习题三

        习题十三、COUNT(*) ... WHERE

combine COUNT(*) with WHERE to return thenumber of rows that matches the WHERE clause.

return the number of rows in friends_of_pickles where the species is a dog?：

select count (*)

            from friends_of_pickles

                    where   species = 'dog';

Result:

count (*)

3

       习题十四、SUM

find the sum of a given value. 

 find the total num_books_read made by thisfamily?

 SELECT SUM(num_books_read)

FROM family_members;

Result:

SUM(num_books_read)

380

           习题十五、AVG

find the average of a given value. 

find the average num_books_read made by eachfamily member?

 select avg (num_books_read)

from family_members ;  

Result:

avg (num_books_read)

126.66666666666667

！！Note: 
- Because of the way computers handle numbers, averages will not always becompletely exact.！！

         习题十六、MIN和MAX

find the maximum or minimum value of a table. 

find the least number of legs in a family member and find the highest num_books_read that afamily member makes?

SELECT MIN(num_legs)

FROM family_members; 

select max (num_books_read)

from family_members;

Result:

max (num_books_read)

200

          习题十七、GROUP BY、

friends_of_pickles

id	name	gender	species	height_cm
1	Dave	male	human	180
2	Mary	female	human	160
3	Fry	male	cat	30
4	Leela	female	cat	25
5	Odie	male	dog	40
6	Jumpy	male	dog	35
7	Sneakers	male	dog	55

             use aggregate functions such as COUNT, SUM, AVG, MAX, and MINwith the GROUP BY clause. splitthe table into different piles based on the value of each row. 

SELECT COUNT(*), species

 FROMfriends_of_pickles

 GROUP BY species;

return the tallest height for each species? Remember toreturn the species name next to the height too, like in the example query.

 select max(height_cm ) ,species

from friends_of_pickles

group by species;

Result:

max(height_cm )	species
30	cat
55	dog
180	human

          习题十八、NESTED QUERIES

put a SQL query inside another SQL query. 

SELECT * FROM family_members

WHERE num_legs =

           (SELECT MIN(num_legs)

          FROM family_members); 

 return the family members that have thehighest num_books_read?

SELECT * from family_members

where num_books_read =

          (select max (num_books_read)

                      from family_members )
Result:

id	name	species	num_books_read	num_legs
1	Dave	human	200	2

Current tables3:

family_members

id	name	gender	species	favorite_book
1	Dave	male	human	To Kill a Mockingbird
2	Mary	female	human	Gone with the Wind
3	Pickles	male	dog	null