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
41k
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
不具合の先にある面白さ~配属3か月目の新卒QAのいま~
cybozuinsideout
PRO
0
91
kintone開発チームの紹介
cybozuinsideout
PRO
1
78k
kintone開発のプラットフォームエンジニアの紹介
cybozuinsideout
PRO
0
160
AIツール開発ワークショップ(Dify)【サイボウズ新人研修2025】
cybozuinsideout
PRO
20
24k
モバイル【サイボウズ新人研修2025】
cybozuinsideout
PRO
3
3.9k
Git/GitHub を使う上で知っておくと嬉しいかも Tips【サイボウズ新人研修2025】
cybozuinsideout
PRO
14
11k
GitHub Copilot活用【サイボウズ新人研修2025】
cybozuinsideout
PRO
15
15k
ソフトウェアライセンス【サイボウズ新人研修2025】
cybozuinsideout
PRO
14
8.5k
エンジニアのためのアウトプット講座 〜知識をシェアするはじめの一歩〜【サイボウズ新人研修2025】
cybozuinsideout
PRO
7
4.9k
Other Decks in Programming
See All in Programming
オープンソースソフトウェアへの解像度🔬
utam0k
17
3.1k
その面倒な作業、「Dart」にやらせませんか? Flutter開発者のための業務効率化
yordgenome03
1
140
Writing Better Go: Lessons from 10 Code Reviews
konradreiche
3
5.6k
デミカツ切り抜きで面倒くさいことはPythonにやらせよう
aokswork3
0
260
The Past, Present, and Future of Enterprise Java
ivargrimstad
0
130
EMこそClaude Codeでコード調査しよう
shibayu36
0
320
Cursorハンズオン実践!
eltociear
2
1.2k
スキーマ駆動で、Zod OpenAPI Honoによる、API開発するために、Hono Takibiというライブラリを作っている
nakita628
0
300
理論と実務のギャップを超える
eycjur
0
170
What's new in Spring Modulith?
olivergierke
1
160
Google Opalで使える37のライブラリ
mickey_kubo
3
130
Developer Joy - The New Paradigm
hollycummins
1
340
Featured
See All Featured
10 Git Anti Patterns You Should be Aware of
lemiorhan
PRO
657
61k
Documentation Writing (for coders)
carmenintech
75
5.1k
How to train your dragon (web standard)
notwaldorf
97
6.3k
[RailsConf 2023] Rails as a piece of cake
palkan
57
5.9k
Optimising Largest Contentful Paint
csswizardry
37
3.5k
Writing Fast Ruby
sferik
629
62k
Automating Front-end Workflow
addyosmani
1371
200k
Measuring & Analyzing Core Web Vitals
bluesmoon
9
630
Connecting the Dots Between Site Speed, User Experience & Your Business [WebExpo 2025]
tammyeverts
10
610
Thoughts on Productivity
jonyablonski
70
4.9k
Designing Dashboards & Data Visualisations in Web Apps
destraynor
231
53k
Building Flexible Design Systems
yeseniaperezcruz
329
39k
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 /