Question

我知道我可以使用这样的url参数：

"myfunction?p=one&p2=two"

并成为

的代码

request.query.p = "one";

但我更喜欢这样（快速路由风格）：

"myfunction/:myvalue"

并使用此网址：

/myfunction/nine

在代码中变成了这个：

request.params.myvalue = "nine"

但我似乎无法找到如何在Azure Functions，任何想法或文档中配置路径路径？

Answer 1

我们在Azure Functions中为HTTP触发器提供了路由支持。您现在可以添加遵循ASP.NET Web API路由命名语法的路由属性。（您可以通过Function.json或通过门户网站UX直接设置它）

"route": "node/products/{category:alpha}/{id:guid}"

Function.json：

{
    "bindings": [
        {
            "type": "httpTrigger",
            "name": "req",
            "direction": "in",
            "methods": [ "post", "put" ],
            "route": "node/products/{category:alpha}/{id:guid}"
        },
        {
            "type": "http",
            "name": "$return",
            "direction": "out"
        },
        {
            "type": "blob",
            "name": "product",
            "direction": "out",
            "path": "samples-output/{category}/{id}"
        }
    ]
}

.NET示例：

public static Task<HttpResponseMessage> Run(HttpRequestMessage request, string category, int? id, 
                                                TraceWriter log)
{
    if (id == null)
       return  req.CreateResponse(HttpStatusCode.OK, $"All {category} items were requested.");
    else
       return  req.CreateResponse(HttpStatusCode.OK, $"{category} item with id = {id} has been requested.");
}

NodeJS示例：

module.exports = function (context, req) {

    var category = context.bindingData.category;
    var id = context.bindingData.id;

    if (!id) {
        context.res = {
            // status: 200, /* Defaults to 200 */
            body: "All " + category + " items were requested."
        };
    }
    else {
        context.res = {
            // status: 200, /* Defaults to 200 */
            body: category + " item with id = " + id + " was requested."
        };
    }

    context.done();
}

官方文档：https://docs.microsoft.com/en-us/azure/azure-functions/functions-bindings-http-webhook#url-to-trigger-the-function

<击> 今天，您必须使用Azure API Management等服务来自定义路线。

~~正在进行PR以向Azure Functions本身添加自定义路由。该团队希望它能在下一个版本中登陆。 https://github.com/Azure/azure-webjobs-sdk-script/pull/490~~

~~注意：我是Azure Functions团队的PM~~

Answer 2

azure-function-express

使用azure-function-express，您可以使用所有expressjs功能，包括routing;）

const createHandler = require("azure-function-express").createAzureFunctionHandler;
const express = require("express");

// Create express app as usual
const app = express();
app.get("/api/:foo/:bar", (req, res) => {
  res.json({
    foo  : req.params.foo,
    bar  : req.params.bar
  });
});

// Binds the express app to an Azure Function handler
module.exports = createHandler(app);

查看更多示例here，其中包括处理all routes within a single Azure Function handler的方法。

PS：随意为该项目做出贡献！

如何在Azure功能中进行路由？

2 个答案:

azure-function-express