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Martin Hope
Jeid
Asked: 2024-10-02 15:39:42 +0000 UTC
  • 6

如何进行正确的 SQL 查询以通过 JSON 字段进行搜索示例表 table1

ID 部分
1 [{"key":"xxx", "name":"命名一"},{"key":"yyy", "name":"命名二"}]
2 [{"key":"zzz", "name":"命名 Tre"},{"key":"xxx", "name":"命名 One"}]
3 [{"key":"aaa", "name":"名称为"},{"key":"zzz", "name":"名称 Tre"}]

你需要提出一个请求，比如

SELECT id FROM table1 WHERE sections->key IN('xxx','zzz')

因此，它应该返回 1,2,3

此外，如何使用此 JSON 字段在 Laravel 中组织正确的关系将是一个优势。

sql
Martin Hope
hhhscvx
Asked: 2024-10-02 14:09:28 +0000 UTC
  • 5

需要pyrogram根据对 API 的请求创建和授权会话。

get-code有效，但sign-in由于代码过期而失败。我很确定这是因为我Client发布了不同代码并执行操作sign_in，但我不知道如何使用代码。

关于可以做什么的任何想法？

代码：

from fastapi import APIRouter, Body
from pyrogram import Client
from pyrogram.errors import SessionPasswordNeeded, BadRequest
from pyrogram.types import SentCode

from core import settings
from .redis_db import save_to_redis, get_from_redis, delete_from_redis

router = APIRouter(tags=['Telegram'])

@router.post('/telegram/get-code')
async def get_code(
    data=Body()
):
    phone = str(data['phone'])
    phone = phone.replace('+', '')

    client = Client(name=f"{phone}",
                    api_id=settings.API_ID,
                    api_hash=settings.API_HASH,
                    workdir=settings.WORKDIR
                    )

    await client.connect()

    try:
        code_data: SentCode = await client.send_code(
            phone_number=phone,
        )

        await save_to_redis(key=f"tg_client_{phone}", value={
            'phone': phone,
            'phone_code_hash': code_data.phone_code_hash
        })

        return {'success': True}
    except Exception as err:
        await client.disconnect()
        raise err

@router.post('/telegram/sign-in')
async def sign_in(
    data=Body()
):
    phone = str(data['phone'])
    phone = phone.replace('+', '')
    phone_code = data['phone_code']

    session_data = await get_from_redis(key=f"tg_client_{phone}")

    if not session_data:
        return {'success': False, 'message': 'Session not found'}

    client = Client(
        name=f"{phone}",
        api_id=settings.API_ID,
        api_hash=settings.API_HASH,
        workdir=settings.WORKDIR
    )

    await client.connect()

    phone_code_hash = session_data['phone_code_hash']

    try:
        await client.sign_in(phone_number=phone,
                             phone_code_hash=phone_code_hash,
                             phone_code=phone_code)

        await client.disconnect()
        await delete_from_redis(key=f"tg_client_{phone}")
        return {'success': True}
    except SessionPasswordNeeded:
        password = data.get('password', None)

        try:
            await client.check_password(password=password)
        except BadRequest:
            await client.disconnect()
            return {'success': False, 'message': 'Incorrect password'}
        else:
            await client.disconnect()
            return {'success': True}

错误：

INFO:     127.0.0.1:41094 - "POST /api/v1/telegram/sign-in HTTP/1.1" 500 Internal Server Error
pyrogram.errors.exceptions.bad_request_400.PhoneCodeExpired: Telegram says: [400 PHONE_CODE_EXPIRED] - The confirmation code has expired (caused by "auth.SignIn")
python
Martin Hope
Biteclipse
Asked: 2024-10-02 05:51:32 +0000 UTC
  • 7

我有一个网站，我想以文本形式显示时间表。我似乎能够处理发送请求，但我似乎无法获取时间表本身。告诉我怎样做才是正确的

import requests
from bs4 import BeautifulSoup

group_id = "93"
date_from = "02.10.2024"
date_to = "03.10.2024"

url = 'https://www.tolgas.ru/services/raspisanie/?id=0'
payload = {
    'vr': group_id,  # Идентификатор группы
    'from': date_from,  # Начальная дата
    'to': date_to,  # Конечная дата
    'submit_button': 'Показать'
}

response = requests.post(url, data=payload)
soup = BeautifulSoup(response.text, 'html.parser')

假设我至少想参加这个日期日期

使用此代码

soup.find_all('div', class_="div.timetable-frame__row:nth-child(1) > div:nth-child(1) > div:nth-child(1) > div:nth-child(2)")
print(soup.find_all)

但最终我显然得到了一个完整的网站，因为......显示大量文本，不仅是 div 标签，还包括所有其他标签

python
Martin Hope
Steven Kirke
Asked: 2024-10-02 04:24:41 +0000 UTC
  • 6

我无法理解任务的逻辑。当顾客在特定日期购买食物时，会有一张桌子（在示例中有几个）。并且有一个在同一时间段内访问过或未访问过的商店的列表。

第一个条件是显示在一段时间内访问过和未访问该商店的所有人员。第二个条件是显示在一段时间内访问过或未访问过的商店。

我无法连接这些数据。问题是如何连接第 1 列和第 3 列。也就是说，第 3 列应该显示未访问过或访问过的商店，同时链接人员，反之亦然，如果该人没有访问该商店。根据条款，禁止使用运营商。

NOT IN, IN, NOT EXISTS, EXISTS, UNION, EXCEPT, INTERSECT

我是这样实现的，但是不知道如何选择。

有三个表，Buers 与表关联，Visit、Stores 与 Hanging 关联，Visit 是有客户 ID 和商店 ID 的地方。

买家

create table person
( id bigint primary key ,
  name varchar(MAX) not null,
  age integer not null default 10,
  gender varchar(6) default 'female' not null ,
  address varchar(MAX)
  );

alter table person add constraint ch_gender check ( gender in ('female','male') );

insert into person values (1, 'Anna', 16, 'female', 'Moscow');
insert into person values (2, 'Andrey', 21, 'male', 'Moscow');
insert into person values (3, 'Kate', 33, 'female', 'Kazan');
insert into person values (4, 'Denis', 13, 'male', 'Kazan');
insert into person values (5, 'Elvira', 45, 'female', 'Kazan');

商店

create table pizzeria
(id bigint primary key ,
 name varchar(50) not null ,
 rating numeric not null default 0);

alter table pizzeria add constraint ch_rating check ( rating between 0 and 5);

insert into pizzeria values (1,'Pizza Hut', 4.6);
insert into pizzeria values (2,'Dominos', 4.3);
insert into pizzeria values (3,'DoDo Pizza', 3.2);
insert into pizzeria values (4,'Papa Johns', 4.9);
insert into pizzeria values (5,'Best Pizza', 2.3);
insert into pizzeria values (6,'DinoPizza', 4.2);

买家来访

create table person_visits
(id bigint primary key ,
 person_id bigint not null ,
 pizzeria_id bigint not null ,
 visit_date date not null default GETDATE(),
 constraint uk_person_visits unique (person_id, pizzeria_id, visit_date),
 constraint fk_person_visits_person_id foreign key  (person_id) references person(id),
 constraint fk_person_visits_pizzeria_id foreign key  (pizzeria_id) references pizzeria(id)
 );

insert into person_visits values (1, 1, 1, '2022-01-01');
insert into person_visits values (2, 2, 2, '2022-01-01');
insert into person_visits values (3, 2, 1, '2022-01-02');
insert into person_visits values (4, 3, 5, '2022-01-03');
insert into person_visits values (5, 3, 6, '2022-01-04');
insert into person_visits values (6, 4, 5, '2022-01-07');
insert into person_visits values (7, 4, 6, '2022-01-08');
insert into person_visits values (8, 5, 2, '2022-01-08');
insert into person_visits values (9, 5, 6, '2022-01-09');
insert into person_visits values (10, 6, 2, '2022-01-09');
insert into person_visits values (11, 6, 4, '2022-01-01');
insert into person_visits values (12, 7, 1, '2022-01-03');
insert into person_visits values (13, 7, 2, '2022-01-05');

第一个请求，我选择在特定日期访问商店的员工

DECLARE @DataMin DATE = '2022-01-01';
DECLARE @DataMax DATE = '2022-01-03';

CREATE TABLE ##ResultVisitPersons
(namePerson NVARCHAR(100), resultVisit BIT, visitDuringPeriod NVARCHAR(100), pizzeriaName NVARCHAR(100), pizzeriaID INT)

INSERT INTO ##ResultVisitPersons
SELECT
    visit.name,
    visit.resultVisit,
    visit.visitDuringPeriod,
    visit.pizzeriaName,
    visit.pizzeria_id
FROM (
  SELECT 
    person.name,
    CASE 
        WHEN person_visits.visit_date BETWEEN @DataMin AND @DataMax THEN 1
        ELSE 0
    END AS resultVisit,
    CASE 
        WHEN person_visits.visit_date BETWEEN @DataMin AND @DataMax THEN LEFT(CONVERT(VARCHAR, person_visits.visit_date, 104), 10)
        ELSE 'NULL'
    END AS visitDuringPeriod,
    person_visits.pizzeria_id,
    pizzeria.name as pizzeriaName
FROM person_visits
JOIN person ON person.id = person_visits.person_id
JOIN pizzeria ON pizzeria.id = person_visits.pizzeria_id
) visit

第二个请求是同一时间段内商店访问的样本。

CREATE TABLE ##ResultVisitPizzeria
(visitDuringPeriod BIT, result NVARCHAR(100), namePizzeria NVARCHAR(100), id INT)

INSERT INTO ##ResultVisitPizzeria
SELECT 
    CASE 
        WHEN person_visits.visit_date BETWEEN @DataMin AND @DataMax THEN 1
        ELSE 0
    END AS visitDuringPeriod,
        CASE 
        WHEN person_visits.visit_date BETWEEN @DataMin AND @DataMax THEN N'Ресторан посещали'
        ELSE N'Ресторан не посещали'
    END AS isVisitDuringPeriod,
    pizzeria.name,
    pizzeria.id
FROM person_visits
JOIN pizzeria ON pizzeria.id = person_visits.pizzeria_id

在此输入图像描述 在此输入图像描述 在此输入图像描述

sql
Martin Hope
Ivan
Asked: 2024-10-02 03:22:26 +0000 UTC
  • 6

垃圾收集器正在尝试清理未签入的连接 <AdaptedConnection <Connection(Thread-3, started daemon 19576)>>

该错误与会话有关，这是代码

使用会话

engine = create_async_engine(os.getenv('DB_LITE'), echo=True)

session_maker = async_sessionmaker(bind=engine, class_=AsyncSession, expire_on_commit=False)

async def create_session():
    async with session_maker() as session:
        return session

函数本身

@post_router.message(F.text)
async def post_delete():
    session= await create_session()
    for post in await orm_get_all_post(session):
        date_today = datetime.now()
        date_meet = post.date
        date_meet = datetime(2024, int(date_meet.split('.')[1]), int(date_meet.split('.')[0]))
        if (date_meet-date_today).days==-2:
            await orm_delete_post(session, post.product_name)

她的挑战

async def main():
    dp.startup.register(on_startup)
    dp.shutdown.register(on_shutdown)

    scheduler=AsyncIOScheduler(timezone='Europe/Moscow')
    scheduler.add_job(post_delete, trigger='interval', seconds=5)
    scheduler.start()

    dp.update.middleware(DataBaseSession(session_pool=session_maker))

    await bot.delete_webhook(drop_pending_updates=True)
    await dp.start_polling(bot, allowed_updates=dp.resolve_used_update_types())

asyncio.run(main())

问题是它是一个后台函数：如果你使用 kwargs，它就会变成一个类。如果有人知道请帮忙！

python
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