Back to marketplace

SQL

No description yet. This curated deck is ready for daily study.

Total cards: 12•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (12)

#1
Front (Question)

INNER JOIN

Back (Answer)

Select all records from Table A and Table B, where the join condition is met. Intersection.

#2
Front (Question)

RIGHT JOIN

Back (Answer)

Select all records from Table B, along with records from Table A for which the join condition is met (if at all).

#3
Front (Question)

FULL JOIN

Back (Answer)

Select all records from Table A and Table B, regardless of whether the join condition is met or not.

#4
Front (Question)

Left outer join

Back (Answer)

A left outer join will give all rows in A, plus any common rows in B

#5
Front (Question)

Right outer join

Back (Answer)

A right outer join will give all rows in B, plus any common rows in A.

#6
Front (Question)

Full outer join

Back (Answer)

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn’t have a corresponding datum in B, then the B portion is null, and vice versa.

#7
Front (Question)

core aggregate functions

Back (Answer)

AVG, COUNT, MIN, MAX, SUM

#8
Front (Question)

normalization

Back (Answer)

To free the collection of relations from undesirable insertion, update and deletion dependencies; To reduce the need for restructuring the collection of relations, as new types of data are introduced, and thus increase the life span of application programs; To make the relational model more informative to users; To make the collection of relations neutral to the query statistics, where these statistics are liable to change as time goes by.

#9
Front (Question)

stored procedure

Back (Answer)

is a subroutine available to applications that access a RDBMS Uses • data-validation (integrated into the database) • access-control mechanisms. • consolidate and centralize logic that was in applications. - execution of several SQL statements One can use nested stored procedures by executing one stored procedure from within another.

#10
Front (Question)

SUBSTRING

Back (Answer)

SUBSTRING(‘hello’, 3, 1) index 1 based!! one letter is substring(‘hello’, 2, 2)

#11
Front (Question)

CASE

Back (Answer)

CASE WHEN blah THEN blah WHEN blah THEN blah ELSE blah END

#12
Front (Question)

group by and getting id

Back (Answer)

joining back with subselect inside join to join columns that match and get id