Question

我在程序中将used = []定义为全局var。现在，我在jimoti中有一个函数while Loop。在函数内部，我遍历了web抓取（bs4）的结果，并将title的web-scarp添加到used列表中。当title中存在used时，我试图不再次打印它，但是一次又一次打印，因为正则表达式在两个或三个关键字上匹配它，所以我的文字相同2 ，打印3次。我该如何更改代码，使其仅打印一次？

这是代码


from bs4 import BeautifulSoup
import requests
from time import sleep
from random import randint
import re
import os

allowed = ["pc", "FUJITSU", "LIFEBOOK", "win" "Windows",
            "PC", "Linux" "linux", "HP", "hp", "notebook", "desktop",
            "raspberry", "NEC", "mac", "Mac", "Core"]
denied = ["philips"]
used = set()

source = requests.get("https://jmty.jp/aichi/sale-pcp").text
soup = BeautifulSoup(source, 'lxml')


def jimoti(sk):
    global used
    for h2 in soup.find_all('div', class_='p-item-content-info'):
        title = h2.select_one('.p-item-title').text
        address = h2.select_one('.p-item-title').a["href"]
        price = (h2.select_one('.p-item-most-important').text).replace("円", "").replace("\n", "").replace(",", "")
        price = int(price)
        town = h2.select_one('.p-item-supplementary-info').text
        if price < 5000000 and title not in used:
            used.add(title)
            for pattern in allowed:
                print(pattern)
                if re.search(pattern, title):
                    second(sk, title, address, price, town)
                    break

def second(sk, title, address, price, town):
    sk = sk
    title = title
    address = address
    price = price
    town = town
    for prh in denied:
        print(prh)
        if re.search(prh, title):
            break
        else:
            send(sk, title, address, price, town)



if __name__ == '__main__':
    while True:
        jimoti(sk)
        sleep(randint(11,20))

Answer 1

最初的问题是关于在设置条件时一次循环打印元素，然后再打印一次-我们避免break在第一次点击后退出循环。

seen = set()
for i in range(10):
    if i not in seen:
        for x in range(10):
            seen.add(i)
            break

内部循环中间有逻辑-

for prh in denied:
        print(prh)
        if re.search(prh, title):
            break
        else:
            send(sk, title, address, price, town)

按照书面形式，它将在prh中寻找title，直到找到为止，然后中断，因此将为prh中不在{{ 1}}。那可能不是您要表达的逻辑-title

send unless any of the prh values is in title?

上一个级别，您对“允许”的逻辑基本相同。我的猜测是正确的逻辑是-

if all([prh not in title for prh in denied]):
    send(sk, title, address, price, town)

我也不确定为什么你在那里睡得乱七八糟，似乎是破坏性的。不确定stackoverflow是您可能需要的支持级别的最佳论坛-对于python初学者有一个Reddit可能会更有用。

由于for循环而感到困惑

1 个答案: