小编Sim*_*ann的帖子

django 调用“listing”函数，即使我告诉它在处理访问 .../watchlist url 时出现的请求时使用“watchlist”函数。我找不到问题。这是错误：

  File "/home/simon/Dokumente/cs50WebProgramming/commerce/auctions/views.py", line 103, in listing
    listing_obj = AuctionListing.objects.get(id=int(listing_id))
ValueError: invalid literal for int() with base 10: 'watchlist'

网址.py

urlpatterns = [
    path("", views.index, name="index"),
    path("login", views.login_view, name="login"),
    path("logout", views.logout_view, name="logout"),
    path("register", views.register, name="register"),
    path("create", views.create_listing, name="create"),
    path("after_create", views.after_create, name="after_create"),
    path("<str:listing_id>", views.listing, name="listing"),
    path("<str:listing_id>/bid", views.after_bid, name="after_bid"),
    path("watchlist", views.watchlist, name="watchlist")
]

视图.py

def listing(request, listing_id):
    listing_obj = AuctionListing.objects.get(id=int(listing_id))

    return render(request, "auctions/listing.html", {
        "listing": listing_obj
    })

def watchlist(request):
    return render(request, "auctions/watchlist.html")

我从nestJS文档（https://docs.nestjs.com/techniques/http-module#http-module）中获取了这个示例，这是我的问题的一个最小示例：

@Injectable()
export class CatsService {
  constructor(private httpService: HttpService) {}

  findAll(): Observable<AxiosResponse<Cat[]>> {
    return this.httpService.get('http://localhost:3000/cats');
  }
}

如何从 Observable<AxiosResponse<Cat[]>> 中提取实际的猫数组？我尝试了以下操作，但它给了我一个订阅者对象，我也不知道如何解开该对象以获取实际数据。

const cats = await this.catsService.findAll().subscribe((val) => val);

我通过使用字符串（class ）和整数（class ）解决了以下leet代码问题https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/。我认为整数解决方案应该更快并且使用更少的内存。但实际上需要更长的时间。当 n=18 和 k=200 时，需要 10.1 秒，而字符串解决方案需要 0.13 秒。SolutionSolution2

import time

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        i = 0
        Sn = "0"
        Sn = self.findR(Sn, i, n)
        return Sn[k-1]

    def findR(self, Sn, i, n):
        if i == n:
            return Sn
        newSn = self.calcNewSn(Sn, i)
        return self.findR(newSn, i+1, n)

    def calcNewSn(self, Sn, i):
        inverted = ""
        for c in Sn:
            inverted += "1" if c == "0" else "0"
        newSn …