Question

Aws Lambda - 如何在Java中实现的lambda函数中从Api Gateway获取查询参数。

我有以下代码段：

package com.amazonaws.lambda.demo;

import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.RequestHandler;

public class LambdaFunctionHandler implements RequestHandler<String, String> {

    @Override
    public String handleRequest(String input, Context context) {
        context.getLogger().log("Input: " + input);

        // TODO: implement your handler
        return "Hello from " + input;
    }
}

如果在上面的handleRequest函数中有任何查询参数，我该如何访问。

Answer 1

在 pom.xml 中，确保在此处指定“ 2.2.5 ”或更高版本

<dependency>
  <groupId>com.amazonaws</groupId>
  <artifactId>aws-lambda-java-events</artifactId>
  <version>2.2.5</version>
</dependency>

您的Java lambda类：

import com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent;
import com.amazonaws.services.lambda.runtime.events.APIGatewayProxyResponseEvent;

public class LambdaFunctionHandler
    implements RequestHandler<APIGatewayProxyRequestEvent, APIGatewayProxyResponseEvent> {
    @Override
    public APIGatewayProxyResponseEvent handleRequest(APIGatewayProxyRequestEvent input,
            Context context) {
        Map<String, String> inputParams = input.getPathParameters();

        APIGatewayProxyResponseEvent responseEvent = new APIGatewayProxyResponseEvent();
        responseEvent.setBody("body text");
        responseEvent.setStatusCode(200);
        return responseEvent;
    }
}

Answer 2

您可以尝试在API网关的集成请求正文映射模板中映射所需的所有内容。一旦你构造了体映射模板，那么在lambda的上下文中你将得到你构造的excate json。

请找到以下链接，我已经为类似问题提供了解决方案，

https://stackoverflow.com/a/46407780/7666972

Answer 3

根据上面 tgk23 的回答，我相信您正在寻找查询参数，而不是路径参数：

public class LambdaFunctionHandler
        implements RequestHandler<APIGatewayProxyRequestEvent, APIGatewayProxyResponseEvent> {
    @Override
    public APIGatewayProxyResponseEvent handleRequest(APIGatewayProxyRequestEvent input, Context context) {
        Map<String, String> inputParams = input.getQueryStringParameters();
        for (Map.Entry<String, String> entry : inputParams.entrySet()) {
            LOGGER.info("param: " + entry.getKey() + "=" + entry.getValue());
        }

        APIGatewayProxyResponseEvent responseEvent = new APIGatewayProxyResponseEvent();
        responseEvent.setBody("body text");
        responseEvent.setStatusCode(200);
        return responseEvent;
    }
}

Aws Lambda：如何在Java中实现的lambda函数中从Api Gateway获取查询参数

3 个答案: