AWS Lambda 读取作为源代码上传的 zip 中的文件内容

Question

I have two files:

MyLambdaFunction.py

config.json

I zip those two together to create MyLambdaFunction.zip . I then upload that through the AWS console to my lambda function.

The contents of config.json are various environmental variables. I need a way to read the contents of the file each time the lambda function runs, and then use the data inside to set run time variables.

How do I get my Python Lambda function to read the contents of a file, config.json , that was uploaded in the zip file with the source code?

Answer 1

Figured it out with the push in the right direction from @helloV.

At the top of the python file put import os

Inside your function handler put the following:

configPath = os.environ['LAMBDA_TASK_ROOT'] + "/config.json"
print("Looking for config.json at " + configPath)
configContents = open(configPath).read()
configJson = json.loads(configContents)
environment = configJson['environment']
print("Environment: " + environment)

That bit right there, line by line, does the following:

• Get the path where the config.json file is stored
• Print that path for viewing in CloudWatch logs
• Open the file stored at that path, read the contents
• Load the contents to a json object for easy navigating
• Grab the value of one of the variables stored in the json
• Print that for viewing in the CloudWatch logs

Here is what the config.json looks like:

{
    "environment":"dev"
}

EDIT AWS lambda now supports use of environmental variables directly in the console UI. So if your use case is the same as mine (ie for a config file) you no longer need a config file.

Answer 2

Try this. The file you uploaded can be accessed like:

import os

os.environ['LAMBDA_TASK_ROOT']/config.json

Answer 3

Actually, I'd rather prefer judging the context of the running lambda to determine the config it should use, instead of uploading different zip files, which is difficult to maintain.

lambda_configs = {
    "function_name_1":{
    },
    "function_name_2":
   {
   }
}

config = lambda_configs[context.function_name]

Answer 4

Just done it in python3.8 lambda with simple:

with open('./dir/file.whatever') as f:

And it works just fine.

Answer 5

import os 

os.environ['LAMBDA_TASK_ROOT'] + '/config.json'

or you can use formatting

{os.environ['LAMBDA_TASK_ROOT']}/config.json"