Amazon web services 无服务器框架如何响应准确的错误状态代码和消息?
我通过下面的命令创建了无服务器框架的函数:Amazon web services 无服务器框架如何响应准确的错误状态代码和消息?,amazon-web-services,aws-lambda,serverless-framework,Amazon Web Services,Aws Lambda,Serverless Framework,我通过下面的命令创建了无服务器框架的函数: sls function create some/api 然后创建了谢尔顿代码: 'use strict'; module.exports.handler = function(event, context, cb) { return cb({ message: 'Go Serverless! Your Lambda function executed successfully!' }); }; 并且,响应模板如下所示: s-fu
sls function create some/api
然后创建了谢尔顿代码:
'use strict';
module.exports.handler = function(event, context, cb) {
return cb({
message: 'Go Serverless! Your Lambda function executed successfully!'
});
};
并且,响应模板如下所示:
s-function.json
"responses": {
"400": {
"statusCode": "400"
},
"default": {
"statusCode": "200",
"responseParameters": {},
"responseModels": {
"application/json;charset=UTF-8": "Empty"
},
"responseTemplates": {
"application/json;charset=UTF-8": ""
}
}
}
但是,当我将error对象返回到回调函数(如cb(err,null)
)时,错误消息被正确显示,但statusCode是200
若我改为调用回调函数,比如cb(“400”,err)
,那个么statusCode正确地返回400,但响应体不好:{“errorMessage”:“400”}
是否有好的设置来显示机器人状态代码(不仅400,还有40140340500…等等)和错误消息?我使用以下响应模板。如果lambda函数返回的消息与响应模板中指定的
selectionPattern
匹配,它将返回正确的状态代码
"responseTemplate": {
"400": {
"selectionPattern": "^\\[BadRequest\\].*",
"statusCode": "400"
},
"401": {
"selectionPattern": "^\\[Unauthorized\\].*",
"statusCode": "401"
},
"403": {
"selectionPattern": "^\\[Forbidden\\].*",
"statusCode": "403"
},
"404": {
"selectionPattern": "^\\[NotFound\\].*",
"statusCode": "404"
},
"409": {
"selectionPattern": "^\\[Conflict\\].*",
"statusCode": "409"
},
"500": {
"selectionPattern": "^\\[Process exited|ServerError\\].*",
"statusCode": "500"
},
"504": {
"selectionPattern": "^\\[Task timed out\\].*",
"statusCode": "504"
},
"default": {
"statusCode": "200",
"responseParameters": {},
"responseModels": {},
"responseTemplates": {
"application/json": ""
}
}
}
假设您使用API网关公开Lambda函数,这里有一个很好的讨论:谢谢!这不是我想象的,但有了它,我可以用另一种方式解决我的问题。稍后我将总结结果。太好了!期待你的发现。