使用Python从JSON API中提取数据

Question

我通过这一部分:

如何从该URL中提取数据？我只想打印出来"networkdiff": 58954.60268219.

from urllib import urlopen

url = urlopen('http://21.luckyminers.com/index.php?page=api&action=getpoolstatus&api_key=8dba7050f9fea1e6a554bbcf4c3de5096795b253b45525c53562b72938771c41').read()

print url

这是打印url命令导致API显示的内容:

{
    "getpoolstatus": {
        "version": "1.0.0",
        "runtime": 16.618967056274,
        "data": {
            "pool_name": "21 Coin Pool @ Luckyminers.com",
            "hashrate": 485426748,
            "efficiency": 98.1,
            "workers": 14,
            "currentnetworkblock": 12025,
            "nextnetworkblock": 12026,
            "lastblock": 12023,
            "networkdiff": 58954.60268219,
            "esttime": 521.61956775542,
            "estshares": 241478052.58625,
            "timesincelast": 427,
            "nethashrate": 485426748
        }
    }
}

Answer 1

您可以使用该json模块解析Python字典并直接获取值,如下所示:

import json
result = json.loads(url)  # result is now a dict
print '"networkdiff":', result['getpoolstatus']['data']['networkdiff']

要多次执行此操作(在评论部分回答您的问题):

import json
import urllib

urls = {'joe': 'url1', 'jack': 'url2', 'jane': 'url3'}
for who in urls.keys():
    url = urllib.urlopen(urls[who])
    result = json.loads(url)  # result is now a dict
    print 'For %s: "networkdiff":' % who, result['getpoolstatus']['data']['networkdiff']

Answer 2

将响应转换为json然后读取

from urllib import urlopen
import simplejson as json

url = urlopen('http://21.luckyminers.com/index.php?page=api&action=getpoolstatus&api_key=8dba7050f9fea1e6a554bbcf4c3de5096795b253b45525c53562b72938771c41').read()
url = json.loads(url)

print url.get('getpoolstatus').get('data').get('networkdiff')