将Ruby中的json响应移植到Python
将Ruby中的json响应移植到Python
提问于 2018-06-16 03:33:39
嘿，我用Ruby编写了一个利用JSON API响应的程序，我想把它移植到python上，但我真的不知道怎么做
JSON响应：
{
"Class": {
"Id": 1948237,
"family": "nature",
"Timestamp": 941439
},
"Subtitles": [
{
"Id":151398,
"Content":"Tree",
"Language":"en"
},
{
"Id":151399,
"Content":"Bush,
"Language":"en"
}
]
}下面是Ruby代码：
def get_word
r = HTTParty.get('https://example.com/api/new')
# Check if the request had a valid response.
if r.code == 200
json = r.parsed_response
# Extract the family and timestamp from the API response.
_, family, timestamp = json["Class"].values
# Build a proper URL
image_url = "https://example.com/image/" + family + "/" + timestamp.to_s
# Combine each line of subtitles into one string, seperated by newlines.
word = json["Subtitles"].map{|subtitle| subtitle["Content"]}.join("\n")
return image_url, word
end
end无论如何，我可以使用请求或者json模块将这段代码移植到Python？我试过了，但惨遭失败。
每个请求；我已经尝试过的内容：
def get_word():
r = requests.request('GET', 'https://example.com/api/new')
if r.status_code == 200:
# ![DOESN'T WORK]! Extract the family and timestamp from the API
json = requests.Response
_, family, timestamp = json["Class"].values
# Build a proper URL
image_url = "https://example.com/image/" + family + "/" + timestamp
# Combine each line of subtitles into one string, seperated by newlines.
word = "\n".join(subtitle["Content"] for subtitle in json["Subtitles"])
print (image_url + '\n' + word)
get_word()响应和_, family, timestamp = json["Class"].values代码无法工作，因为我不知道如何移植它们。
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-06-16 04:22:25
如果您使用的是requests模块，那么可以调用requests.get()来进行GET调用，然后使用json()来获取JSON响应。此外，如果要导入json模块，则不应该使用json作为变量名。
尝试在您的函数中进行以下更改：
def get_word():
r = requests.get("https://example.com/api/new")
if r.status_code == 200:
# Extract the family and timestamp from the API
json_response = r.json()
# json_response will now be a dictionary that you can simply use
...并使用json_response字典获取您的变量所需的任何内容。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50881690
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