[SQL]中級SQL(1)

關聯

• 基礎SQL

正文

這裏咱們會遇到subquery，它能夠出如今select子句中或者where子句或者from子句中。它會產生一個對應的結果表格，咱們能夠給這個表示命名。

數據集

咱們這一篇文章採用PostgreSQL的SQL語法。重點咱們關注select...from...where這種讀操做，分析query　(analytical query)。
數據集在　https://hyper-db.de/interface...　能夠直接使用。另外在這個網頁不容許進行寫操做:insert, update, delete之類的transactional query。固然create table和drop table也不被容許。

架構 Schema:

下載:
https://db.in.tum.de/teaching...

Schma和大部分SQL語句來自Prof. Alfons Kemper, Ph.D.的課件和書。

課件：

• https://db.in.tum.de/teaching...
• https://db.in.tum.de/teaching...

書：　https://db.in.tum.de/teaching...

中級SQL

• 在pruefen中搜索note小於平局值的：

select *
from pruefen
where note < (
    select avg(note)
    from pruefen
    )

• 對每個professoren，對應的vorlesungen的sws求和：

-- correlated sub-query
select p.persnr, p.name, (
    select sum(v.sws) as lehrbelastung
    from vorlesungen v
    where v.gelesenvon = p.persnr
    )
from professoren p

-- no sub-query
select p.persnr, p.name, sum(sws)
from professoren p left outer join vorlesungen v on p.persnr = v.gelesenvon
group by p.name, p.persnr

• 搜索上課數大於２的學生：

select tmp.matrnr, tmp.name, tmp.vorlanzahl
from (select s.matrnr, s.name, count(*) as vorlanzahl
    from studenten s, hoeren h
    where s.matrnr = h.matrnr
    group by s.matrnr, s.name) tmp
where tmp.vorlanzahl > 2

這時候咱們對這個subquery的結果表格進行命名tmp。固然咱們能夠用with子句來作一樣的事情。我主觀上更喜歡用with,它很清晰地把暫時須要的表格寫在最上方，並且對debug也更加友好。固然二者是結果等價，運行時間也等價的。

with tmp as (select s.matrnr, s.name, count(*) as vorlanzahl
    from studenten s, hoeren h
    where s.matrnr = h.matrnr
    group by s.matrnr, s.name) 

select tmp.matrnr, tmp.name, tmp.vorlanzahl
from tmp
where tmp.vorlanzahl > 2

• 計算每個vorlesungen的人數佔比：

select h.vorlnr, h.anzProVorl, g.gesamtAnz, cast(h.anzProVorl as decimal(6, 1)) / g.gesamtAnz as MarkAnteil
from (select vorlnr, count(*) as anzProVorl
    from hoeren
    group by vorlnr) as h,
     (select count(*) as gesamtAnz
    from studenten) g

-- with子句版本
with h as (select vorlnr, count(*) as anzProVorl
    from hoeren
    group by vorlnr),
     g as (select count(*) as gesamtAnz
    from studenten)

select h.vorlnr, h.anzProVorl, g.gesamtAnz, cast(h.anzProVorl as decimal(6, 1)) / g.gesamtAnz as MarkAnteil
from h, g

• 計算每個professoren經過上課認識的studenten個數以及比例：

with kenntSich as (
    select distinct v.gelesenvon as profpersnr, h.matrnr as studmatrnr
    from hoeren h join vorlesungen v on h.vorlnr =v.vorlnr
    ),
     kenntAnzahl as (
    select profpersnr, count(*) as anzstudenten
    from kenntSich
    group by profpersnr),
     wieviel as (
    select count(*) as gesamtanz
    from studenten)

select k.profpersnr, p.name, k.anzstudenten, w.gesamtanz, 1.00 * k.anzstudenten / w.gesamtanz as bekanntheitsgard
from kenntAnzahl k, wieviel w, professoren p
where k.profpersnr = p.persnr
order by bekanntheitsgard desc

• 搜索聽了全部sws=4 vorlesungen的學生：

SELECT s.*
FROM studenten s
where not exists(
    select *
    from vorlesungen v
    where v.sws = 4 and not exists(
        select *
        from hoeren h
        where h.vorlnr = v.vorlnr and h.matrnr = s.matrnr
        )
    )

SQL92中沒有定義for all Quantifier(全稱量詞)。因此咱們只能改寫關係代數：

$$ \{s|s\in studenten \wedge \forall v \in vorlesungen (v.sws = 4 \Rightarrow \\ \exists h \in hoeren (h.vorlnr = v.vorlnr \wedge h.matrnr = s.matrnr)) \} $$

咱們先把$\forall t \in R (P(t))$改寫成$\neg (\exists t \in R(\neg P(t)))$:

$$ \{s|s\in studenten \wedge \neg (\exists v \in vorlesungen \; \neg (v.sws = 4 \Rightarrow \\ \exists h \in hoeren (h.vorlnr = v.vorlnr \wedge h.matrnr = s.matrnr))) \} $$

再把$R \Rightarrow T$改寫成$\neg R \vee T$:

$$ \{s|s\in studenten \wedge \neg (\exists v \in vorlesungen \; \neg (\neg (v.sws = 4) \vee \\ \exists h \in hoeren (h.vorlnr = v.vorlnr \wedge h.matrnr = s.matrnr))) \} $$

再用DeMorgan律簡化一下：

$$ \{s|s\in studenten \wedge \neg (\exists v \in vorlesungen (v.sws = 4) \wedge \\ \neg (\exists h \in hoeren (h.vorlnr = v.vorlnr \wedge h.matrnr = s.matrnr))) \} $$

用中文說：不存在一門sws=4的課，沒有被這個學生聽。這樣咱們能夠對應關係代數到上面的SQL。

另一種trick解法，使用count:

-- 先把hoeren變成sws=4hoeren: hoerenStudentenWith4SWS
with hoerenStudentenWith4SWS (matrnr, vorlnr) as (
    select h.matrnr, v.vorlnr
    from hoeren h, vorlesungen v
    where h.vorlnr = v.vorlnr and v.sws = 4
    )

-- 再看學生是否是聽完了全部hoerenStudentenWith4SWS
select h.matrnr
from hoerenStudentenWith4SWS h
group by h.matrnr
having count(*) = (select count(*) from vorlesungen v where v.sws = 4)

• (對上面的相似練習) 搜索學生全部考過的試對應的科目，都是這個同窗所聽過：

select s.*
from studenten s
where not exists(
    select *
    from pruefen p
    where p.matrnr = s.matrnr and not exists(
        select *
        from hoeren h
        where h.vorlnr = p.vorlnr and h.matrnr = s.matrnr
        )
    )

用中文說：沒有一門被考過的科目，沒有出如今對應學生hoeren表格中。

另外由於這個要求是獨立得應用在每個學生上，每個學生由於考試不一樣，全部要求聽的科目也不一樣。所以上面那題的trick再也不適用。trick應用條件是對全部學生須要廣泛性，而排除獨立性 -- 一視同仁。

• 求至少聽Sokrates一門課的學生們的平均學期數：

with vl_von_sokrates as (
    select *
    from vorlesungen v, professoren p
    where v.gelesenvon = p.persnr and p.name = 'Sokrates'
), studenten_von_sokrates as (
    select distinct s.name, s.matrnr, s.semester
    from studenten s, hoeren h, vl_von_sokrates v
    where s.matrnr = h.matrnr and h.vorlnr = v.vorlnr
)

select avg(semester)
from studenten_von_sokrates;

這題必定要注意,可能一個學生聽了Sokrates的不少課，可是這種同窗不能被重複計數。咱們能夠用distinct。

可是咱們也有一種解法不須要distinct，它不用join,而是帶exists的correlated subquery:

with vl_von_sokrates as (
    select *
    from vorlesungen v, professoren p
    where v.gelesenvon = p.persnr and p.name = 'Sokrates'
), studenten_von_sokrates as (
    select *
    from studenten s
    where exists(
        select *
        from hoeren h, vl_von_sokrates vl
        where h.matrnr = s.matrnr and h.vorlnr = vl.vorlnr
    )
)

select avg(semester)
from studenten_von_sokrates;

• 求每一個學生聽幾節課，須要考慮不放任何課的學生：

select count(*) as hcount
    from hoeren
    ),
     s as (
    select count(*) as scount
    from studenten
)

select hcount / (scount * 1.00) as avg_vl
from h, s

或者

with h as (
    select count(*) as hcount
    from hoeren
    ),
     s as (
    select count(*) as scount
    from studenten
)

select hcount / (cast(scount as decimal(10, 4))) as avg_vl
from h, s

1. 搜索學生經過上課能認識的其餘學生名字：

select s1.name, s2.name
from studenten s1, hoeren h1, hoeren h2, studenten s2
where h1.vorlnr = h2.vorlnr and h1.matrnr = s1.matrnr  and h2.matrnr = s2.matrnr and s1.matrnr != s2.matrnr

1. 對每個同窗認識的人進行計數：

with bekannte as (
    select s1.matrnr as student, s2.matrnr as sein_bekannte
    from studenten s1,
         hoeren h1,
         hoeren h2,
         studenten s2
    where h1.vorlnr = h2.vorlnr
      and h1.matrnr = s1.matrnr
      and h2.matrnr = s2.matrnr
      and s1.matrnr != s2.matrnr
)

select s.matrnr, s.name, count(b.sein_bekannte) as num_friends
from studenten s, bekannte b
where s.matrnr = b.student
group by s.matrnr, s.name
order by num_friends desc

1. 在2.的基礎上再考慮：不上課(也就不認識同窗)的人

with bekannte as (
    select s1.matrnr as student, s2.matrnr as sein_bekannte
    from studenten s1,
         hoeren h1,
         hoeren h2,
         studenten s2
    where h1.vorlnr = h2.vorlnr
      and h1.matrnr = s1.matrnr
      and h2.matrnr = s2.matrnr
      and s1.matrnr != s2.matrnr
)

select s.matrnr, s.name, count(b.sein_bekannte) as num_friends
from studenten s left outer join bekannte b
on s.matrnr = b.student
group by s.matrnr, s.name
order by num_friends desc

這裏用了一個left outer join。右邊的表格bekannte b只含有上課的同窗(即出如今hoeren表格中的同窗)，可是左邊的表格studenten s含有全部學生。

• 搜索選課超過學生選課sws平均數的學生，須要考慮不上課的學生：

with num_stu as (
    select count(*) as count_stu
    from studenten),
     num_sws as (
    select sum(vor.sws) as count_sws
    from hoeren h, vorlesungen vor
    where h.vorlnr = vor.vorlnr)

select s.*
from studenten s
where s.matrnr in (
    select h.matrnr
    from hoeren h, vorlesungen v
    where h.vorlnr = v.vorlnr
    group by h.matrnr
    having sum(sws) > (select cast(num_sws.count_sws as decimal (5, 2)) / num_stu.count_stu from num_sws, num_stu)
    )

或者

with num_stu as (
    select count(*) as count_stu
    from studenten),
     num_sws as (
    select sum(vor.sws) as count_sws
    from hoeren h, vorlesungen vor
    where h.vorlnr = vor.vorlnr),
     avg_sws as (
    select cast(num_sws.count_sws as decimal(5, 2)) / num_stu.count_stu as sws
    from num_stu, num_sws),
     stu_sws as (
    select s.matrnr, s.name, s.semester, sum(v.sws) as sum_sws
    from studenten s, hoeren h, vorlesungen v
    where s.matrnr = h.matrnr and h.vorlnr = v.vorlnr
    group by s.matrnr, s.name, s.semester)

select s.*
from stu_sws s, avg_sws
where s.sum_sws > avg_sws.sws

或者

with swsProStudent as (
    select s.matrnr, s.name,
        cast((case when sum(v.sws) is null then 0
                                   else sum(v.sws) end) as real) as anzSWS
    from studenten s
    left outer join hoeren h on s.matrnr = h.matrnr
    left outer join vorlesungen v on h.vorlnr = v.vorlnr
    group by s.matrnr, s.name
)

select s.*
from studenten s
where s.matrnr in (
    select sws.matrnr
    from swsProStudent sws
    where sws.anzSWS > (
        select avg(anzSWS)
        from swsProStudent
        )
    )

• 比較聽課並考試的同窗的成績和不聽課只考試的同窗成績：

with no_lec as (
    select avg(note) as avg_note
    from pruefen p
    where not exists (
        select *
        from hoeren h
        where h.matrnr = p.matrnr
        )),
     with_lec as (
    select avg(note) as avg_note
    from pruefen p
    where exists (
        select *
        from hoeren h
        where h.matrnr = p.matrnr
        ))

select *
from no_lec, with_lec;

假設咱們的schema變成上圖(SQL不能運行　數據集不對應上圖)：

• 求每個FakName對應的女性佔比：

with anz(Fakname,AnzStudenten) as (
    select s.FakName, count(*)
    from StudentenGF s
    group by s.FakNAme),
     anzw(Fakname,AnzWeiblich) as (
    select sw.FakName,count(*) as AnzWeiblich
    from StudentenGF sw
    where sw.Geschlecht ='W'
    group by sw.FakName)

select anz.FakName, anz.AnzStudenten, anzw.AnzWeiblich, (cast(anzw.AnzWeiblich as decimal(5,2))/anz.AnzStudenten * 100) as ProzentWeiblich
from anz, anzw
where anz.FakName = anzw.FakName

• 求每個FakName對應的男性佔比：

with anz(Fakname, AnzStudenten) as (
    select s.FakName, count(*)
    from StudentenGF s
    group by s.FakNAme),
     anzm(Fakname, AnzMaenner) as (
         select sw.FakName, count(*) as AnzWeiblich
         from StudentenGF sw
         where sw.Geschlecht = 'M'
         group by sw.FakName)

select anz.FakName,
       anz.AnzStudenten,
       anzm.AnzMaenner,
       (case when anzm.AnzMaenner is null then 0 else anzm.AnzMaenner end) / anz.AnzStudenten * 100.00 as ProzentMaenner
from anz left outer join anzm
on anz.FakName = anzm.FakName

這裏並非女性版直接更改爲男性。一個重點是:存在系沒有任何男性。
case也能夠被替換爲: COALESCE(anzm.AnzMaenner, 0) / anz.AnzStudenten * 100.00 as ProzentMaenner

或者再換一種：

select fakname, 
       (sum(case when geschlechte = 'M' then 1.00 else 0.00 end)) / count(*)
from studentenFG
group by fakname

• 搜索全部學生把本身系教授提供的課都聽完了：

select s.*
from studentenFG s
where not exists(
    select *
    from vorlesungen v, professorenF p
    where v.gelesenvon = p.persnr and p.fakname = s.fakname and not exists(
        select *
        from hoeren h
        where h.vorlnr = v.vorlnr and h.matrnr = s.matrnr
        )
    )

用中文就是：對這個學生，不存在一門他系裏教授的課，這個學生沒有聽過。

或者

select s.*
from studentenFG s
where (
    select count(*)
    from vorlesungen v, professorenF p
    where v.gelesenvon = p.persnr and p.fakname = s.fakname
          )
=
      (
    select count(*)
    from hoeren h, vorlesungen v, professorenF p
    where h.matrnr = s.matrnr and h.vorlnr = v.vorlnr and p.persnr = v.gelesenvon and p.fakname= s.fakname
          )