如何解决但是,无论if语句还是else打印出来,都没有得到url数据库
此程序是一个控制台应用程序,它为用户显示应用程序标题和菜单选项,显示配方类别(即牛肉,鸡肉,素食主义者,侧面,沙漠等)的列表,并显示基于膳食的列表在用户选择的类别上显示,并显示基于用户选择的用餐的用餐信息(路线)。我只需要显示用户所选类别的食物清单的帮助即可。
我正在使用一个网站,该网站为此程序提供免费的JSON API,并且一切正常。该程序显示类别和食谱,但不会显示每个类别的单独餐食。 这很奇怪,因为我在函数的if语句之后放置了print()语句,该语句按类别获取进餐以尝试对其进行调试,但是它同时打印了pint(1)和print(2)语句,因此它可以完成获取工作正确,但仍会打印else语句。这部分代码是search_meal_by_categories函数中的recipes.py文件(位于以下)的一部分。 这就是我调试它的目的,以弄清楚为什么它同时打印出if和else:
def search_meal_by_category(categories):
lookup_category = input("Enter a category: ")
found = False
for i in range(len(categories)):
category = categories[i]
if category.get_category().lower() == lookup_category.lower():
found = True
print(1)
if found:
meals = requests.get_meals_by_category(lookup_category)
list_meals_by_category(lookup_category,meals)
print(2)
else:
print("Invalid Category,please try again")
这很奇怪,因为如果您输入的是食谱,它会显示食谱,但不会显示该类别中的食物。在命令菜单中,您可以: (“ 1-列出所有类别”) (“ 2-列出一个类别的所有餐点”) (“ 3-按名称搜索餐点”) (“ 0-退出程序”) 到目前为止,我可以输入命令1和3,但不能输入2或0。为此,有3个不同的python文件,我将这3个文件中的每一个都包含在每个文件中,仅按类别列出了特定的进餐功能,以简化操作(希望?)。
基本上,目标是要求程序在您键入的类别中显示每餐。到目前为止,如果您输入食谱的名称,则仅显示类别和食谱说明。但是,我可以考虑到这一点,因为它可以正常工作。
(object.py文件)
class Category:
def __init__(self,category):
self.__category = category
def get_category(self):
return self.__category
def set_category(self,category):
self.__category = category
class Meal:
def __init__(self,meal_id,meal,meal_thumb):
self.__meal_id = meal_id
self.__meal = meal
self.__meal_thumb = meal_thumb
def get_meal_id(self):
return self.__meal_id
def set_meal_id(self,meal_id):
self.__meal_id = meal_id
def get_meal(self):
return self.__meal
def set_meal(self,meal):
self.__meal = meal
def get_meal_thumb(self):
return self.__meal_thumb
def set_meal_thumb(self,meal_thumb):
self.__meal_thumb = meal_thumb
(requests.py文件)
from urllib import request,parse
import json
from objects import Category,Meal
def get_categories():
url = 'https://www.themealdb.com/api/json/v1/1/list.php?c=list'
f = request.urlopen(url)
categories = []
try:
data = json.loads(f.read().decode('utf-8'))
for category_data in data['meals']:
category = Category(category_data['strCategory'])
categories.append(category)
except (ValueError,KeyError,TypeError):
print("JSON format error")
return categories
def get_meals_by_category(category):
url = 'https://www.themealdb.com/api/json/v1/1/filter.php?c=Seafood' + category
f = request.urlopen(url)
meals = []
try:
data = json.loads(f.read().decode('utf-8'))
for meal_data in data['meals']:
category = Meal(meal_data['idMeal'],meal_data['strMeal'],meal_data['strMealThumb'])
meals.append(category)
except (ValueError,TypeError):
print("JSON Format error")
return meals
这是3个文件中的最后一个文件,我认为问题出在哪里。我在每个if之后都放置了print(1),print(2)语句,并打印出了所有AND否则else语句?基本上,现在显示的内容是“无效类别,请重试”,例如10次。
import requests
def show_title():
print("My recipes Program")
print()
def show_menu():
print("COMMAND MENU")
print("1 - List all Categories")
print("2 - List all Meals for a Category")
print("3 - Search Meal by Name")
print("0 - Exit the program")
print()
def list_categories(categories):
print("CATEGORIES")
for i in range(len(categories)):
category = categories[i]
print(category.get_category())
print()
def list_meals_by_category(category,meals):
print(category.upper() + " MEALS ")
for i in range(len(meals)):
meal = meals[i]
print(meal.get_meal())
print()
def search_meal_by_category(categories):
lookup_category = input("Enter a category: ")
found = False
for i in range(len(categories)):
category = categories[i]
if category.get_category().lower() == lookup_category.lower():
found = True
break
if found:
meals = requests.get_meals_by_category(lookup_category)
list_meals_by_category(lookup_category,meals)
else:
print("Invalid Category,please try again")
如果您需要更多信息,请发表评论,这是我第一次进行如此大的python项目(我知道这对每个人似乎都很容易,因为这里的每个人都是天才)。我真的已经盯着这个呆了两天了。谢谢大家的帮助!
解决方法
您是要写这个吗?
def search_meal_by_category(categories):
lookup_category = input("Enter a category: ")
found = False
for i in range(len(categories)):
category = categories[i]
if category.get_category().lower() == lookup_category.lower():
found = True
break
if found:
meals = requests.get_meals_by_category(lookup_category)
list_meals_by_category(lookup_category,meals)
else:
print("Invalid Category,please try again")
我没有缩进if found:
部分,因此当您获得found = True; break
时,它将跳过其余的for
并检测到found
。
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