Question

我有一个JSON对象，想知道如何迭代该对象以提取“ id”的值。

{
"totalSize": 5,
"done": true,
"records": [
    {
        "attributes": {
            "type": "EventLogFile",
            "url": "/services/data/v38.0/sobjects/EventLogFile/0AT1U000003kk7dWAA"
        },
        "Id": "0AT1U000003kk7dWAA"
    },
    {
        "attributes": {
            "type": "EventLogFile",
            "url": "/services/data/v38.0/sobjects/EventLogFile/0AT1U000003kk7eWAA"
        },
        "Id": "0AT1U000003kk7eWAA"

我正在尝试以下操作。

sub_data = s["records"]["id"]
for i in sub_data:
        print(sub_data['id'])

Answer 1

您可以遍历列表中的records键，然后访问每个子字典的Id键：

for i in s["records"]:
    print(i['Id'])

Answer 2

s = """{ "totalSize": 5, 
        "done": true, "records": [ 
            { "attributes": { 
                "type": "EventLogFile", 
                "url": "/services/data/v38.0/sobjects/EventLogFile/0AT1U000003kk7dWAA" }, 
                "Id": "0AT1U000003kk7dWAA" } 
        ] 
    }"""
s = json.loads(s)

[r['Id'] for r in s['records']]

['0AT1U000003kk7dWAA']

使用Python迭代JSON对象

2 个答案: