Lambda 在匹配操作中引发语法错误

Question

我正在尝试创建一个可以在我的机器上运行的 Lambda 函数，但是当我测试它时，我收到以下错误：

Response
{
  "errorMessage": "Syntax error in module 'lambda_function': invalid syntax (lambda_function.py, line 64)",
  "errorType": "Runtime.UserCodeSyntaxError",
  "requestId": "ae5f2481-a836-4b4e-8c8d-a82ccd12c630",
  "stackTrace": [
    "  File \"/var/task/lambda_function.py\" Line 64\n        match operation:\n"
  ]
}

Function Logs
START RequestId: ae5f2481-a836-4b4e-8c8d-a82ccd12c630 Version: $LATEST
[ERROR] Runtime.UserCodeSyntaxError: Syntax error in module 'lambda_function': invalid syntax (lambda_function.py, line 64)
Traceback (most recent call last):
  File "/var/task/lambda_function.py" Line 64
        match operation:END RequestId: ae5f2481-a836-4b4e-8c8d-a82ccd12c630
REPORT RequestId: ae5f2481-a836-4b4e-8c8d-a82ccd12c630  Duration: 1.34 ms   Billed Duration: 2 ms   Memory Size: 128 MB Max Memory Used: 35 MB  Init Duration: 117.69 ms

Request ID
ae5f2481-a836-4b4e-8c8d-a82ccd12c630

所以在浪费了几个小时之后：

我看不到任何语法错误。
我试过捆绑枚举包（不知道是否有必要）。
我还评论了任何使用外部参数的行。
我在 Lambda 中使用 Python 3.9，在本地使用 python 3.10。
日志始终指向函数 get_url

match operation:

这是错误函数之前的代码：

import requests
import json
from enum import Enum
import urllib.parse
import boto3
import base64

max_number_of_digit = 1
base_url = "https://somesite.io"
app_version = "version-test/"
bubble_api_token = "some api token"
pdf_creator_api_token = 'some other api token'


class APP_ENTITY(str, Enum):
    PROJECT = "project"
    COMPANY = "company"
    PAYMENT = "payment"
    TRANCHE = "tranche"
    COST_ITEMS = "payment"
    DOCUMENT = "document"


class OPERATIONS(int, Enum):
    CREATE = 1
    DELETE = 2
    PATCH = 3
    READ = 4
    GET = 5


document = dict()
customer = dict()
base_assets = [
    'bank.png',
    'calculator.png',
    'card.png',
    'caret.png',
    'jetbrains-mono-v11-latin-500.woff2',
    'jetbrains-mono-v11-latin-regular.woff2',
    'line.png',
    'logo.svg',
    'poppins-v19-latin-300.woff2',
    'poppins-v19-latin-500.woff2',
    'poppins-v19-latin-600.woff2',
    'poppins-v19-latin-regular.woff2',
    'style.css'
]


def set_app_version(event):
    global app_version
    result = "version-test/"
    try:
        result = event['queryStringParameters']['appversion']
        result = "" if (result.lower() != 'test') else "version-test/"
    except:
        result = "version-test/"
    app_version = result
    return result


def get_url(operation: OPERATIONS, entity: APP_ENTITY, cursor: int = 0, record_id: str = "", project_id: str = ""):
    match operation:
        case OPERATIONS.CREATE:
            result = base_url + \
                "/{}api/1.1/obj/{}/bulk".format(app_version, entity)
        case OPERATIONS.DELETE:
            result = base_url + \
                "/{}api/1.1/obj/{}/".format(app_version, entity)
        case OPERATIONS.PATCH:
            result = base_url + \
                "/{}api/1.1/obj/{}/{}".format(app_version, entity, record_id)
        case OPERATIONS.READ:
            match entity:
                case APP_ENTITY.COST_ITEMS:
                    result = base_url + \
                        "/{}api/1.1/obj/{}/?api_token={}&cursor={}&constraints=".format(
                            app_version, entity, bubble_api_token, cursor)
                    constraints = [
                        {
                            "key": "type",
                            "constraint_type": "equals",
                            "value": "development"
                        },
                        {
                            "key": "project",
                            "constraint_type": "equals",
                            "value": project_id
                        }]
                    constraints = urllib.parse.quote_plus(
                        json.dumps(constraints))
                    result = result+constraints
        case OPERATIONS.GET:
            result = base_url + \
                "/{}api/1.1/obj/{}/{}".format(app_version, entity, record_id)
    return result

下面是 lambda 控制台的屏幕截图。 Lambda console

这里是错误的屏幕截图： Lambda error

Answer 1

match 和 case 是 Python 3.10 中新增的内容。 Lambda 不支持 Python 3.10， 因此您遇到了错误。您必须移植回您的代码才能使用 Python 3.9

Lambda 在匹配操作中引发语法错误

Lambda raise a syntax error on a match operation

python

amazon-web-services

aws-lambda