Question

所以我有下一个json，我只想获取github链接和twitter，而不必获取foursquare和gravatar一个。

这还不是全部，有时json数据可以更改，如果找不到Foursquare网址，则github将成为第一个，而twitter将成为第二个。与其他人（github或twitter）相同。

如果github和twitter网址与json中的位置不同，如何获取它们？

{ "socialProfiles": [
    {
        "type": "foursquare",
        "typeId": "foursquare",
        "typeName": "Foursquare",
        "url": "https://foursquare.com/user/somerandomuser",
        "id": "554225246246"
    },
    {
        "type": "github",
        "typeId": "github",
        "typeName": "Github",
        "url": "https://github.com/somerandomuser",
        "username": "somerandomuser"
    },
    {
        "type": "gravatar",
        "typeId": "gravatar",
        "typeName": "Gravatar",
        "url": "https://gravatar.com/somerandomuser",
        "username": "somerandomuser",
        "id": "132341667"
    },
    {
        "bio": " This is a bio of a random user",
        "followers": 543,
        "following": 222,
        "type": "twitter",
        "typeId": "twitter",
        "typeName": "Twitter",
        "url": "https://twitter.com/somerandomuser",
        "username": "somerandomuser",
        "id": "41414515335"
    }
]

}

Answer 1

使用简单的迭代。

例如：

int main() {
   cpu_set_t mask;
   int temp[FILENAME_MAX]
   for(int i = 0; i < CPU_SETSIZE; i++)
   {
      if(CPU_ISSET(i, &mask))
      {
         temp[i] = i; 
      }
   }
   for(int i: temp)
      os << i;
   string str(os.str());
   cout << str;
   return 0;
}

输出：

checkList = ["twitter", "github"]
for i in data["socialProfiles"]:
    if i["typeId"] in checkList:    #Check if typeid is in your check-list
        print(i["url"])

https://github.com/somerandomuser https://twitter.com/somerandomuser是您的字典。

Answer 2

您只能使用dict comprehension和dict的网址来使用twitter创建另一个github

search = ['github', 'twitter']

urls = {dct['type']: dct['url'] for dct in data.get('socialProfiles', []) if dct['type'] in search}
print(urls)

输出

{
    'github': 'https://github.com/somerandomuser',
    'twitter': 'https://twitter.com/somerandomuser'
}

然后，您将获得所需的url。

print(urls['github'])

# Output
# https://github.com/somerandomuser

Answer 3

ES6: let newArr = arr.filter(e => e);

Python：从具有相同名称的列表中解析json数据

3 个答案: