python - 迭代python字典以将值分配给表的字段

使用API​​，我以以下形式获得json字典作为响应：

{"transacDet":[{"field1":6933434802,"field2":323499903,"field3":293483842},
{"field1":693433448,"field2":333400300,"field3":298334300}],"responseStatus"
:{"statusMessage":"success","statusCode":0}}

我正在寻找在具有与json响应相同的对应字段的表中创建记录：field1，field2，field3。

从上面的示例中，我将获得2条记录：

field1 = 6933434802
field2 = 333400298
field3 = 298334842

接着：

field1 = 693433448
field2 = 333400300
field3 = 298334300

我必须反复回答，但不清楚如何做。

我尝试使用：

for k in resp_json.items()[0][1][0]:
print k

但我被困在那里。
TIA

最佳答案

仅使用标准库中的json模块，您可以轻松地按每个“行”的字段名称对它们进行迭代，如下所示：

#! /usr/bin/env
from __future__ import print_function
import json


JSON_TEXT = '''
{"transacDet":[{"field1":6933434802,"field2":323499903,"field3":293483842},
{"field1":693433448,"field2":333400300,"field3":298334300}],"responseStatus"
:{"statusMessage":"success","statusCode":0}}'''

resp_json = json.loads(JSON_TEXT)
for d in resp_json['transacDet']:
    for k in sorted(d.keys()):
        print(k, d[k])

产生：

field1 6933434802
field2 323499903
field3 293483842
field1 693433448
field2 333400300
field3 298334300

更新以概述如何将这些值推入INSERT调用：

# inside the loop instead of iterating over the fields in the
# dict / row at hand, one could easily for instance do:
resp_json = json.loads(JSON_TEXT)
for d in resp_json['transacDet']:
    insert_row(**d)

在这里，insert_row将是一个定义如下的函数（示例）：

def insert_row(**kwargs):
    """Not production ready ;-)."""
    # pick up the row key-value pairs from dict
    list_of_tuples = [
        (k, kwargs[k] for k in sorted(kwargs.keys())]
    pass  # here you might use this list of (k, v) tuples