university; 

SELECT prof.dept 
FROM prof, student 
GROUP BY prof.dept 
HAVING count(prof.*) > count(student.*); 

我該怎麼做？我不是在尋找完整的答案，而是提示如何構建查詢。我在Windows 7上使用PostgreSQL 9.1.3。

編輯：感謝您的幫助。正確的答案是：

 
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.*); 

編輯2：有一個更好的解決方案。我已經接受了包含此解決方案的答案。

Answer 1

Find all departments with more professors than students.

考慮這樣一個：

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; 

應該快得多比你現在的解決方案，因爲學生和教授們，才數較少（按數量級）由dept加入製成。

它也觀察部門有沒有學生，您的解決方案目前錯過的情況。你的交叉聯合消除了沒有學生（或教授）的所有部門。

Answer 2

你的第一次查詢的一部分，這是造成錯誤：

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，你必須通過professor表迭代（它可以選擇不）和每個教授計算沒有教授在該部門，然後相應找到與上述教授屬於同一部門的學生。 事情是這樣的：

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

Answer 3

另一種解決方案使用的觀點：

 
CREATE OR REPLACE TEMPORARY VIEW p AS (
    SELECT dept, count(*) num 
    FROM prof 
    GROUP BY dept 
); 
CREATE OR REPLACE TEMPORARY VIEW s AS (
    SELECT dept, count(*) num 
    FROM student 
    GROUP BY dept 
); 
SELECT DISTINCT p.dept 
FROM p, s 
WHERE p.num > s.num; 

此代碼正確地處理那裏有一個給定的部門沒有教授或學生的情況。