Aggregation, GROUP BY, HAVING

4-bob 13 kod bloki 3388 so'z

Hozirgacha har bir query allaqachon mavjud row larni qaytardi. Aggregation boshqacha ish qiladi: u ko'p row ni bittaga siqadi. "Nechta student bor?", "O'rtacha baho qancha?", "Qaysi course eng ko'p daromad keltiradi?" — analytics aslida shunday savollarni beradi, va aynan shu bobdan boshlab SQL qidiruv tilidan analytics tiliga aylanadi.

COUNT, SUM, AVG, MIN, MAX

Deyarli hamma narsani qoplaydigan beshta function:

Function	Nima qaytaradi	NULL ni e'tiborsiz qoldiradimi?
`COUNT(*)`	Row lar soni	Yo'q — har bir row ni sanaydi
`COUNT(col)`	`col` dagi NULL bo'lmagan qiymatlar soni	Ha
`SUM(col)`	Yig'indi	Ha
`AVG(col)`	O'rtacha arifmetik	Ha
`MIN(col)` / `MAX(col)`	Eng kichik / eng katta	Ha

sql8 qator

SELECT
    COUNT(*)      AS total_enrollments,
    COUNT(grade)  AS graded,
    ROUND(AVG(grade), 2) AS avg_grade,
    MIN(grade)    AS lowest,
    MAX(grade)    AS highest,
    SUM(grade)    AS total_points
FROM enrollments;

total_enrollments	graded	avg_grade	lowest	highest	total_points
2 500	1 826	72.98	38	100	133 255

1 qator

1 ms

total_enrollmentsgradedavg_gradelowesthighesttotal_points

201479.43471001112

Yigirmata row kirdi, bitta row chiqdi. Aggregate ni belgilaydigan asosiy xususiyat shu: u row lar to'plamini iste'mol qilib, bitta qiymat hosil qiladi.

COUNT(*) vs COUNT(col) vs COUNT(DISTINCT col)

Deyarli bir xil ko'rinadigan va uchta har xil savolga javob beradigan uchta function. Bu intervyularda doim so'raladi.

sql5 qator

SELECT
    COUNT(*)             AS rows_total,      -- 10
    COUNT(city)          AS cities_present,  -- 9   (Sevara ning NULL i tashlandi)
    COUNT(DISTINCT city) AS unique_cities    -- 6   (NULL umuman sanalmaydi)
FROM students;

rows_total	cities_present	unique_cities
300	292	8

1 qator

0 ms

IfodaNatijaQaysi savolga javob beradi

COUNT(*)10Nechta student bor? COUNT(city)9Nechta student ning shahri yozilgan? COUNT(DISTINCT city)6Biz nechta har xil shaharga xizmat qilamiz?

:::tip 💡 Tayyor data-quality tekshiruvi COUNT(*) - COUNT(col) — o'sha column dagi NULL lar soni. Bitta query bilan har bir column ning to'liqligini bir vaqtda olasiz — notanish table ga birinchi qarash uchun juda foydali:

sql4 qator

SELECT COUNT(*) AS rows,
       COUNT(*) - COUNT(city)   AS missing_city,
       COUNT(*) - COUNT(gender) AS missing_gender
FROM students;

:::

Aggregate lar NULL bilan qanday ishlaydi

COUNT(*) dan boshqa har bir aggregate NULL larni indamay tashlab ketadi. Odatda bu to'g'ri — lekin u maxrajni o'zgartiradi, va talabalar buni bilishi shart.

20 ta enrollment ustida AVG(grade)mavjud baholar : 14 ta qiymat, yig'indisi 1112 yo'q baholar : 6 ta NULL AVG(grade) = 1112 / 14 = 79.43 ← NULL lar butunlay tashlandi SUM(grade)/COUNT() = 1112 / 20 = 55.60* ← NULL lar nol deb hisoblandi

Bir xil data, ikkita javob, ikkalasi ham asoslanadi — lekin ular har xil narsani anglatadi. Avval biznes savolini so'rang: "bizda bor baholarning o'rtachasi" (79.43) mi, yoki "har bir enrollment ga to'g'ri keladigan o'rtacha ball, baholanmagani hech narsa deb hisoblanadi" (55.60) mi? Birinchisi deyarli doim kerak bo'lgan narsa; ikkinchisi deyarli doim xato.

⚠️ Hech narsaning SUM i 0 emas, NULL

SELECT SUM(grade) FROM enrollments WHERE status = 'nonexistent'; 0 emas, NULL qaytaradi — qo'shadigan hech narsa bo'lmagan. Dashboard ga raqam kerak bo'lsa, o'rang: COALESCE(SUM(grade), 0). COUNT bu yerda istisno: u bo'sh to'plam uchun to'g'ri 0 qaytaradi.

GROUP BY

GROUP BY "butun table uchun bitta son" g'oyasini olib, uni har bir guruhga qo'llaydi. "O'rtacha baho" o'rniga "har bir course uchun o'rtacha baho" olasiz.

sql9 qator

SELECT
    course_id,
    COUNT(*)             AS enrollments,
    COUNT(grade)         AS graded,
    ROUND(AVG(grade), 2) AS avg_grade,
    MAX(grade)           AS best
FROM enrollments
GROUP BY course_id
ORDER BY course_id;

course_id	enrollments	graded	avg_grade	best
1	193	132	79.15	100
2	186	146	74.27	96
3	174	128	63.59	86
4	167	119	74.18	100
5	167	125	69.37	94
6	174	118	75.17	94
7	165	122	79.89	100
8	170	131	75.15	100
9	161	118	76.1	97
10	179	147	73.52	100
11	200	140	68.34	92
13	187	127	73.86	97
14	189	135	74.65	100
15	188	138	65.59	91

14 qator

1 ms

course_idenrollmentsgradedavg_gradebest

15468.092 25587.2100 33276.589 43280.081 54191.091

:::gotcha ⚠️ 6-course yo'q — 5-course esa chalg'ituvchi Bitta natija jadvalida ikkita saboq:

"Deep Learning" (6-course) umuman ko'rinmaydi. GROUP BY faqat mavjud row larni guruhlay oladi, 6-course da esa enrollment yo'q. Shu tarzda qurilgan "course lar hisoboti" hech kim yozilmagan har bir course ni indamay tashlab ketadi — holbuki rahbariyat aynan shu haqda eshitmoqchi bo'lgan. Yechim — courses dan LEFT JOIN (5-bob).
5-course to'rtta enrollment dan bitta baho asosida 91.0 ko'rsatyapti. Kichik namunadagi o'rtachalar ishonchli ko'rinadi, lekin unday emas. Hisobotda AVG yoniga doim COUNT qo'ying, shunda o'quvchi son qanchalik qimmatga ega ekanini ko'radi. :::

Matnli column bo'yicha guruhlash ham xuddi shunday ishlaydi:

sql4 qator

SELECT status, COUNT(*) AS n
FROM enrollments
GROUP BY status
ORDER BY n DESC;

status	n
completed	1 634
active	472
dropped	202
failed	192

4 qator

1 ms

:::tip 💡 Bu — value_counts() pandas dan kelgan talabalar buni darrov taniydi: GROUP BY col + COUNT(*) aynan df['col'].value_counts(). Bu bog'lanishni erta ko'rsatish butun bobni tushunarli qiladi. :::

NULL esa o'z guruhini hosil qiladi — SQL barcha NULL larni teng deb hisoblaydigan yagona joy:

sql5 qator

SELECT COALESCE(city, '(unknown)') AS city, COUNT(*) AS students
FROM students
GROUP BY city
ORDER BY students DESC, city;
-- Tashkent 3, Samarkand 2, keyin Andijan/Bukhara/Fergana/Namangan/(unknown) 1 tadan

city	students
Tashkent	82
Samarkand	40
Bukhara	38
Fergana	36
Namangan	28
Andijan	24
Karshi	22
Nukus	22
(unknown)	8

9 qator

0 ms

GROUP BY ning oltin qoidasi :::info 📌 Qoida SELECT dagi har bir column yo GROUP BY da bo'lishi, yo aggregate function ichida bo'lishi shart. Boshqasiga ruxsat yo'q. :::

Sababi yuqoridagi diagrammaga qarasangiz oddiy. Guruhlashdan keyin 1-course to'rtta kirish row idan qurilgan bitta chiqish row i. student_id so'rasangiz, savolning javobi yo'q — u yerda to'rtta har xil student bor edi. Qaysi birini chop etsin?

python7 qator

-- XATO: student_id na guruhlangan, na aggregate qilingan
SELECT course_id, student_id, AVG(grade)
FROM enrollments
GROUP BY course_id;

-- PostgreSQL: XATO — "student_id" column i GROUP BY da bo'lishi kerak
-- SQLite/MySQL: xato YO'Q — ular tasodifiy qiymat o'ylab topadi. Bu ancha xavfli.

course_id	student_id	AVG(grade)
1	59	79.152
2	241	74.267
3	85	63.586
4	211	74.176
5	281	69.368
6	21	75.169
7	130	79.893
8	33	75.145
9	215	76.102
10	102	73.524
11	216	68.343
13	221	73.858
14	188	74.652
15	95	65.594

14 qator

1 ms

:::gotcha ⚠️ SQLite va MySQL bunga yo'l qo'yadi PostgreSQL va SQL Server query ni rad etadi. SQLite va eski MySQL esa hech qanday ogohlantirishsiz qandaydir student_id ni qaytaradi — engine oxirgi tegib o'tgan row nikini. Talabalar keyin real ko'rinadigan, lekin ma'nosiz qiymatlar bilan to'la hisobotni topshiradi. Engine nimaga yo'l qo'yishidan qat'i nazar, qoidani mutlaq deb o'rgating. :::

Aslida kerak bo'lgan narsani olishning uchta to'g'ri yo'li:

python9 qator

-- (a) uni ham guruhlash — har bir (course, student) uchun bitta row
SELECT course_id, student_id, AVG(grade) FROM enrollments
GROUP BY course_id, student_id;

-- (b) uni aggregate qilish — masalan nechta har xil student
SELECT course_id, COUNT(DISTINCT student_id) AS students, AVG(grade) FROM enrollments
GROUP BY course_id;

-- (c) aslida "har bir course dagi eng yaxshi student" kerak edi → window function, 7-bob

1-natija

course_id	student_id	AVG(grade)
1	3	77
1	6	84
1	8	66
1	10	75
1	11	NaN
1	12	93
1	13	NaN
1	14	77
1	19	69
1	20	77
1	21	76
1	22	77
1	23	74.333
1	24	100
1	26	96

… 1866 qatordan birinchi 15 tasi

1866 qator

2-natija

course_id	students	AVG(grade)
1	134	79.152
2	134	74.267
3	124	63.586
4	121	74.176
5	130	69.368
6	133	75.169
7	129	79.893
8	127	75.145
9	129	76.102
10	137	73.524
11	152	68.343
13	136	73.858
14	148	74.652
15	132	65.594

14 qator

3 ms

Bir nechta column bo'yicha guruhlash Bir nechta column sanab o'tish har bir birikma uchun bitta guruh hosil qiladi. Chiqish row lari soni nazariy juftliklar soniga emas, haqiqatan uchraydigan birikmalar soniga teng.

python8 qator

-- "Har bir shaharda qaysi jinsdan nechta student bor?"
SELECT
    COALESCE(city, '(unknown)') AS city,
    gender,
    COUNT(*) AS students
FROM students
GROUP BY city, gender
ORDER BY city, gender;

city	gender	students
(unknown)	F	2
(unknown)	M	6
Andijan	F	10
Andijan	M	14
Bukhara	F	18
Bukhara	M	20
Fergana	F	16
Fergana	M	20
Karshi	F	11
Karshi	M	11
Namangan	F	15
Namangan	M	13
Nukus	F	15
Nukus	M	7
Samarkand	F	20

… 18 qatordan birinchi 15 tasi

18 qator

0 ms

💡 Uni haqiqiy cross-tab ga aylantirish

gender ni row emas, column qilib olish — ya'ni pivot table — uchun SUM ni CASE WHEN bilan qo'shing (8-bob):

sql run SELECT COALESCE(city,'(unknown)') AS city, SUM(CASE WHEN gender = 'M' THEN 1 ELSE 0 END) AS male, SUM(CASE WHEN gender = 'F' THEN 1 ELSE 0 END) AS female, COUNT(*) AS total FROM students GROUP BY city;

Bu SUM(CASE WHEN ...) pattern i real analytics ishida eng ko'p ishlatiladigan konstruksiyalardan biri. Mashq qildirishga arziydi.

HAVING va WHERE

1-bobdagi qoida, endi sababi ko'rinib turibdi: WHERE guruhlashdan oldin row larni, HAVING guruhlashdan keyin guruhlarni filtrlaydi.

Har biri qayerda turadiFROM → row larni oladi WHERE → row larni tashlaydi (aggregate yo'q — hali guruh mavjud emas) GROUP BY → guruhlarni quradi HAVING → butun guruhlarni tashlaydi (aggregate bor — guruhlar mavjud) SELECT → chiqishni quradi ORDER BY → saralaydi

python10 qator

-- "Faqat tugatilgan enrollment lar orasida qaysi course larning o'rtachasi 80 dan yuqori?"
SELECT
    course_id,
    COUNT(*)             AS completed_count,
    ROUND(AVG(grade), 2) AS avg_grade
FROM enrollments
WHERE status = 'completed'      -- ROW larni filtrlaydi: active/failed/dropped tashlanadi
GROUP BY course_id
HAVING AVG(grade) > 80          -- GURUH larni filtrlaydi: past o'rtachali course lar tashlanadi
ORDER BY avg_grade DESC;

course_id	completed_count	avg_grade
7	120	80.25

1 qator

1 ms

course_idcompleted_countavg_grade

5191.0 2587.2 1283.5

4-course yo'q: uning tugatilgan baholari 81 va 79, o'rtachasi aynan 80.0, > 80 esa qat'iy. 3-course ning o'rtachasi 76.5. Shuningdek e'tibor bering: 1-course ning o'rtachasi 68.0 dan 83.5 ga ko'tarildi — chunki WHERE status = 'completed' avval ikkita yiqilgan bahoni olib tashladi. WHERE ni o'zgartirish guruhlar ichidagi narsani o'zgartiradi.

⚠️ Row shartini HAVING ga qo'ymang

Bu ikkalasi bir xil javob beradi:

WHERE status = 'completed' GROUP BY course_id -- tez GROUP BY course_id HAVING status = 'completed' -- sekin / noto'g'ri Ikkinchisi har bir row dan har bir guruhni quradi va shundan keyingina ishni tashlab yuboradi — qat'iy database da esa u shunchaki xato, chunki status guruhlanmagan. Iloji boricha erta filtrlang. Agar shartda aggregate ishtirok etmasa, u WHERE ga tegishli.

💡 Foydali bitta qator: dublikatlarni topish

sql run SELECT student_id, course_id, COUNT(*) AS times FROM enrollments GROUP BY student_id, course_id HAVING COUNT(*) > 1;

Qaytgan har bir row — bitta course ga ikki marta yozilgan student. Bu GROUP BY ... HAVING COUNT(*) > 1 pattern i dublikat topishning standart query si va juda ko'p uchraydigan intervyu topshirig'i. Bizning toza data da u hech narsa qaytarmaydi.

ROUND va butun bo'lish tuzog'i

2-bobdagi tuzoq shu yerda qaytadi, va aggregation da u haqiqiy zarar keltiradi.

python11 qator

-- "Enrollment larning necha foizi tugatilgan?"

-- XATO: butun arifmetikada 12 / 20 = 0
SELECT COUNT(*) / (SELECT COUNT(*) FROM enrollments) * 100 AS pct
FROM enrollments WHERE status = 'completed';
-- 0 qaytaradi

-- TO'G'RI: bo'lishdan oldin * 1.0 bilan kasrga majburlash
SELECT ROUND(COUNT(*) * 100.0 / (SELECT COUNT(*) FROM enrollments), 1) AS pct
FROM enrollments WHERE status = 'completed';
-- 60.0 qaytaradi

1-natija

pct
0

1 qator

2-natija

pct
65.4

1 qator

1 ms

"Shartga mos row larning foizi" uchun ixcham idiom, yodlashga arziydi:

Guruhning shartga mos foiziROUND(100.0 * SUM(CASE WHEN shart THEN 1 ELSE 0 END) / COUNT(*), 1)

python9 qator

-- har bir course bo'yicha o'tish foizi, bitta query da
SELECT
    course_id,
    COUNT(*) AS graded,
    ROUND(100.0 * SUM(CASE WHEN grade >= 60 THEN 1 ELSE 0 END) / COUNT(*), 1) AS pass_rate_pct
FROM enrollments
WHERE grade IS NOT NULL
GROUP BY course_id
ORDER BY pass_rate_pct DESC;

course_id	graded	pass_rate_pct
1	132	98.5
7	122	98.4
13	127	96.9
6	118	96.6
9	118	94.9
10	147	94.6
4	119	93.3
14	135	92.6
2	146	91.8
8	131	90.1
5	125	88
11	140	82.9
15	138	73.2
3	128	63.3

14 qator

1 ms

course_idgradedpass_rate_pct

25100.0 32100.0 42100.0 51100.0 1450.0

WHERE grade IS NOT NULL ga e'tibor bering: usiz baholanmagan enrollment lar maxrajda qolib, har bir o'tish foizini pastga tortadi. Maxrajga nima kirishini hal qilish — aslida asosiy analitik ish; SQL esa oson qismi.

Mashqlar to'plami

python41 qator

-- 1. Har bir shaharda nechta student? (noma'lum aniq ko'rsatilsin)
SELECT COALESCE(city, '(unknown)') AS city, COUNT(*) AS n
FROM students GROUP BY city ORDER BY n DESC;

-- 2. Kategoriya bo'yicha course larning o'rtacha narxi
SELECT category, COUNT(*) AS courses, ROUND(AVG(price)) AS avg_price
FROM courses GROUP BY category ORDER BY avg_price DESC;

-- 3. Har bir student ning o'rtacha bahosi, eng yaxshilari birinchi
SELECT student_id,
       COUNT(grade)         AS graded,
       ROUND(AVG(grade), 1) AS avg_grade
FROM enrollments
GROUP BY student_id
HAVING COUNT(grade) > 0
ORDER BY avg_grade DESC;

-- 4. 3 tadan ko'p enrollment i bor course lar
SELECT course_id, COUNT(*) AS n
FROM enrollments GROUP BY course_id HAVING COUNT(*) > 3;

-- 5. Oylar bo'yicha enrollment soni
SELECT strftime('%Y-%m', enrolled_at) AS month, COUNT(*) AS n
FROM enrollments GROUP BY month ORDER BY month;

-- 6. Har bir course bo'yicha tugatish foizi
SELECT course_id,
       COUNT(*) AS total,
       SUM(CASE WHEN status = 'completed' THEN 1 ELSE 0 END) AS completed,
       ROUND(100.0 * SUM(CASE WHEN status = 'completed' THEN 1 ELSE 0 END) / COUNT(*), 1) AS pct
FROM enrollments GROUP BY course_id ORDER BY pct DESC;

-- 7. Data-quality tekshiruvi: column lardagi NULL lar
SELECT COUNT(*) AS rows_total,
       COUNT(*) - COUNT(city)       AS missing_city,
       COUNT(*) - COUNT(birth_date) AS missing_birth_date
FROM students;

-- 8. Dublikat enrollment bormi? (hech narsa qaytmasligi kerak)
SELECT student_id, course_id, COUNT(*) AS times
FROM enrollments GROUP BY student_id, course_id HAVING COUNT(*) > 1;

1-natija

city	n
Tashkent	82
Samarkand	40
Bukhara	38
Fergana	36
Namangan	28
Andijan	24
Nukus	22
Karshi	22
(unknown)	8

9 qator

2-natija

category	courses	avg_price
Business	3	1 366 667
Math	3	1 216 667
Data	4	1 175 000
Programming	3	1 166 667
Design	2	900 000

5 qator

3-natija

student_id	graded	avg_grade
197	3	87.3
1	3	86.3
116	2	84
107	2	84
6	8	82.8
270	2	82.5
176	4	82.3
103	5	82.2
276	4	81.8
248	9	81.6
247	3	81
130	6	81
199	4	80.8
167	6	80.8
298	3	80.3

… 300 qatordan birinchi 15 tasi

300 qator

4-natija

course_id	n
1	193
2	186
3	174
4	167
5	167
6	174
7	165
8	170
9	161
10	179
11	200
13	187
14	189
15	188

14 qator

5-natija

month	n
2025-02	1 077
2025-03	131
2025-09	1 188
2025-10	104

4 qator

6-natija

course_id	total	completed	pct
10	179	139	77.7
7	165	120	72.7
2	186	134	72
9	161	112	69.6
8	170	118	69.4
1	193	130	67.4
4	167	111	66.5
14	189	125	66.1
5	167	110	65.9
13	187	123	65.8
6	174	114	65.5
11	200	116	58
15	188	101	53.7
3	174	81	46.6

14 qator

7-natija

rows_total	missing_city	missing_birth_date
300	8	0

1 qator

8-natija

student_id	course_id	times
1	3	2
1	11	2
2	11	2
3	6	2
3	11	2
4	7	2
4	15	2
5	2	2
5	5	2
6	4	2
6	14	2
7	9	2
8	1	2
10	14	2
11	7	2

… 514 qatordan birinchi 15 tasi

514 qator

5 ms

Intervyu savollari

S: COUNT(*), COUNT(column) va COUNT(1) farqi?

COUNT(*) row larni sanaydi, hatto butunlay NULL row larni ham. COUNT(column) faqat o'sha column NULL bo'lmagan row larni sanaydi. COUNT(1) zamonaviy engine larda COUNT(*) bilan aynan bir xil — uning tezroq degan eski qarash afsona.

S: AVG NULL larni hisobga oladimi?

Yo'q. U NULL bo'lmagan qiymatlar soniga bo'ladi. NULL lar nol deb hisoblanishini xohlasangiz, buni aniq yozishingiz kerak: AVG(COALESCE(grade, 0)) — lekin biznes haqiqatan shuni nazarda tutayaptimi, so'rang.

S: Nol row ustida SUM nima qaytaradi?

0 emas, NULL. Raqam kerak bo'lsa, COALESCE(SUM(x), 0) ga o'rang. COUNT esa to'g'ri 0 qaytaradi.

S: WHERE va HAVING?

WHERE GROUP BY dan oldin row larni filtrlaydi va aggregate ishlata olmaydi. HAVING GROUP BY dan keyin guruhlarni filtrlaydi va ishlata oladi. Aggregate talab qilmaydigan har qanday shart WHERE ga tegishli, chunki ertaroq filtrlash kamroq ish degani.

S: GROUP BY siz HAVING ishlatsa bo'ladimi?

Ha. U holda butun table bitta guruh deb qaraladi: SELECT AVG(grade) FROM enrollments HAVING AVG(grade) > 70; bitta row yoki hech narsa qaytaradi.

S: Nega SELECT course_id, student_id, AVG(grade) ... GROUP BY course_id noto'g'ri?

Chunki guruhda ko'p har xil student_id bor, chop etish uchun esa faqat bittasi mumkin. PostgreSQL xato beradi; SQLite va MySQL tasodifiy qiymat qaytaradi, bu esa battarroq — natija to'g'ridek ko'rinadi, lekin ma'nosiz.

S: Dublikat row larni qanday topasiz?

Unique bo'lishi kerak bo'lgan column lar bo'yicha GROUP BY, keyin HAVING COUNT(*) > 1.

S: Nega "mahsulot bo'yicha savdo" hisobotida ba'zi mahsulotlar yo'q bo'lishi mumkin?

Chunki GROUP BY faqat mavjud row larni ko'radi. Savdosi nol bo'lgan mahsulotning guruhlanadigan row i yo'q. Uni 0 bilan hisobotda qoldirish uchun products table idan LEFT JOIN kerak.

Xulosa

Aggregate lar ko'p row ni bittaga siqadi: COUNT, SUM, AVG, MIN, MAX.
COUNT(*) = row lar · COUNT(col) = NULL bo'lmagan qiymatlar · COUNT(DISTINCT col) = har xil NULL bo'lmagan qiymatlar. Ularning farqi — bepul NULL tekshiruvi.
COUNT(*) dan boshqa har bir aggregate NULL larni tashlaydi, bu esa AVG ning maxrajini o'zgartiradi. Nol row ustida SUM 0 emas, NULL.
GROUP BY har bir guruh uchun bitta chiqish row i yasaydi. U row i yo'q guruhlarni ko'rsata olmaydi — buning uchun LEFT JOIN kerak.
Oltin qoida: tanlangan har bir column yo GROUP BY da, yo aggregate ichida. SQLite/MySQL buzishga yo'l qo'yadi; buzmang.
WHERE guruhlashdan oldin, HAVING keyin. Iloji boricha erta filtrlang.
GROUP BY ... HAVING COUNT(*) > 1 — standart dublikat topuvchi.
* 1.0 yoki * 100.0 bilan kasrga majburlang; SUM(CASE WHEN ... THEN 1 ELSE 0 END) — foiz idiomi.
AVG yoniga doim COUNT ni chiqaring, shunda hech kim bitta qiymatdan yasalgan o'rtachaga ishonmaydi.

Keyingi: 5-bob — JOIN lar. Hozirgacha har bir hisobot bitta table ichida qamalgan edi; JOIN lar esa bitta natijada "student ismi va course nomi" deyish imkonini beradi.

oldingi3. Chuqur filtering (IN, LIKE, NULL)keyingi5. JOIN lar

Bu bobni bepul o'qidingiz.

Kitob o'rgatadi, lekin sizning kodingiz nega ishlamaganini ayta olmaydi. Kursda uy vazifangiz tekshiriladi.

Telegramda yozish Narxlar

course_id	n
1	193
2	186
3	174
4	167
5	167
6	174
7	165
8	170
9	161
10	179
11	200
13	187
14	189
15	188

course_id	n
1	193
2	186
3	174
4	167
5	167
6	174
7	165
8	170
9	161
10	179
11	200
13	187
14	189
15	188

course_id	n
1	193
2	186
3	174
4	167
5	167
6	174
7	165
8	170
9	161
10	179
11	200
13	187
14	189
15	188