【发布时间】:2018-09-05 01:39:32
【问题描述】:
我有一个嵌套字典,我正在尝试解析它,但似乎不知道如何访问第三级项目。帮助表示赞赏 这是我的字典
{
"FunctionName": "RDSInstanctStart",
"LastModified": "2018-03-24T07:19:56.792+0000",
"MemorySize": 128,
"Environment": {
"Variables": {
"DBInstanceName": "test1234"
}
},
"Version": "$LATEST",
"Role": "arn:aws:iam::xxxxxxx:role/lambda-start-RDS",
"Timeout": 3,
"Runtime": "python2.7",
"TracingConfig": {
"Mode": "PassThrough"
},
"CodeSha256": "tBdB+UDA9qlONGb8dgruKc6Gc82gvYLQwdq432Z0118=",
"Description": "",
"VpcConfig": {
"SubnetIds": [],
"SecurityGroupIds": []
},
"CodeSize": 417,
"FunctionArn": "arn:aws:lambda:us-east-1:xxxxxxxx:function:RDSInstanctStart",
"Handler": "lambda_function.lambda_handler"
}
我正在尝试访问键“变量”的值 到目前为止,这是我的代码:
try:
for evnt in funcResponse['Environment']['Variables']['DBInstanceName']:
print (evnt[0])
except ClientError as e:
print(e)
我得到的结果是
t
e
s
t
1
2
3
4
如果我没有给出envt 变量的索引,我会得到一个类型错误。
【问题讨论】:
-
另一方面,您可能对
dpath或其竞争对手之一感兴趣,将嵌套集合视为具有复杂路径的平面集合,例如用于键的'Environment/Variables/DBInstanceName'。
标签: python json dictionary nested