Question

全新的Python并尝试解析如下所示的JSON数组：

[
  {"Event":"start","EventDateTime":"2015-09-15T03:45:16.681428Z"},
  {"Event":"process","EventDateTime":"2015-09-15T03:45:16.681428Z"},
  {"Event":"end","EventDateTime":"2015-09-15T03:45:16.681428Z"}
]

我需要将输出放在字符串制表符分隔的字段和新行分隔的行上，如下所示：

start \t 2015-09-15T03:45:16.681428Z \n
process \t 2015-09-15T03:45:16.681428Z \n
end \t 2015-09-15T03:45:16.681428Z

这是我到目前为止的代码：

          import json
      if not j or not i:
        return None
      try:
        arr = json.loads(j)
      except ValueError:
        return None
      if len(arr) <= 0:
        return None
      row=i
      for li in arr
        elem = json.loads(li, object_pairs_hook=collections.OrderedDict)
        row=row + '\t' + elem.Key + '\t' + elem.Value + \n
      return row

首先我得到，缩进错误，修复了缩进，但收到了关于未定义的'集合'的错误。

有没有办法在不使用该集合的情况下完成我需要的工作。当我删除集合对象时，我得到其他错误。

谢谢！

Answer 1

import json    

j = """[
  {"Event":"start","EventDateTime":"2015-09-15T03:45:16.681428Z"},
  {"Event":"process","EventDateTime":"2015-09-15T03:45:16.681428Z"},
  {"Event":"end","EventDateTime":"2015-09-15T03:45:16.681428Z"}
]"""

j = json.loads(j)
for item in j:
    print '%s\t%s' % (item['Event'], item['EventDateTime'])

Answer 2

如下：

    j = '''[
{"Event":"start","EventDateTime":"2015-09-15T03:45:16.681428Z"},
    {"Event":"process","EventDateTime":"2015-09-15T03:45:16.681428Z"},
      {"Event":"end","EventDateTime":"2015-09-15T03:45:16.681428Z"}
      ]
      '''
import json
content = json.loads(j)
keys = content[0].keys()
for i in content:
    print(' \t '.join([i[k] for k in keys]) + '\n')

Answer 3

鉴于OP正在使用OrderedDict钩子，那么你可以在不参考名称的情况下做到这一点：

arr = json.loads(j, object_pairs_hook=collections.OrderedDict)
for item in arr:
    print(' \t '.join(item.values()))

在Python中解析JSON数组以创建分隔的字符串

3 个答案: