通过代码打开json文件链接 [英] Open json file link through a code

问题描述

我正在创建插件，并且正在修改py文件中的某些函数. 我打算做的是以下代码，我有此代码:

I'm creating an addon and I'm modifying some functions that come within a py file. What I intend to do is the following, I have this code:

def channellist():
    return json.loads(openfile('lib.json',pastafinal=os.path.join(tugapath,'resources')))

此代码可访问资源子文件夹中tugapath文件夹内的lib.json文件.我所做的就是将lib.json文件放入保管箱，并希望将其替换为lib.json文件中的保管箱链接，而不是调用文件夹.

This code gives access to a lib.json file that is inside the tugapath folder in the resources subfolder. What I did was put the lib.json file in the dropbox and wanted to replace it with the dropbox link from the lib.json file instead of calling the folders.

我试图更改代码，但没有成功.

I tried to change the code but without success.

def channellist():
    return json.loads(openfile('lib.json',pastafinal=os.path.join("https://www.dropbox.com/s/sj1246qtiodm6qd/lib.json?dl=1')))

如果有人可以帮助我，我将不胜感激！ 首先谢谢你.

If someone can help me, I'm grateful! Thank you first.

推荐答案

鉴于您的链接包含有效的json-您发布的内容并非如此-您可以使用请求.

Given that your link holds valid json - which is not the case with the content you posted - you could use requests.

如果Dropbox上的内容看起来像这样:

If the content at dropbox looked liked this:

{"tv": 
    {"epg": "tv",
    "streams": 
        [{"url": "http://topchantv.net:3456/live/Stalker/Stalker/838.m3u8", 
        "name": "IPTV", 
        "resolve": False, 
        "visible": True}], 
    "name": "tv", 
    "thumb": "thumb_tv.png"
    }
}

然后获取内容就是这样

import requests

url = 'https://www.dropbox.com/s/sj1246qtiodm6qd/lib.json?dl=1'
r = requests.get(url)
json_object = r.json()

因此，如果您需要在函数中使用它，我想您应该输入url并返回json，如下所示:

So if you needed it inside a function, I guess you'd input the url and return the json like so:

def channellist(url):
    r = requests.get(url)
    json_object = r.json()
    return json_object 

这篇关于通过代码打开json文件链接的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！