Java AWS Lambda 處理程序 - 深度嵌套的 Json

Question

我有一個深度嵌套的 json 對象，我想發送到 Java AWS Lambda。

我為請求處理程序嘗試了不同的類型。

public class LambdaHandler implements RequestHandler<Map<String, ?>, String>
public class LambdaHandler implements RequestHandler<JsonNode, String>
public class LambdaHandler implements RequestHandler<String, String>

在所有情況下，以下簡單的單級 json 都可以正常工作：

{
  "hello": "world"
}

但是任何嵌套的屬性都失敗了：

{
  "hello": "world",
  "oh": {
      "no": "it's dead"
  }
}

隨着錯誤：

An error occurred during JSON parsing: java.lang.RuntimeException
java.lang.RuntimeException: An error occurred during JSON parsing
Caused by: java.io.UncheckedIOException: com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `java.lang.String` out of START_OBJECT token
 at [Source: (ByteArrayInputStream); line: 3, column: 9] (through reference chain: java.util.LinkedHashMap["oh"])
    at com.amazonaws.services.lambda.runtime.serialization.factories.JacksonFactory$InternalSerializer.fromJson(JacksonFactory.java:184)
Caused by: com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `java.lang.String` out of START_OBJECT token
 at [Source: (ByteArrayInputStream); line: 3, column: 9] (through reference chain: java.util.LinkedHashMap["oh"])
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:59)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.DeserializationContext.reportInputMismatch(DeserializationContext.java:1442)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.DeserializationContext.handleUnexpectedToken(DeserializationContext.java:1216)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.DeserializationContext.handleUnexpectedToken(DeserializationContext.java:1126)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:63)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:10)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.deser.std.MapDeserializer._readAndBindStringKeyMap(MapDeserializer.java:527)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:364)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:29)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.ObjectReader._bindAndClose(ObjectReader.java:1719)
    at com.amazonaws.lambda.thirdparty.com.fasterxml.jackson.databind.ObjectReader.readValue(ObjectReader.java:1228)
    at com.amazonaws.services.lambda.runtime.serialization.factories.JacksonFactory$InternalSerializer.fromJson(JacksonFactory.java:182)

如何編寫允許深度嵌套 json 屬性的 AWS Java Lambda？

Answer 1

選項1：
將event類型更改為Map<String, Object>並使用 Gson 轉換為您的數據模型。

例子：

public class Handler implements RequestHandler<Map<String, Object>, String> {

    @Override
    public String handleRequest(Map<String, Object> event, Context context) {
        Gson gson = new Gson();
        MyDataModel request = gson.fromJson(gson.toJson(event), MyDataModel.class);

        // use request

        return "";
    }
}

---

選項 2：
對於復雜的 json/data 模型，您可以使用RequestStreamHandler類，它不需要您提供數據類型。 它將為您提供InputStream ，您可以將其轉換為您喜歡的任何數據模型。

例子：

public class HandlerStream implements RequestStreamHandler {

    @Override
    public void handleRequest(InputStream input, OutputStream output, Context context) {
        BufferedReader reader = new BufferedReader(new InputStreamReader(input));
        JsonElement eventElement = new JsonParser().parse(reader);
        JsonObject event = (JsonObject) eventElement;
        
        MyDataModel request = new Gson().fromJson(event, MyDataModel.class);
        
        // use request
    }
}