即使两个选择是 none 但第一个选项是 python 列表中的正确选择，输出也不会打印结果

Question

晚上好。

我想问一下为什么我下面的代码没有结果，即使 3 是 celfone 的价格，我输入了 none 作为上限价格， none 作为上限价格搜索品牌。输出是 "You did not enter any search parameters."

我的代码片段如下：

def search_celfone():
templist = []
temp = []
count = 0

lower_price = input("Enter lower bound of price: ")
upper_price = input("Enter upper bound of price: ")
search_brand = str(input("Enter brand name: ")).upper()

#IF CHOICES 1 AND 2 ARE NONE
if lower_price == "none" and upper_price == "none": 
    for cp in celfones:
        temp = celfones[count]
        count += 1

    #IF CHOICES 1 AND 2 ARE NONE AND CHOICE 3 IS IN THE LIST
        if temp[1] == search_brand:
            templist.append(temp)
            break

    #IF CHOICES 1 AND 2 ARE NONE AND CHOICE 3 IS NOT IN THE LIST INCLUDING NONE
        else:
            temp[1] != search_brand
            continue
        break   

#IF CHOICE 1 IS INTEGER AND CHOICES 2 AND 3 ARE NONE    
elif lower_price != "none" and upper_price == "none" and search_brand == "none":
    lower_price = float(lower_price)
    for cp in celfones:
        temp = celfones[count]
        count += 1


        if temp[1] == search_brand and temp[2] >= lower_price:
            templist.append(temp)
            break
        #elif temp[1] == search_brand and temp[2] >= lower_price:
            #templist.append(temp)
            #break
        #elif 

        #elif temp[1] != search_brand or temp[2] >= lower_price:
        elif temp[1] != search_brand and temp[2] >= lower_price:
            continue
        break               


else:
    print ("you did not enter any search parameter") 
count += 1
print (templist)
buyer_menu()

BUYER MODE

============================== [1] 显示所有celfone [2] 根据搜索celfone 价格和品牌

[3]返回上一级菜单

选择 >>> 2 输入价格下限：3 输入价格上限：none 输入品牌名称：none 您没有输入任何搜索参数

[]

CHOICE >>> 1 There are 2 celfones in the list. 1. Lumia, NOKIA, Php3.0 Features: 2. Xperia, SONY, Php4.0 Features:

================================

Answer 1

深入了解这些代码行：

    if temp[1] == search_brand and temp[2] >= lower_price:
        templist.append(temp)
        break
    elif temp[1] != search_brand and temp[2] >= lower_price:
        continue
    break

search_brand 是，凭借片段 and search_brand == "none": none。因此，在添加 "none" 品牌 phone 之前，您不会找到任何匹配项。此外，它第一次运行循环 where temp[2] < lower_price 时，它会跳出循环，忽略任何后来的匹配对象。

我建议删除该片段中的两个 break 语句，并完全忽略 if 语句中的 search_brand。

Answer 2

感谢您的投入。我已经找到了解决这个问题的方法。我刚刚对该功能进行了重大修改。

主要区别是：

我在三个选择之后立即放置了for循环。
我删除了里面的几个 for 循环。
我也很难理解和应用 if 和 elif 语句的真相 table。（我只是编程新手）
此外，我还删除了所有语句中的中断。感谢您的输入@Fawful
此外，我将某些语句的 "none" 更改为 "NONE"。感谢您的输入@PM 2Ring

感兴趣的朋友，下面是我修改后的代码：

for cp in celfones:
    temp = celfones[count]
    tempf = temp[3]
    count += 1

    if lower_price == "none" and upper_price == "none": 

        if temp[1] == search_brand:
            templist.append(temp)

    elif lower_price != "none" and upper_price == "none":
        lower_price = float(lower_price)

        if temp[1] == search_brand and temp[2] >= lower_price:
            templist.append(temp)

        elif search_brand == "NONE" and temp[2] >= lower_price:
            templist.append(temp)

    elif lower_price == "none" and upper_price != "none":
        upper_price = float(upper_price)

        if temp[1] == search_brand and temp[2] <= upper_price:
            templist.append(temp)

        elif search_brand == "NONE" and temp[2] <= upper_price:
            templist.append(temp)

    elif lower_price != "none" and upper_price != "none":
        lower_price = float(lower_price)
        upper_price = float(upper_price)

        if search_brand == "NONE" and temp[2] >= lower_price and temp[2] <= upper_price:
            templist.append(temp)

        elif temp[1] == search_brand and temp[2] >= lower_price and temp[2] <= upper_price:
            templist.append(temp)

    else:
        print ("you did not enter any search parameter") 


print (len(templist), "celfone(s) found.")  ### counting of results

即使两个选择是 none 但第一个选项是 python 列表中的正确选择，输出也不会打印结果

Output does not print result even though the two choices are none but the first option is the right choice in python list

python

nested

list

[3]返回上一级菜单

[]