4

我正在使用aws-sdk-go和aws-lambda-go构建一个 aws lambda ，但遇到了一个小问题。

我想测试我的 lambda 处理程序。为此，我需要模拟一个有效的context.Context ，其中包含lamdacontext.LambdaContext的有效属性并满足lambdacontext.FromContext。

我似乎无法找到一种方法来构建这样的模拟，因为lambdacontext.FromContext总是返回 me _, false。

这是我的主要内容，在events.SNSEvent事件上有一个简单的处理程序：

package main

import (
    "context"
    "github.com/aws/aws-lambda-go/events"
    "github.com/aws/aws-lambda-go/lambda"
    "github.com/aws/aws-lambda-go/lambdacontext"
)

func main() {
    lambda.Start(handleRequest)
}

func handleRequest(ctx context.Context, snsEvent events.SNSEvent) error {
    lc, ok := lambdacontext.FromContext(ctx); if !ok {
        // Always false
        ...
        return someErr
    }
    . . .
    return nil
}

这是我的测试handleRequest：

package main

import (
    "context"
    "github.com/aws/aws-lambda-go/events"
    "github.com/aws/aws-lambda-go/lambdacontext"
    "github.com/stretchr/testify/assert"
    "gitlab.easy-network.it/meg/aml-rekognition/testdata"
    "testing"
)

const imgMock = `
{
  \"some_parameter\": \"some_value\"
}`

func TestHandleRequest(t *testing.T) {

    c := context.Background()

    ctxV := context.WithValue(c, "", map[string]interface{}{
        "AwsRequestID" : "some_aws_id",
        "InvokedFunctionArn" : "some_arn",
        "Identity" : lambdacontext.CognitoIdentity{},
        "ClientContext" : lambdacontext.ClientContext{},
    })

    snsEventMock := events.SNSEvent{
        Records: []events.SNSEventRecord{
            {
                SNS: events.SNSEntity{
                    Message: imgMock,
                },
            },
        },
    }

    err := handleRequest(ctxV, snsEventMock)
    assert.NoError(t, err)

}

我还尝试了其他模拟，例如将带有这些参数的结构传递给它等，但我总是得到false. 例如，我也尝试过：

type TestMock struct {
    AwsRequestID string
    InvokedFunctionArn string
    Identity lambdacontext.CognitoIdentity
    ClientContext lambdacontext.ClientContext
}

func TestHandleRequest(t *testing.T) {

    c := context.Background()

    testMock := TestMock{
        AwsRequestID : "some_aws_id",
        InvokedFunctionArn : "some_arn",
        Identity : lambdacontext.CognitoIdentity{},
        ClientContext : lambdacontext.ClientContext{},
    }

    ctxV := context.WithValue(c, "", testMock)

    . . .

}

我检查了来源，FromContext我一直在摸不着头脑。

// LambdaContext is the set of metadata that is passed for every Invoke.
type LambdaContext struct {
    AwsRequestID       string
    InvokedFunctionArn string
    Identity           CognitoIdentity
    ClientContext      ClientContext
}

// An unexported type to be used as the key for types in this package.
// This prevents collisions with keys defined in other packages.
type key struct{}

// The key for a LambdaContext in Contexts.
// Users of this package must use lambdacontext.NewContext and 
lambdacontext.FromContext

// instead of using this key directly.
var contextKey = &key{}

// FromContext returns the LambdaContext value stored in ctx, if any.
func FromContext(ctx context.Context) (*LambdaContext, bool) {
    lc, ok := ctx.Value(contextKey).(*LambdaContext)
    return lc, ok
}

当然，false即使我只是将 a 传递context.Background()给它，它也会返回。

关于我应该如何建立一个有效context.Context的让lambdacontext.FromContext返回的任何想法true？

4

1 回答 1

7

lambda.FromContext()检查传递context.Context的值是否包含一个在包内具有“私有”键的值lambdacontext：

// An unexported type to be used as the key for types in this package.
// This prevents collisions with keys defined in other packages.
type key struct{}

// The key for a LambdaContext in Contexts.
// Users of this package must use lambdacontext.NewContext and lambdacontext.FromContext
// instead of using this key directly.
var contextKey = &key{}

// FromContext returns the LambdaContext value stored in ctx, if any.
func FromContext(ctx context.Context) (*LambdaContext, bool) {
    lc, ok := ctx.Value(contextKey).(*LambdaContext)
    return lc, ok
}

您无法访问此密钥，也无法“复制”它（您无法创建与此“私有”密钥相等的值）。

但是有一个简单的方法，只需使用lambdacontext.NewContext()来派生一个将具有此键的上下文：

// NewContext returns a new Context that carries value lc.
func NewContext(parent context.Context, lc *LambdaContext) context.Context {
    return context.WithValue(parent, contextKey, lc)
}

所以解决方案：

ctx := context.Background()
// Add keys to your liking, then:
lc := new(lambdacontext.LambdaContext)
ctx = lambdacontext.NewContext(ctx, lc)
于 2018-05-23T14:29:07.480 回答