Upgrade to Pro
— share decks privately, control downloads, hide ads and more …
Speaker Deck
Features
Speaker Deck
PRO
Sign in
Sign up for free
Search
Search
2018-11-データベース / 2018-11 database
Search
Cybozu
PRO
July 17, 2018
Programming
6
40k
2018-11-データベース / 2018-11 database
データベースの概要を知る。
データベースに求められる特性を知る。
SQLの概要を把握する。
自分でSQLを書けるようになる。
Cybozu
PRO
July 17, 2018
Tweet
Share
More Decks by Cybozu
See All by Cybozu
サイボウズフロントエンドエキスパートチームについて / FrontendExpert Team
cybozuinsideout
PRO
6
39k
2024/11/25 ReDesigner Online Meetup 会社紹介
cybozuinsideout
PRO
0
380
サイボウズ 開発本部採用ピッチ / Cybozu Engineer Recruit
cybozuinsideout
PRO
9
49k
テクニカルライティング
cybozuinsideout
PRO
4
580
サイボウズのアジャイルクオリティ2024
cybozuinsideout
PRO
3
460
モブに早く慣れたい人のためのガイド2024
cybozuinsideout
PRO
3
630
モバイル
cybozuinsideout
PRO
3
340
ソフトウェアライセンス
cybozuinsideout
PRO
4
310
ソフトウェアテスト
cybozuinsideout
PRO
3
520
Other Decks in Programming
See All in Programming
お前もAI鬼にならないか?👹Bolt & Cursor & Supabase & Vercelで人間をやめるぞ、ジョジョー!👺
taishiyade
6
4k
Unity Android XR入門
sakutama_11
0
160
Lottieアニメーションをカスタマイズしてみた
tahia910
0
130
Amazon S3 TablesとAmazon S3 Metadataを触ってみた / 20250201-jawsug-tochigi-s3tables-s3metadata
kasacchiful
0
170
pylint custom ruleで始めるレビュー自動化
shogoujiie
0
120
もう僕は OpenAPI を書きたくない
sgash708
5
1.8k
2024年のWebフロントエンドのふりかえりと2025年
sakito
3
250
Amazon Bedrock Multi Agentsを試してきた
tm2
1
290
『GO』アプリ データ基盤のログ収集システムコスト削減
mot_techtalk
0
120
苦しいTiDBへの移行を乗り越えて快適な運用を目指す
leveragestech
0
620
GitHub Actions × RAGでコードレビューの検証の結果
sho_000
0
270
データベースのオペレーターであるCloudNativePGがStatefulSetを使わない理由に迫る
nnaka2992
0
150
Featured
See All Featured
The Cult of Friendly URLs
andyhume
78
6.2k
Build your cross-platform service in a week with App Engine
jlugia
229
18k
GitHub's CSS Performance
jonrohan
1030
460k
The Language of Interfaces
destraynor
156
24k
Fashionably flexible responsive web design (full day workshop)
malarkey
406
66k
Designing Dashboards & Data Visualisations in Web Apps
destraynor
231
53k
Faster Mobile Websites
deanohume
306
31k
Fireside Chat
paigeccino
34
3.2k
Producing Creativity
orderedlist
PRO
344
39k
Imperfection Machines: The Place of Print at Facebook
scottboms
267
13k
10 Git Anti Patterns You Should be Aware of
lemiorhan
PRO
656
59k
How GitHub (no longer) Works
holman
314
140k
Transcript
! !
%! $'" #( &
()
EL C l
X l C N D P l AQ S C I I I AQ
& %# ( "'"$ ! '
l l l
l l / l /
!%0 ! %0
l 75) +42 l # 1-&( l ' *3 $,6. / "
A C C l
l l D I
l
23 23 23 23 23 23 51 .
0 4
0 0 0 0 0 0 2.34 1 10
! = A l ( ) A )
1 B2 A B l 5 B
l 9 0 l 9 l 9 9 5 93
9 541 54 0 12 9 541 54
. 13 9 = 0 6 2 3 53 54 0 1 2
= l ) ) ) ) l
( I
- l ) ) ) l
l l ) ( l ( l (
V PfdcK S l CD B .)
,B 4 .) B B l 4G 4 OR G .) aMN , 4D D C l eQ , 4 Rb , 4D D C aMN L B D C l ( G D B l C 4 4 GB 4 C 4 D D B l 4
)( 3 4 1 2
- 0 7
)( 2 l 0 7 - 1 l
0 43 l 2
:- 3 0 1 l 47 2 0
D D I
n l o D1B l l a
o ( ) ( B R a R i e R B 1
)( (
C ) ( ) ) l l
L ) ) ) (/ ( l D (/ ( l
E C l E C L , ( R
l E C PS ), R l E C IN R l E C ( R E C D : A :
0/. -73394 l e atQ0/.
qL QF d S c i l 6 ## 9 73394 2 :# # 1 l k 0/. s Q q p ! ! ! ! n o ! ! fhm 7 4l belong_to organization employee
- l
SELECT [] FROM [ ]; SELECT name FROM employee;
- * l * SELECT * FROM employee;
- l SELECT [ 21 ] FROM [
0 ] WHERE []; SELECT * FROM employee WHERE joined_at >= '2010/1/1';
- ] a l [ _ d 1 ./
5 75 .7 :/05 6 /6 / :/ 276 1 5 joined_at > '2007/1/1' 726/. _ ' name = '' 6 5/ _ c e id <> 10001 2. _
- iNO f ( rL ) rL !=
ds ) nI K )( ( c , tvap . ul moh q eT ./ . ul moh q eT E6 < = > 5 : : 5 : : 6
name LIKE '%%'; % ' % l l
l l l l
) ) ( % l name LIKE
'%¥%%' ( '
- l id IN (10001, 10003) l
- 41 l l l 0 2
0 SELECT * FROM employee WHERE joined_at >= '2010/1/1' AND joined_at <= '2014/1/1';
1 1
- D E l E ) E l AE E
l C C( E D l SELECT [ ] FROM [ ] ORDER BY [ ] [ASC/DESC]; SELECT * FROM employee ORDER BY joined_at ASC;
l l 2 SELECT [ ] FROM [
ORDER BY [ ] [ASC/DESC] LIMIT []; SELECT * FROM employee ORDER BY joined_at ASC LIMIT 2; SELECT * FROM employee ORDER BY joined_at ASC LIMIT 1, 2;
#$ % ) l "
! SELECT [&'] FROM [(] GROUP BY []; SELECT organization_id, COUNT(*) FROM belong_to GROUP BY organization_id;
) * ( * ) * ( * ) *
( * 3 4 12 GROUP BY org_id SELECT org_id, COUNT(*) 0 0 2 3 4 1 ) *
l ) ( )
15 0 I a D l 678 047 2
a l 3 66 2 l 506837 d_ 9 : l _ l g d l ed l ced grade
SELECT class_id, COUNT(*), AVG(result), MAX(result), MIN(result) FROM grade GROUP BY
class_id;
B O RO l U H E l G
P
l SELECT * FROM [ ] GROUP BY
[ 2 ] HAVING []; SELECT organization_id, COUNT(*) AS num FROM belong_to GROUP BY organization_id HAVING num >= 2;
GROUP BY org_id SELECT org_id, COUNT(*) 0 1 3 HAVING num >= 2 2 4
" $ ! #%
- 3 4 1 2
- 0 7 - 0 3 - 0 4 7 1 2
- SELECT organization.name, employee.name FROM belong_to INNER JOIN organization ON
belong_to.organization_id = organization.id INNER JOIN employee ON belong_to.employee_id = employee.id;
-
belong_to organization " $%"!# … belong_to INNER JOIN organization ON belong_to.organization_id = organization.id …
- . . .
… belong_to INNER JOIN organization ON belong_to.organization_id = organization.id …
- . . . 143 2 0 … belong_to INNER
JOIN organization ON belong_to.organization_id = organization.id INNER JOIN employee ON belong_to.employee_id = employee.id;
-
SELECT organization.name, employee.name FROM belong_to INNER JOIN organization ON belong_to.organization_id = organization.id INNER JOIN employee ON belong_to.employee_id = employee.id;
- ( "!&%$ ' %#& l "!&%#&
l "!&%$ ' %#& SELECT MAX(result) FROM grade; SELECT AVG(result) FROM grade WHERE result <> [];
- 1 SELECT AVG(result) FROM grade WHERE result <>
(SELECT MAX(result) FROM grade);
- > l < l < l <
SELECT class_id, MAX(result) FROM grade GROUP BY class_id; SELECT class_id, AVG(result) FROM grade WHERE (class_id = 1 AND result <> [1 ]) OR (class_id = 2 AND result <> [2 ]) OR … GROUP BY class_id;
- 1 1 SELECT class_id, AVG(result) FROM grade
AS g1 WHERE g1.result <> ( SELECT MAX(result) FROM grade AS g2 WHERE g1.class_id = g2.class_id ) GROUP BY class_id;
SELECT class_id, AVG(result) FROM grade AS g1 WHERE [] GROUP BY class_id;
6 63
8 3 9 7 0 2 1 5 4 g1 SELECT class_id, AVG(result) FROM grade AS g1 WHERE [] GROUP BY class_id;
1 g1
WHERE g1.result <> ( SELECT MAX(result) FROM grade AS g2 WHERE g1.class_id = g2.class_id )
g1 g2 WHERE g1.result <> ( SELECT MAX(result) FROM grade AS g2 WHERE g1.class_id = g2.class_id )
.
g2 g1 WHERE g1.result <> ( SELECT MAX(result) FROM grade AS g2 WHERE g1.class_id = g2.class_id )
MAX(result)
g2 g1 WHERE g1.result <> ( SELECT MAX(result) FROM grade AS g2 WHERE g1.class_id = g2.class_id )
36 4215 0
g2 g1 WHERE g1.result <> 100
g1 1 89 B 4 3 0 2 6 753
& Q & E C L
& Q
INSERT INTO [
] ([ ]) VALUES ([]); INSERT INTO employee (id, name, joined_at) VALUES (10001, '', '2013/04/01'), (10002, '', '2014/04/01'), (10003, '', '2007/04/01');
$# " %
! & UPDATE [ ] SET [ ] = [ ] WHERE [ ]; UPDATE employee SET joined_at = '2007/4/2' WHERE id = 10003;
DELETE
FROM [] WHERE []; TRUNCATE []; DELETE FROM [];
! !
Q l e j
c i l D P E S IC A l PX A L N N N D j
3 L : 88
Q . 4 666 0 721 2 /
& h M 1 0 S d L Q
M c a .774 888 1 : 2 /4 4
U B Ca B JF 2 26
N S 3 : / / 0 4: 1: .
9Q S : a L c
d : h .55 4 0 721 2 /