Python-从列表中的每个词典]，包含2所列出

-1

{ 
"error": null, 
"page": "1", 
"per_page": "1000", 
"results": [ 
    { 
     "cves": [ 
      { 
       "cve_id": "CVE-2016-1583", 
       "href": "https://www.redhat.com/security/data/cve/CVE-2016-1583.html" 
      }, 
      { 
       "cve_id": "CVE-2016-5195", 
       "href": "https://www.redhat.com/security/data/cve/CVE-2016-5195.html" 
      } 
     ], 
     "description": "The kernel packages contain the Linux kernel, the core of any Linux operating\nsystem.\n\nSecurity Fix(es):\n\n* A race condition was With this update, a set of patches has been applied that fix\nthese problems. As a result, the time stamps of GFS2 files are now handled\ncorrectly. (BZ#1374861)", 
     "errata_id": "RHSA-2016:2124", 
     "hosts_applicable_count": 0, 
     "hosts_available_count": 0, 
     "id": "81ee41e6-2a3a-4475-a88e-088dee956787", 
     "issued": "2016-10-28", 
     "packages": [ 
      "kernel-2.6.18-416.el5.i686", 

     ], 
     "reboot_suggested": true, 
     "severity": "Important", 
     "solution": "For details on how to apply this update, which includes the changes described in\nthis advisory, refer to:\n\nhttps://access.redhat.com/articles/11258\n\nThe system must be rebooted for this update to take effect.", 
     "summary": "An update for kernel is now available for Red Hat Enterprise Linux 5.\n\nRed Hat Product Security

我想提取errata_id和总结（只是RHEL版本）的值，其我想放置的重新解释，即：RHSA-2016：2098：红帽企业Linux 5

我能提取errats名单，但不与总结作为一个字典只是作为一个列表：

ERRATA_ID_LIST = [] 
for errata_ids in erratas_by_cve_dic['results']: 
    ERRATA_ID = errata_ids['errata_id'] 
    ERRATA_ID_LIST.append(ERRATA_ID

来源

2017-05-29 Max_il

你在寻找什么样的输出？我无法理解你在这里想问什么 –

你在这里发布的json不是有效的json。请检查它 –

如果我知道你想创建{ID，总结}一本字典，所以你可以做：

ERRATA_ID_DICT = {} 
for element in erratas_by_cve_dic['results']: 
    ERRATA_ID_DICT[element['errata_id']] = element['summary']

来源

2017-05-29 12:06:26

有两种可能的方法：

1）使用for循环，像你这样的，但将数据放入一个词典：

errata = {} 
for errata_ids in erratas_by_cve_dic['results']: 
    errata_id = errata_ids['errata_id'] 
    summary = errata_ids['summary'] 
    errata[errata_id] = summary

，或者更短但可能不太清楚：

errata = {} 
for errata_ids in erratas_by_cve_dic['results']: 
    errata[errata_ids['errata_id']] = errata_ids['summary']

使用字典解析10

2）：

errata = { errata_ids['errata_id']: errata_ids['summary'] for errata_ids in erratas_by_cve_dic['results'] }

我喜欢的第二方法更多，因为它不仅是短，但也更Python（即。使用Python成语）。

来源

2017-05-29 12:14:14

Python-从列表中的每个词典]，包含2所列出

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