java 通过 Amazon API Gateway 将 POST 请求正文传递给 Lambda
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passing POST request body through Amazon API Gateway to Lambda
提问by GameDroids
I have a AWS Lambda function written in Java, that gets triggered by an AWS API Gateway call.
我有一个用 Java 编写的 AWS Lambda 函数，它由 AWS API Gateway 调用触发。
I am trying to make a POSTrequest to one of the endpoints with a JSON as payload.
我正在尝试POST使用 JSON 作为有效负载向其中一个端点发出请求。
curl -H "Content-Type: application/json" -X POST -d '{"firstName":"Mr", "lastName":"Awesome"}' https://someexample.execute-api.eu-central-1.amazonaws.com/beta/MethodHandlerLambda
The gateway will then detectthe Content-Typeand pass all request parameters (including the body) through a default template. The interesting part is this one
然后网关将检测的Content-Type，并通过一个缺省传递所有请求参数（包括人体）的模板。有趣的部分是这个
#set($allParams = $input.params())
{
"body-json" : $input.json('$'),
....
It is supposed to present me with a Map<String, Object>that is passed to my Java method:
它应该向我展示Map<String, Object>传递给我的 Java 方法的一个：
public void myHandler(Map<String, Object> input, Context context){
input.keySet().forEach((key) -> {
System.out.println(key + " : " + input.get(key));
});
}
And the result shouldbe something like this:
结果应该是这样的：
body-json : {"firstName":"Mr", "lastName":"Awesome"}
...
But what I am getting is this:
但我得到的是：
body-json : {firstName=Mr, lastName=Awesome}
Another possibility would be to pass the whole body as string:
另一种可能性是将整个主体作为字符串传递：
"body" : $input.body
but that again just "converts" to key=valueinstead of key:value
但这又只是“转换”为key=value而不是key:value
How do I have to configure the template to simply pass me the body so I can use it in a JSON parser?
我如何配置模板以简单地将主体传递给我，以便我可以在 JSON 解析器中使用它？
采纳答案by GameDroids
And again - just posting a question here on SO helps to find the answer on your own :)
再一次 - 只是在 SO 上发布一个问题有助于您自己找到答案:)
In the AWS Api Gateway template I set the body to
在 AWS Api Gateway 模板中，我将正文设置为
"body-json" : $input.body
which shouldreturn the complete payload as a String.
它应该将完整的有效负载作为字符串返回。
But more importantly I read Greggs answer to his own questionand changed my methodto
但更重要的是，我阅读了Greggs 对他自己问题的回答并将我的方法改为
public void myHandler(InputStream inputStream, OutputStream outputStream, Context context) throws IOException{
final ObjectMapper objectMapper = new ObjectMapper();
JsonNode json = objectMapper.readTree(inputStream);
System.out.println(json.toString());
}
So, it is sufficient to have a simple InputStreamand read it as a JsonNodewith whatever JSON library one prefers (I am using Hymanson FasterXML). And voila, it packs all possible parameters in a single JSON (as specified by the template)
因此，拥有一个简单的InputStream并将其作为JsonNode一个喜欢的任何 JSON 库阅读就足够了（我使用的是 Hymanson FasterXML）。瞧，它将所有可能的参数打包在一个 JSON 中（由模板指定）
{
"body-json": {
"firstName": "Mr",
"lastName": "Awesome"
},
"params": {
...
},
"stage-variables": {
...
},
"context": {
...
}
}
