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问AWS SAM启动本地api返回“函数名称是必需的”错误
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Stack Overflow用户

提问于 2020-08-22 21:45:05

回答 1查看 780关注 0票数 3

我们使用CDK来构建我们的基础设施配置。此外，如果有帮助的话，我会用cdk synth <stack_name> --no-staging > template.yml为SAM创建我的template.yml。我使用AWS Toolkit在Intellij上调用/调试我的lambda函数，它工作得很好。然而，如果我在终端上运行sam local start-api并向我的一个函数发送一个请求，那么它会返回一个堆栈跟踪错误；

Traceback (most recent call last):
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/app.py", line 2317, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/app.py", line 1840, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/app.py", line 1743, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/_compat.py", line 36, in reraise
    raise value
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/app.py", line 1838, in full_dispatch_request
    rv = self.dispatch_request()
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/flask/app.py", line 1824, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/samcli/local/apigw/local_apigw_service.py", line 203, in _request_handler
    self.lambda_runner.invoke(route.function_name, event, stdout=stdout_stream_writer, stderr=self.stderr)
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/samcli/commands/local/lib/local_lambda.py", line 84, in invoke
    function = self.provider.get(function_name)
  File "/usr/local/Cellar/aws-sam-cli/0.53.0/libexec/lib/python3.7/site-packages/samcli/lib/providers/sam_function_provider.py", line 65, in get
    raise ValueError("Function name is required")
ValueError: Function name is required

这是我运行的命令

sam local start-api --env-vars env.json --docker-network test

它给出了输出

Mounting None at http://127.0.0.1:3000/v1 [GET, OPTIONS, POST]
Mounting None at http://127.0.0.1:3000/v1/user [GET, OPTIONS, POST]
You can now browse to the above endpoints to invoke your functions. You do not need to restart/reload SAM CLI while working on your functions, changes will be reflected instantly/automatically. You only need to restart SAM CLI if you update your AWS SAM template
2020-08-22 16:32:46  * Running on http://127.0.0.1:3000/ (Press CTRL+C to quit)
2020-08-22 16:33:03 Exception on /v1/user [OPTIONS]

下面是我用作函数环境变量的env.json

{
  "tenantGetV1Function54F63CB9": {
    "db": "alpha",
    "connectionString": "mongodb://mongo"
  },
  "tenantPostV1FunctionA56822D0": {
    "db": "alpha",
    "connectionString": "mongodb://mongo"
  },
  "userGetV1Function7E6E55C2": {
    "db": "alpha",
    "connectionString": "mongodb://mongo"
  },
  "userPostV1FunctionEB035EB0": {
    "db": "alpha",
    "connectionString": "mongodb://mongo"
  }
}

我还在macOS操作系统上运行Docker Desktop。

编辑:您可以在这里找到只有一个端点(一个函数定义)的简化函数，它是用于tenantGetV1Function54F63CB9 template.yml的。它将映射到GET /v1端点。我不想包含4个函数的整个模板，这使得大约1000行.yml代码。

https://gist.github.com/flexelem/d887136484d508e313e0a745c30a2d97

amazon-web-services

aws-sam

aws-sam-cli

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-09-03 05:25:51

如果我在CDK中通过传递Function实例而不是它的Alias实例来创建LambdaIntegration，这个问题就解决了。因此，我们正在创建一个带有别名的lambdas。然后，我们将别名从Api网关传递到与其关联的Resource实例。

这就是我们创造的方式；

Function tenantGetV1Function = Function.Builder.create(this, "tenantGetV1Function")
    .role(roleLambda)
    .runtime(Runtime.JAVA_8)
    .code(lambdaCode)
    .handler("com.yolda.tenant.lambda.GetTenantHandler::handleRequest")
    .memorySize(512)
    .timeout(Duration.minutes(1))
    .environment(environment)
    .description(Instant.now().toString())
    .build();

Alias tenantGetV1Alias = Alias.Builder.create(this, "tenantGetV1Alias")
    .aliasName("live")
    .version(tenantAdminGetV1Function.getCurrentVersion())
    .provisionedConcurrentExecutions(provisionedConcurrency)
    .build();

Resource v1Resource = v1Resource.addResource("{tenantId}");
v1Resource.addMethod("GET", LambdaIntegration.Builder.create(tenantGetV1Alias).build(), options);

如果我用tenantGetV1Function替换tenantGetV1Alias，那么sam build命令成功地构建了所有函数，这将使sam local start-api旋转它们。

Resource v1Resource = v1Resource.addResource("{tenantId}");
v1Resource.addMethod("GET", LambdaIntegration.Builder.create(tenantGetV1Function).build(), options);

不知何故，如果我们分配别名，SAM就不能从CloudFormation模板中获得函数名属性。

票数 2

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/63536861

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