Basic SQL statement - only record

------------------------------- Operation on the table ------------ ------------------

1. Create a table:

create table student(
    id int,
    name varchar(32),
    age int ,
    score double(4,1),
    birthday date,
    insert_time timestamp
);

2. View table structure
desc table name;
3. Modify table name
alter table table name rename to new table name;
4. Add a column
alter table table name add column name data type;
5. Delete column
alter table table name drop column name ;
6. Delete table
drop table table name;
drop table if exists table name;

-------------------------------Data Manipulation Language------------- -------------------

1. Increase insert into
without writing column names (all columns are added)
insert into table name values ​​(value 1, value 2, ... value n);
insert some data
insert into table name (column name 1, column name 2) values(value1, value2);

2. delete delete 

Delete data in the table
delete from table name where column name = value;
delete all data in the table
delete from table name;

3. Modify update

Unconditional modification (all rows will be modified)
update table name set column name = value;
conditional modification
update table name set column name = value where column name = value;

---------------------------------Query class --------------- -----------------

1.between ... and ...

Query age greater than or equal to 20 and less than or equal to 30                

SELECT * FROM student WHERE age >= 20 &&  age <=30;
SELECT * FROM student WHERE age >= 20 AND  age <=30;
SELECT * FROM student WHERE age BETWEEN 20 AND 30;

Query age information of 22, 18, and 25

SELECT * FROM student WHERE age = 22 OR age = 18 OR age = 25
SELECT * FROM student WHERE age IN (22,18,25);

2. is null (not a null value)

Query English score is not null
SELECT * FROM student WHERE english IS NOT NULL;

3. like (fuzzy query) _: single arbitrary character  %: multiple arbitrary characters Who are the ones with the surname Ma? like SELECT * FROM student WHERE NAME LIKE '马%'; - query the person whose name the second word is Hua             SELECT * FROM student WHERE NAME LIKE "_化%";                 query the person whose name is 3 characters SELECT * FROM student WHERE NAME LIKE '___';                     Query people whose name contains Germany SELECT * FROM student WHERE NAME LIKE '%德%';
 

4.distinct (remove duplicate values)
The keyword DISTINCT is used to return uniquely different values.
Syntax: SELECT DISTINCT column name FROM table name

The first deduplication method:
SELECT DISTINCT name FROM student ;

The second deduplication method:
SELECT name FROM student GROUP BY name;

5. sort order by

Default ascending
SELECT * FROM person ORDER BY math;
ascending second method
SELECT * FROM users ORDER BY username asc;

降序
SELECT * FROM person ORDER BY math desc; 

--------------------------------- Computing class --------------- -----------------

1.count: Calculate the number

//使用 COUNT(*)对表中行的数目进行计数时，不管表列中包含的是空值（NULL）还是非空值都会被计数，因为*代表了所有
SELECT COUNT(*) FROM oderlist
//使用 COUNT(column)对特定列中具有值的行进行计数，忽略 NULL 值。
SELECT COUNT(item_price) FROM oderlist

2.max: Calculate the maximum value min: Calculate the minimum value

//返回表中的最大单价
SELECT MAX(item_price) AS maxPrice FROM oderlist
//返回表中的最小单价
SELECT MIN(item_price) AS maxPrice FROM oderlist

3.sum: calculate and

//SUM()函数忽略列值为 NULL 的行
//返回某一种水果goodsName“火龙果”的销售数量quantity
SELECT SUM(quantity) AS totalQuantity FROM oderlist WHERE goodsName = '火龙果'

4.avg: Calculate the average

//求oderlist表格内 item_price 的平均值,返回为avgPrice
SELECT AVG(item_price) avgPrice FROM oderlist 
//或者：
SELECT AVG(item_price) AS avgPrice FROM oderlist
//加上条件，返回item_price<10 的 item_price的平均值
SELECT AVG(item_price) AS avgPrice FROM oderlist WHERE item_price <10

5. Combination of aggregate functions:

//查询订单列表中订单量（totalOrders）、最低单价（minItemPrice）、最高单价（maxItemPrice）、均价（avgItemPrice）
SELECT 
COUNT(*) AS totalOrders,
MIN(item_price) AS minItemPrice,
MAX(item_price) AS maxItemPrice,
AVG(item_price) AS avgItemPrice
FROM "oderlist" 

------------------------------------------------example: -----------------------------------------------

Answer:

select
    count(DISTINCT device_id) as did_cnt,
    count(id) as question_cnt
from
    question_practice_detail
where
    date like '2021-08%'