从json python中提取键值
我需要帮助从中提取“url”值-从json python中提取键值,json,angularjs,python-2.7,parsing,Json,Angularjs,Python 2.7,Parsing,我需要帮助从中提取“url”值- 它在上面不起作用。请帮助使用此代码可以获取第一个数组的url import urllib2, json response = urllib2.urlopen('https://www.udemy.com/api-2.0/channels/1640/courses?is_angular_app=true&p=2') data = json.load(response) print data['results'][0]['url'] 使用for循环从每个
它在上面不起作用。请帮助使用此代码可以获取第一个数组的url
import urllib2, json
response = urllib2.urlopen('https://www.udemy.com/api-2.0/channels/1640/courses?is_angular_app=true&p=2')
data = json.load(response)
print data['results'][0]['url']
使用for循环从每个数组获取所有url
import urllib2, json
response = urllib2.urlopen('https://www.udemy.com/api-2.0/channels/1640/courses?is_angular_app=true&p=2')
data = json.load(response)
i = 0;
for data["url"] in data['results']:
print data['results'][i]['url']
i+=1
试试这个:
import urllib2, json
response = urllib2.urlopen('https://www.udemy.com/api-2.0/channels/1640/courses?is_angular_app=true&p=2')
data = json.load(response)
for i in range(len(data['results'])):
print data['results'][i]['url']
检查此项:响应返回一个JSON字典,没有数组,也没有属性“url”-但是,响应包含属性“result”,该属性存储一个包含url属性的course对象数组。因此,您可以在整个响应上迭代results属性。我将继续&接受你们的答案,它帮助我获得了迭代“结果”数组的想法。
import urllib2, json
response = urllib2.urlopen('https://www.udemy.com/api-2.0/channels/1640/courses?is_angular_app=true&p=2')
data = json.load(response)
for i in range(len(data['results'])):
print data['results'][i]['url']