使用Python获取json对象的名称

Question

如何获取json对象的名称，以便在if语句中匹配它？

def test():
    device_config = api_get_conf()
    routing_protocol = device_config['Cisco-IOS-XE-native:native']
    if 'ospf' in routing_protocol:
        print('It worked!')
    else:
        print('dunno')

routing_protocol变量中包含以下信息：

        {"Cisco-IOS-XE-ospf:router-ospf": {
            "ospf": {
                "process-id": [
                    {
                        "id": 100,
                        "area": [
                            {
                                "area-id": 0,
                                "authentication": {
                                    "message-digest": [
                                        null
                                    ]
                                }
                            }
                        ],
                        "network": [
                            {
                                "ip": "192.168.200.200",
                                "wildcard": "0.0.0.0",
                                "area": 0
                            }
                        ]
                    }
                ]
            }
        }
    }

我只想匹配'Cisco-IOS-XE-ospf:router-ospf‘或'ospf’。任何关于我如何做到这一点的帮助将不胜感激。

Answer 1

我不明白你的意思，但是看起来api_get_conf()返回字典，其中"Cisco-IOS-XE-ospf:router-ospf“是第一个键，它的值是另一个字典，其中键是"ospf”。如果这是您想要比较的，那么您可以简单地使用routing_protocol dict上的.keys()方法。

def test():
    device_config = api_get_conf()
    routing_protocol = device_config['Cisco-IOS-XE-native:native']
    if 'ospf' in routing_protocol.keys():
        print('It worked!')
    else:
        print('dunno')

Answer 2

def test():
    device_config = api_get_conf()
    # since we use get here, if we dont find it we set routing_protocol as False, easier to use on if
    routing_protocol = device_config.get('Cisco-IOS-XE-native:native', False)
    if routing_protocol:
        if 'ospf' in routing_protocol:
            print('It worked!')
        print("ospf not a key")
    else:
        print('routing protocol not a key')