error "fork/exec /var/task/main: no such file or directory" while executing aws-lambda function

Question

Getting error fork/exec /var/task/main: no such file or directory while executing lambda function.

I am using windows platform to run and build code in Go.

I have done following steps to deploy go aws-lambda handler:

1. Written code in go language with VSCode in windows platform
2. Build project with : go build main.go
3. Convert main.exe to main.zip
4. Uploaded main.zip with handler name main aws-lambda fuction using aws console account
5. Created test event to test lambda function
6. Got error "fork/exec /var/task/main: no such file or directory while executing lambda function"

package main

import (
    "fmt"

    "github.com/aws/aws-lambda-go/lambda"
)

// Request represents the requested object
type Request struct {
    ID    int    `json:"ID"`
    Value string `json:"Value"`
}

// Response represents the Response object
type Response struct {
    Message string `json:"Message"`
    Ok      bool   `json:"Ok"`
}

// Handler represents the Handler of lambda
func Handler(request Request) (Response, error) {
    return Response{
        Message: fmt.Sprint("Process Request Id %f", request.ID),
        Ok:      true,
    }, nil
}

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

build command

go build main.go

Detail Error in AWS console

{
  "errorMessage": "fork/exec /var/task/main: no such file or directory",
  "errorType": "PathError"
}

Log Output in AWS console

START RequestId: 9ef206ed-5538-407a-acf0-06673bacf2d7 Version: $LATEST
fork/exec /var/task/main: no such file or directory: PathError
null
END RequestId: 9ef206ed-5538-407a-acf0-06673bacf2d7
REPORT RequestId: 9ef206ed-5538-407a-acf0-06673bacf2d7  Duration: 0.64 ms   Billed Duration: 100 ms Memory Size: 512 MB Max Memory Used: 31 MB  Init Duration: 1.49 ms

Solution 1

Run following commands in command prompt

set GOOS=linux
set GOARCH=amd64
set CGO_ENABLED=0

After this , build your project and upload zip file to aws console lambda

like this

GOOS=linux GOARCH=amd64 CGO_ENABLED=0 go build -o main main.go

Reference Link : https://github.com/aws/aws-lambda-go

Solution 2

In my case the problem was default setted handler to 'hello' function.

Needed to change it to 'main' via AWS Lambda view panel -> Basic Settings -> Edit.

Solution 3

There are two reasons can happen:

1. You did't use GOOS=linux GOARCH=amd64 with go build, so try:

   GOOS=linux GOARCH=amd64 go build -o main main.go

2. You used to build this programm some CI function with golang-alpine base image, so try to use full golang image instead.

Solution 4

Recently I was facing similar issue, and I solved it. As the error says it s finding executable with handle name, so you should name your executable same as handler.

Follow these steps and your will not get this error, I am using PowerShell

> go.exe get -u github.com/aws/aws-lambda-go/cmd/build-lambda-zip # If you do not this have utility earlier
> $env:GOOS = "linux"
> $env:GOARCH = "amd64"
> $env:CGO_ENABLED = "0"
> go build -o .\Handler .\main.go # considering your are in the same directory where your main.go or handler code is present
> ~\Go\Bin\build-lambda-zip.exe -o .\Handler.zip .\Handler

Upload this code, and change handler name to Handler(name of the executable that you created above)

Let me know if this helps.

Solution 5

In my embarrasing case I was zipping the folder from outside so the path in the zip I was uploading was 'folder/main'

View more solutions

Comments

[David Fulton]

While it wasn't a problem in this example, I think it might be worth pointing out that the executable has to be called "main", not just the function in the code. I found this out the hard way when following the instructions above.

[Pooja K Bhatt]

Yes you are right, that the executable has to be called "main", not just the function in the code.

[Poh Zi How]

this can now be found under "Runtime Settings"

[Conor]

using the standard serverless framework boiler code, this solves the error

[coder kemp]

Naming the Handler in AWS lambda function 'Handler' is a important step, once the steps mentioned above have been followed.

[Nigel Savage]

you did your best but this does not rely add anything that is not in the other answers

[Admin]

Please provide additional details in your answer. As it's currently written, it's hard to understand your solution.

[Charlie Araya]

After the first step I only had to do 'go build -o main main.go', zipped the main file that appeared and uploaded to AWS. That solved my issue at least

[pyFiddler]

why in the F is the default handler set to "hello" on a new Lambda function? Even the basic tutorial has you name everything "main"!

[colm.anseo]

Yes the executable name is key - but the name doesn't strictly need to be main. One can change the Handler name on the AWS side to another executable name.