peewee

Posted 2021-02-17 escapist

query = Facility.select()   等价于SQL   SELECT * FROM facilities

Facility.select(Facility.name, Facility.membercost)  等价于  SELECT name, membercost FROM facilities;

Facility.select().where(Facility.membercost > 0) 等价于 SELECT * FROM facilities WHERE membercost > 0

Facility.select().where(Facility.name.contains(‘tennis‘))  等价  SELECT * FROM facilities WHERE name ILIKE ‘%tennis%‘;

　或者   Facility.select().where(Facility.name ** ‘%tennis%‘)

 

query = Facility.select().where(Facility.facid.in_([1, 5]))   in操作

 

SELECT name,
CASE WHEN monthlymaintenance > 100 THEN ‘expensive‘ ELSE ‘cheap‘ END
FROM facilities;

　对应ORM语句

cost = Case(None, [(Facility.monthlymaintenance > 100, ‘expensive‘)], ‘cheap‘)
query = Facility.select(Facility.name, cost.alias(‘cost‘))

 

比较日期

SELECT memid, surname, firstname, joindate FROM members
WHERE joindate >= ‘2012-09-01‘;

Member.select(Member.memid, Member.surname, Member.firstname, Member.joindate).where(Member.joindate >= datetime.date(2012, 9, 1))

 

SELECT DISTINCT surname FROM members ORDER BY surname LIMIT 10;

Member.select(Member.surname).order_by(Member.surname).limit(10).distinct()

　UNION 操作  把查询到的数据 合并 到一起  前提：列数相同

SELECT surname FROM members UNION SELECT name FROM facilities;

lhs = Member.select(Member.surname)
rhs = Facility.select(Facility.name)
query = lhs | rhs

• |   代表   UNION        相同的只显示一次
• +   代表   UNION ALL    都显示
• &   代表   INTERSECT    只显示相同的
• -  代表   EXCEPT       

 

SELECT MAX(join_date) FROM members;

query = Member.select(fn.MAX(Member.joindate))