与json.loads的Python3奇怪的错误

时间:2016-05-03 16:38:14

标签: python json python-3.x flask

我在使用service api的Web应用程序中使用生成JSON响应。该函数的以下部分工作正常并返回JSON文本输出:

def get_weather(query = 'london'):
    api_url = "http://api.openweathermap.org/data/2.5/weather?q={}&units=metric&appid=XXXXX****2a6eaf86760c"
    query = urllib.request.quote(query)
    url = api_url.format(query)
    response = urllib.request.urlopen(url)
    data = response.read()    
    return data

返回的输出是:

{"coord":{"lon":-0.13,"lat":51.51},"weather":[{"id":803,"main":"Clouds","description":"broken clouds","icon":"04d"}],"base":"cmc stations","main":{"temp":12.95,"pressure":1030,"humidity":68,"temp_min":12.95,"temp_max":12.95,"sea_level":1039.93,"grnd_level":1030},"wind":{"speed":5.11,"deg":279.006},"clouds":{"all":76},"dt":1462290955,"sys":{"message":0.0048,"country":"GB","sunrise":1462249610,"sunset":1462303729},"id":2643743,"name":"London","cod":200}

这意味着data是一个字符串,不是吗?

然而,评论return data,然后添加以下两行:

jsonData = json.loads(data)
return jsonData

生成以下错误:

  

TypeError:JSON对象必须是str,而不是'bytes'

怎么了? data JSON对象,以前作为字符串返回!我需要知道错误在哪里?

1 个答案:

答案 0 :(得分:2)

request库返回的数据是二进制字符串,而json.loads接受str,因此您需要使用编码将数据(decode)转换为字符串您的请求返回(通常可以认为它是UTF-8)。

您应该只需将代码更改为:

return json.loads(data.decode("utf-8"))

PS:在返回变量之前存储变量是多余的,所以我简化了事情