Error executing "Hello World" for AWS Lambda in Java

Question

I have written following Hello World Lambda which I am executing by uploading on AWS via AWS toolkit for Eclipse.

public class HelloWorldLambdaHandler implements RequestHandler<String, String> {
    public String handleRequest(String input, Context context) {
        System.out.println("Hello World! executed with input: " + input);
        return input;
    }
}

I am getting following error when executing above code. Any idea what I maybe doing wrong here? BTW Maven project which have this handler, doesn't have any other class and only dependency is aws-lambda-java-core version 1.1.0.

Skip uploading function code since no local change is found...
Invoking function...
==================== FUNCTION OUTPUT ====================
{"errorMessage":"An error occurred during JSON parsing","errorType":"java.lang.RuntimeException","stackTrace":[],"cause":{"errorMessage":"com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@2f7c7260; line: 1, column: 1]","errorType":"java.io.UncheckedIOException","stackTrace":[],"cause":{"errorMessage":"Can not deserialize instance of java.lang.String out of START_OBJECT token\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@2f7c7260; line: 1, column: 1]","errorType":"com.fasterxml.jackson.databind.JsonMappingException","stackTrace":["com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:148)","com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:835)","com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:59)","com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:12)","com.fasterxml.jackson.databind.ObjectReader._bindAndClose(ObjectReader.java:1441)","com.fasterxml.jackson.databind.ObjectReader.readValue(ObjectReader.java:1047)"]}}}

What payload are you using to test the invocation? – Will Hayworth Feb 24 '16 at 07:44 — Will Hayworth, Feb 24 '16 at 07:44
@WillHayworth Payload is empty JSON object ({}). – Hemant Feb 24 '16 at 17:53 — Hemant, Feb 24 '16 at 17:53

score 94 · Accepted Answer · answered Mar 02 '16 at 23:01

94

For some reason Amazon can't deserialize json to a String. You would think String would be as general as input parameter as you can get but rightly or wrongly it's not compatible.

To handle JSON you can either use a Map or a custom POJO.

public class HelloWorldLambdaHandler {
    public String handleRequest(Map<String,Object> input, Context context) {
        System.out.println(input);
        return "Hello";
    }
}

answered Mar 02 '16 at 23:01

Lionel Port

3,492
23
26

2

Indeed you are right - I replaced String with a "Person" class which doesn't have any property and it worked fine. – Hemant Mar 04 '16 at 02:22
9

This might be right, but why is it like that? Is there any explanation, since documentation itself suggests to use a String in their example. Can anybody from AWS explain that? – Ondrej Tokar May 10 '16 at 12:05
5

I agree that the documentation should be improved to explain that a handler input parameter declared as `String` only accepts a valid JSON-string (i.e. a Unicode sequence surrounded by quotation marks, e.g. `"foo"`) and not a string representation of a valid JSON object ( e.g. `{"foo": "bar"}`) – fspinnenhirn Jul 01 '16 at 02:15
The wording of the answer is rather strange. JSON is a String. You can't deserialize JSON to a String: it's already a string. – Shannon Jul 21 '17 at 15:14
Can anyone give an example of a pojo? I create a pojo and it still won't work. – EGHDK Aug 08 '17 at 00:15
Thank you! Wish they'd just use gson instead – Nathan Adams Oct 04 '18 at 22:26
1

String do work, just wrap your json in quotes and escape all the quotes in the json – sbnarra Oct 19 '18 at 13:53
Thanks, I had no use of input and hence this works for me to put a dummy parameter. This is so weird I need some json as input – Ajak6 Mar 13 '20 at 06:45
1

Having an input of `Map` and _not_ `Map` fixed it for me, even though I was sending in a JSON map of String to String... =/ – Matt Klein Apr 10 '20 at 19:00
The error message is trying to say that "Since `input` is a String, I am expecting it to start with a `"`. But I am seeing a `{` instead". That is why changing `input` type to `Object`, `Map` or `Person` made the parser happy. If `input` should truly be a String, the payload value itself must start with a `"`. For instance, `String payload = "\"string input\""`. – Big Pumpkin Jan 18 '21 at 17:37

score 16 · Answer 2 · answered Jun 01 '16 at 17:48

16

Read the error from the stack trace. It says "Can not deserialize instance of java.lang.String out of START_OBJECT token". The "START_OBJECT" token is '{'.

The problem was simply that you need to pass an actual String as input, e.g., "A String". This is your json input. Not {}. {} is not a String. You don't need any braces, just a string (in quotes). On the other hand, {} is a valid Person object, so it worked once you changed it to handle a Person as the input.

answered Jun 01 '16 at 17:48

Kelly Denehy

207
2
5

1

@Cenxui This actually is the correct answer to the question: _"Any idea what @Hemnat is doing wrong when calling that particular lambda function with an empty JSON object (`{}`)"_ – fspinnenhirn Jul 01 '16 at 01:58
This is the correct answer - if you specify `"{}"` as your Lambda input, Lambda will try to parse it (even though your RequestHandler input type is String). If you specify `""` (empty string) instead, it works. – Luke May 03 '17 at 16:05
I tied to use string, and it didnot work. The above solution to of Map or person worked for me – user93796 Dec 20 '17 at 00:56

score 15 · Answer 3 · edited Jul 20 '18 at 13:56

15

I tried with the following value in the test :

"TestInput"

instead of :

{ Input : "TestInput"}

and it seems to have worked fine.

edited Jul 20 '18 at 13:56

Samuel Hulla

6,617
7
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answered Jul 20 '18 at 12:42

Anit Mohanty

151
1
4

1

This is really helped me. I was struggling due to this small issue 2 days. – skvp Feb 20 '19 at 08:03
This solution worked for me. Apparently AWS tries to parse the JSON format as an object. – kl27driver May 13 '19 at 10:33

score 3 · Answer 4 · answered Sep 29 '18 at 19:34

The complete working solution is

public class HelloWorldLambdaHandler implements RequestHandler<String, String> {
    public String handleRequest(String input, Context context) {
        System.out.println("Hello World! executed with input: " + input);
        return input;
    }
}

then input has to be in double quotes as a String - "Test Input"

score 0 · Answer 5 · answered May 22 '19 at 19:33

The input window for the test configurator takes raw json, or strings.

If you pass raw json, AWS converts the json into a Map where the variable names are keys that map the respective values.
If you wrap the json in double quotes and delimit inner quotes, this is an acceptable Java string representation of a json object and can be parsed as usual.

score 0 · Answer 6 · answered Apr 08 '21 at 05:06

Workaround 1:

Instead of

{
  "key1": "value1",
  "key2": "value2",
  "key3": "value3"
}

use some String as input, as shown below

"anyString"

It'll work fine.

Workaround 2: please refer the answer of @Lionel Port.

Reason:

As @Lionel Port suggested, you must be having String as inputType in your handleRequest() method!

public String handleRequest(String input, Context context) {}

you can see here, input type is String and you're passing Json, that's why giving this error!

Error executing "Hello World" for AWS Lambda in Java

6 Answers6

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