Question

当我尝试搜索列表时收到 TypeError: cannot unpack non-iterable bool object。当我返回一个布尔值和我正在搜索的项目的索引值时。

def name_ser(name):
    found = False
    for i in range(len(names)):
        if names[i] == name:
            found = True
            return found, names.index(name)
        else:
            found = False
            return found,None


def main_menu():
    print('Welcome!\nPlease Choose from the following options...')
    print('1: Create an account\n2: Login ')
    opt = int(input('Enter Your Choice: '))
    if opt == 1:
        name_search = input('Enter Name... ')
        found, _ = name_ser(name_search)
        if found == True:
            print("Account Already excites!")
        elif found == False & len(names) == 0:
            acc_creation(name_search)
            print('Account created!')

错误：

    Traceback (most recent call last):
  File "/Users/darkmbs/VS-Code/FirstPythonProject/accounts.py", line 100, in <module>
    main_menu()
  File "/Users/darkmbs/VS-Code/FirstPythonProject/accounts.py", line 77, in main_menu
    found, _ = name_ser(name_search)
TypeError: cannot unpack non-iterable NoneType object

Answer 1

import React from 'react'; import { Header } from 'components/header/Header'; import { SEO } from 'components/seo/SEO'; import { SliceZone } from 'components/slice-zone/SliceZone'; import { client } from 'lib/prismic'; import { RichText } from 'prismic-reactjs'; const Index = ({ document }: any) => { const slices = document; if (!slices) return null; return ( <> <SEO title={RichText.asText(document.page_title)} description={document.meta_description} banner={document.social_image.url} /> <SliceZone slices={slices} /> </> ); }; export const getServerSideProps = async () => { const options = { lang: 'en-us', }; const { data: document } = await client.getSingle('homepage', options); return { props: { document, }, }; }; export default Index; 可以返回 3 个不同的对象。

name_ser

您需要一致的返回类型。但实际上，这个功能根本不应该存在。它试图返回 def name_ser(name): found = False for i in range(len(names)): if names[i] == name: found = True return found, names.index(name) <== returns 2-tuple else: <== if name[0] doesn't match, found = False executes immediately return found <== returns boolean <== if names is empty, returns None 或 name 中 names 的索引。 False 差不多就是这样。

str.find

在添加帐户之前，我还删除了 def main_menu(): print('Welcome!\nPlease Choose from the following options...') print('1: Create an account\n2: Login ') opt = int(input('Enter Your Choice: ')) if opt == 1: name_search = input('Enter Name... ') if name_search in names: print("Account Already exists!") else: acc_creation(name_search) print('Account created!') 为空的检查。我觉得如果只有一个人可以创建一个帐户会很奇怪。

Answer 2

正如之前的一些评论中提到的，您不能同时返回一个或两个值。我的建议是返回名称的索引。或者，如果未找到，则返回值 -1。

def name_ser(name):
    # Check if element is in list
    if name in names:
        # Return index of element
        return names.index(name)
    else:
        # Name was not found in the list
        return -1

def main_menu():
    print('Welcome!\nPlease Choose from the following options...')
    print('1: Create an account\n2: Login ')

    opt = int(input('Enter Your Choice: '))

    # Create account
    if opt == 1:
        name_search = input('Enter Name... ')
        found = name_ser(name_search)

        if found >= 0:
            print("Account Already exists!")

        elif found == -1 & len(names) == 0:
            acc_creation(name_search)
            print('Account created!')
    
    # Login to Account
    if opt == 2:
        # Update to use index instead of boolean

Python TypeError：无法解包不可迭代的 bool 对象

2 个答案: