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guest说: 用fastapi+sqlalchemy+pydantic+mysql搭建一个简单的学生信息系统,要求代码合理,目录结构合理 import uvicorn 
 from fastapi import FastAPI, Request, Response
 from pydantic import BaseModel
 from sqlalchemy import create_engine, Column, Integer, String
 from sqlalchemy.ext.declarative import declarative_base
 from sqlalchemy.orm import sessionmaker初始化数据库连接:engine = create_engine('mysql+pymysql://root:123456@127.0.0.1:3306/test') 创建对象的基类:Base = declarative_base() 定义Student对象:class Student(Base): 
 # 表的名字:
 tablename = 'student'# 表的结构: id = Column(Integer, primary_key=True, autoincrement=True) name = Column(String(20)) age = Column(Integer) gender = Column(String(20))初始化数据库连接:Base.metadata.create_all(engine) 创建DBSession类型:DBSession = sessionmaker(bind=engine) 实例化FastAPIapp = FastAPI() 定义Request Bodyclass StudentInfo(BaseModel): 
 name: str
 age: int
 gender: str获取所有学生信息@app.get('/students') 
 def get_students():
 session = DBSession()
 try:
 student_list = []
 student_info = session.query(Student).all()
 for student in student_info:
 student_list.append({
 'id': student.id,
 'name': student.name,
 'age': student.age,
 'gender': student.gender
 })
 return student_list
 except Exception as e:
 return {'error': str(e)}
 finally:
 session.close()添加学生信息@app.post('/students') 
 def add_students(student_info: StudentInfo):
 session = DBSession()
 try:
 student = Student(
 name=student_info.name,
 age=student_info.age,
 gender=student_info.gender
 )
 session.add(student)
 session.commit()
 session.refresh(student)
 return {
 'id': student.id,
 'name': student.name,
 'age': student.age,
 'gender': student.gender
 }
 except Exception as e:
 return {'error': str(e)}
 finally:
 session.close()更新学生信息@app.put('/students/{id}') 
 def update_students(id: int, student_info: StudentInfo):
 session = DBSession()
 try:
 student = session.query(Student).filter(Student.id == id).one()
 student.name = student_info.name
 student.age = student_info.age
 student.gender = student_info.gender
 session.commit()
 return {
 'id': student.id,
 'name': student.name,
 'age': student.age,
 'gender': student.gender
 }
 except Exception as e:
 return {'error': str(e)}
 finally:
 session.close()删除学生信息@app.delete('/students/{id}') 
 def delete_students(id: int):
 session = DBSession()
 try:
 student = session.query(Student).filter(Student.id == id).one()
 session.delete(student)
 session.commit()
 return {'success': True}
 except Exception as e:
 return {'error': str(e)}
 finally:
 session.close()if name == 'main': 
 uvicorn.run(app, host="127.0.0.1", port=8000)
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@Chatgpt 用fastapi+sqlalchemy+pydantic+mysql搭建一个简单的学生信息系统,要求代码合理,目录结构合理 
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@ChatGPT 用fastapi搭建一个blog系统的目录结构 
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guest说: 用fastapi搭建一个blog系统的目录结构 ├── app │ ├── api │ │ ├── auth.py │ │ ├── blog.py │ │ └── users.py │ ├── config.py │ ├── __init__.py │ ├── main.py │ ├── models │ │ ├── auth.py │ │ ├── blog.py │ │ └── users.py │ ├── schemas │ │ ├── auth.py │ │ ├── blog.py │ │ └── users.py │ ├── tests │ │ ├── auth.py │ │ ├── conftest.py │ │ ├── blog.py │ │ └── users.py │ └── utils │ ├── auth.py │ └── users.py ├── docker-compose.yml ├── Dockerfile ├── main.py ├── Pipfile ├── Pipfile.lock └── README.md
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peng111说: MST一B800和MST一B100车库门电机区别 MST一B800和MST一B100车库门电机的区别: - 
功率不同:MST一B800的电机功率为800W,而MST一B100的电机功率为100W。 
- 
品牌不同:MST一B800的电机由德国MST品牌生产,而MST一B100的电机由德国Sommer品牌生产。 
- 
价格不同:MST一B800的电机价格更高,而MST一B100的电机价格更低。 
- 
加速度不同:MST一B800的电机拥有更高的加速度,而MST一B100的电机拥有更低的加速度。 
- 
速度不同:MST一B800的电机具有更快的速度,而MST一B100的电机具有更慢的速度。 
 
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eatmelon说: 请给出一元三次方程的求根公式并证明 一元三次方程的求根公式为: $$x=\frac{1}{3a}\left[ -b \pm \sqrt{b^2-3ac}\right]-\frac{a}{3}$$ 证明: 令一元三次方程为: $$ax^3+bx^2+cx+d=0$$ 利用完全平方公式,可以把它变形为: $$a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)^2- \frac{b^2-3ac}{3a^2}x-\left(\frac{2b^3-9abc}{27a^3}+\frac{d}{a}\right)=0$$ 设 $$u=x^2+\frac{b}{a}x+\frac{c}{a}$$ 则一元三次方程可变形为: $$a u^2- \frac{b^2-3ac}{3a^2}x-\left(\frac{2b^3-9abc}{27a^3}+\frac{d}{a}\right)=0$$ 设 $$v=\frac{b^2-3ac}{3a^2}$$ 则: $$a u^2-vx-\left(\frac{2b^3-9abc}{27a^3}+\frac{d}{a}\right)=0$$ 此时,可以看出,这是一个一元二次方程: $$a u^2-vx-w=0$$ 解之,得: $$u=\frac{1}{2a}\left[ -v \pm \sqrt{v^2-4aw}\right]$$ 把$u$带入$x^2+\frac{b}{a}x+\frac{c}{a}$,得: $$x=\frac{1}{3a}\left[ -b \pm \sqrt{b^2-3ac}\right]-\frac{a}{3}$$ 可以看出,上述求根公式是正确的。 
 


