我如何在我的 kivy 程序中使用 json api

问题描述

您好，我目前正在阅读 O'Reilly 的《在 Kivy 中创建应用程序》，有一个示例我无法使其正常工作，因为在他写这本书时 openWeatherMap 不需要 api 密钥 (APPID)，但现在它需要，而且我是一名新手程序员，不知道如何更改代码以使其正常工作.

hi I'm currently reading O'Reilly's Creating Apps in Kivy and there's an example that I can't get it to work correctly because by the time he wrote the book openWeatherMap didn't require api key (APPID) but now it does and I'm a novice programmer and don't know how to change the code so it would work.

这是main.py源代码:

this is the main.py source code:

from kivy.app import App
from kivy.uix.boxlayout import BoxLayout
from kivy.properties import ObjectProperty
from kivy.network.urlrequest import UrlRequest
import json

class AddLocationForm(BoxLayout):
    search_input = ObjectProperty()

def search_location(self):
    search_template = "http://api.openweathermap.org/data/2.5" + "find?q={}&type=like"
    search_url = search_template.format(self.search_input.text)
    request = UrlRequest(search_url, self.found_location)

def found_location(self, request, data):
    data = json.loads(data.decode()) if not isinstance(data, dict) else data
    cities = ["{} ({})".format(d['name'], d['sys']['country'])
        for d in data['list']]
    self.search_results.item_strings = cities
    print("
".join(cities))

class WeatherApp(App):
pass

if __name__ == '__main__':
WeatherApp().run()

这是weather.kv源代码:

and this is weather.kv source code:

AddLocationForm:

<AddLocationForm>:
    orientation: "vertical"
    search_input: search_box
    search_results: search_results_list
    BoxLayout:
    height: "40dp"
    size_hint_y: None
    TextInput:
        id: search_box
        size_hint_x: 50
    Button:
        text: "Search"
        size_hint_x: 25
        on_press: root.search_location()
    Button:
        text: "Current Location"
        size_hint_x: 25
ListView:
    id: search_results_list
    item_strings: []

代码很简单，您在文本框中输入城市名称并点击搜索，它会通过显示收到的名称来确认.

the code's simple you put a city name in textbox and hit search and it confirms it by showing the name it recieved.

推荐答案

好吧，不知道是不是迟到了，但是最近买了这本书，我也发现自己被这个问题卡住了.在谷歌搜索这个问题时，我偶然发现了你的问题以及这本书的 O'Reilly 链接.这就是作者对这个问题的看法:

OK, so I don't know if I'm late or not but having bought this book recently, I too found myself stuck exactly in this problem. Upon Googling this issue, I happened to stumble upon your question as well as O'Reilly link for this book. This is what the author had to say about this problem:

我已确认问题；openweathermap 已更改他们的查询过程和书中的网址现在都坏了.这将彻底破坏所有新读者的读者体验；我们需要做一个更新，也许应该谈谈第二版."

"I've confirmed the issue; openweathermap has changed their query process and the urls in the book are now all broken. This is going to utterly ruin the reader experience for all new readers; we'll need to do an update and should maybe talk about a second edition."

幸运的是，一位好心人找到了解决这个问题的方法.但为了做到这一点，您必须首先创建一个免费的开放天气帐户.创建帐户后，您将获得一个 API 密钥.它会在您的个人资料中.

Luckily a good Samaritan found out the solution to this problem. But in order to do so you must first create a free Open Weather account. After creating the account, you'll get an API key. It'll be in your profile.

所以，现在这段代码:

search_template = "http://api.openweathermap.org/data/2.5" + "find?q={}&type=like"

变成:

search_template = "http://api.openweathermap.org/data/2.5/find?q={}&type=like&APPID=" + "YOUR_API_KEY"

这对我有用.我知道我迟到了 3 个月，现在你可能已经得到了答案，但我认为这对遇到类似问题的人很有用，他们的谷歌结果会将他们带到这个地方.

This worked for me. I know I'm 3 months late and probably by now you'd have gotten your answer, but I thought this will be useful for those who run into similar problems and their Google result will bring them to this place.

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