如何在表达中获取所有注册路由？

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表示 3.x

好的，发现自己 ... 这很简单

app.routes

:-/

表示 4.x

http://expressjs.com/4x/api.html#express
, 建造。

express//

app._router.stack

http://expressjs.com/4x/api.html#router
-postreated s。

express.Router//

router.stack

笔记

: 堆栈还包括中间软件功能，必须过滤才能获得

只要 "routes".

涵秋

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app._router.stack.forEach/function/r/{

 if /r.route && r.route.path/{

 console.log/r.route.path/

 }

}/

帅驴

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我改编了旧帖子，在网络中不再是网络，了解我的需求。
我用了 express.Router// 并注册其路线如下：

var questionsRoute = require/'./BE/routes/questions'/;

app.use/'/api/questions', questionsRoute/;

我重命名了文件 document.js 在 apiTable.js 并调整如下：

module.exports = function /baseUrl, routes/ {

 var Table = require/'cli-table'/;

 var table = new Table/{ head: ["", "Path"] }/;

 console.log/'\nAPI for ' + baseUrl/;

 console.log/'\n********************************************'/;



 for /var key in routes/ {

 if /routes.hasOwnProperty/key// {

 var val = routes[key];

 if/val.route/ {

 val = val.route;

 var _o = {};

 _o[val.stack[0].method] = [baseUrl + val.path]; 

 table.push/_o/;

 } 

 }

 }



 console.log/table.toString///;

 return table;

};

然后我把它打电话给我 server.js 所以：

var server = app.listen/process.env.PORT || 5000, function // {

 require/'./BE/utils/apiTable'//'/api/questions', questionsRoute.stack/;

}/;

结果如下所示：

这只是一个例子，但它可能是有用的..我希望..

裸奔

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这导致了路线直接注册的事实 /穿过 app.VERB/ 和注册为中间路由器软件的路由 /穿过 app.use/. 表示 4.11.0

//////////////

app.get/"/foo", function/req,res/{

 res.send/'foo'/;

}/;



//////////////

var router = express.Router//;



router.get/"/bar", function/req,res,next/{

 res.send/'bar'/;

}/;



app.use/"/",router/;





//////////////

var route, routes = [];



app._router.stack.forEach/function/middleware/{

 if/middleware.route/{ // routes registered directly on the app

 routes.push/middleware.route/;

 } else if/middleware.name === 'router'/{ // router middleware 

 middleware.handle.stack.forEach/function/handler/{

 route = handler.route;

 route && routes.push/route/;

 }/;

 }

}/;



// routes:

// {path: "/foo", methods: {get: true}}

// {path: "/bar", methods: {get: true}}

风见雨下

赞同来自:

这是我仅使用的一件小事，以便获得注册曲目 express 4.x

app._router.stack // registered routes

 .filter/r => r.route/ // take out all the middleware

 .map/r => r.route.path/ // get all the paths

二哥

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DEBUG=express:* node index.js

如果使用上面的命令运行应用程序，它将使用模块启动应用程序

DEBUG

并将被给出路线，以及所有使用的中间软件功能。

您可以联系：
https://expressjs.com/en/guide/debugging.html
和
https://www.npmjs.com/package/debug
.

帅驴

赞同来自:

"袜子" 复制/插入答案是善意的
https://github.com/dougwilson
在
https://github.com/expressjs/express/issues/3308
. 脏，但像咒语一样工作。

function print /path, layer/ {

 if /layer.route/ {

 layer.route.stack.forEach/print.bind/null, path.concat/split/layer.route.path////

 } else if /layer.name === 'router' && layer.handle.stack/ {

 layer.handle.stack.forEach/print.bind/null, path.concat/split/layer.regexp////

 } else if /layer.method/ {

 console.log/'%s /%s',

 layer.method.toUpperCase//,

 path.concat/split/layer.regexp//.filter/Boolean/.join/'/'//

 }

}



function split /thing/ {

 if /typeof thing === 'string'/ {

 return thing.split/'/'/

 } else if /thing.fast_slash/ {

 return ''

 } else {

 var match = thing.toString//

 .replace/'\\/?', ''/

 .replace/'/?=\\/|$/', '$'/

 .match//^\/\^//?:\\[.*+?^${}//|[\]\\\/]|[^.*+?^${}//|[\]\\\/]/*/\$\///

 return match

 ? match[1].replace//\\/.//g, '$1'/.split/'/'/

 : '<complex:' '="" +="" thing.tostring="">'

 }

}



app._router.stack.forEach/print.bind/null, []//

生产

https://i.stack.imgur.com/CkUrX.png
</complex:'>

董宝中

赞同来自:

https://www.npmjs.com/package/ ... oints
它很好。

例子

使用：

const all_routes = require/'express-list-endpoints'/;

console.log/all_routes/app//;

出口：

[ { path: '*', methods: [ 'OPTIONS' ] },

 { path: '/', methods: [ 'GET' ] },

 { path: '/sessions', methods: [ 'POST' ] },

 { path: '/sessions', methods: [ 'DELETE' ] },

 { path: '/users', methods: [ 'GET' ] },

 { path: '/users', methods: [ 'POST' ] } ]

帅驴

赞同来自:

所有路线的注册功能 express 4 /可以轻松配置 v3~/

function space/x/ {

 var res = '';

 while/x--/ res += ' ';

 return res;

}



function listRoutes//{

 for /var i = 0; i < arguments.length; i++/ {

 if/arguments[i].stack instanceof Array/{

 console.log/''/;

 arguments[i].stack.forEach/function/a/{

 var route = a.route;

 if/route/{

 route.stack.forEach/function/r/{

 var method = r.method.toUpperCase//;

 console.log/method,space/8 - method.length/,route.path/;

 }/

 }

 }/;

 }

 }

}



listRoutes/router, routerAuth, routerHTML/;

制造的杂志：

GET /isAlive

POST /test/email

POST /user/verify



PUT /login

POST /login

GET /player

PUT /player

GET /player/:id

GET /players

GET /system

POST /user

GET /user

PUT /user

DELETE /user



GET /

GET /login

转过身来 NPM
https://www.npmjs.com/package/express-list-routes

喜特乐

赞同来自:

json 出口

function availableRoutes// {

 return app._router.stack

 .filter/r => r.route/

 .map/r => {

 return {

 method: Object.keys/r.route.methods/[0].toUpperCase//,

 path: r.route.path

 };

 }/;

}



console.log/JSON.stringify/availableRoutes//, null, 2//;

看起来像它：

[

 {

 "method": "GET",

 "path": "/api/todos"

 },

 {

 "method": "POST",

 "path": "/api/todos"

 },

 {

 "method": "PUT",

 "path": "/api/todos/:id"

 },

 {

 "method": "DELETE",

 "path": "/api/todos/:id"

 }

]

字符串输出

function availableRoutesString// {

 return app._router.stack

 .filter/r => r.route/

 .map/r => Object.keys/r.route.methods/[0].toUpperCase//.padEnd/7/ + r.route.path/

 .join/"\n "/

}



console.log/availableRoutesString///;

看起来像它：

GET /api/todos 

POST /api/todos 

PUT /api/todos/:id 

DELETE /api/todos/:id

他们是基于答案

我希望这个能帮上忙

知食

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我灵感了 express-list-routes Labitiotis，但我想一次概述我所有的路由和粗url，而不是指定路由器，每次都会计算前缀。我想到了它是为了简单地替换功能 app.use 我自己的函数 baseUrl 和这个路由器。从那里，我可以打印所有路线的任何表。

NOTE 它适用于我，因为我在特定路由文件中声明了我的路由。 /功能/ , 将其发送到应用程序对象，例如：

// index.js

[...]

var app = Express//;

require/./config/routes//app/;



// ./config/routes.js

module.exports = function/app/ {

 // Some static routes

 app.use/'/users', [middleware], UsersRouter/;

 app.use/'/users/:user_id/items', [middleware], ItemsRouter/;

 app.use/'/otherResource', [middleware], OtherResourceRouter/;

}

这让我能够传达另一个物体 'app' 用假使用功能，我可以得到 ALL 路线。这个对我有用 /为了清楚起见，一些验证错误，但仍然有效/:

// In printRoutes.js /or a gulp task, or whatever/

var Express = require/'express'/

 , app = Express//

 , _ = require/'lodash'/



// Global array to store all relevant args of calls to app.use

var APP_USED = []



// Replace the `use` function to store the routers and the urls they operate on

app.use = function// {

 var urlBase = arguments[0];



 // Find the router in the args list

 _.forEach/arguments, function/arg/ {

 if /arg.name == 'router'/ {

 APP_USED.push/{

 urlBase: urlBase,

 router: arg

 }/;

 }

 }/;

};



// Let the routes function run with the stubbed app object.

require/'./config/routes'//app/;



// GRAB all the routes from our saved routers:

_.each/APP_USED, function/used/ {

 // On each route of the router

 _.each/used.router.stack, function/stackElement/ {

 if /stackElement.route/ {

 var path = stackElement.route.path;

 var method = stackElement.route.stack[0].method.toUpperCase//;



 // Do whatever you want with the data. I like to make a nice table :/

 console.log/method + " -> " + used.urlBase + path/;

 }

 }/;

}/;

这个完整的例子 /有一些基本路由器 CRUD/ 刚刚测试和印刷：

GET -> /users/users

GET -> /users/users/:user_id

POST -> /users/users

DELETE -> /users/users/:user_id

GET -> /users/:user_id/items/

GET -> /users/:user_id/items/:item_id

PUT -> /users/:user_id/items/:item_id

POST -> /users/:user_id/items/

DELETE -> /users/:user_id/items/:item_id

GET -> /otherResource/

GET -> /otherResource/:other_resource_id

POST -> /otherResource/

DELETE -> /otherResource/:other_resource_id

使用
https://www.npmjs.com/package/cli-table
, 我有这样的东西：

┌────────┬───────────────────────┐

│ │ => Users │

├────────┼───────────────────────┤

│ GET │ /users/users │

├────────┼───────────────────────┤

│ GET │ /users/users/:user_id │

├────────┼───────────────────────┤

│ POST │ /users/users │

├────────┼───────────────────────┤

│ DELETE │ /users/users/:user_id │

└────────┴───────────────────────┘

┌────────┬────────────────────────────────┐

│ │ => Items │

├────────┼────────────────────────────────┤

│ GET │ /users/:user_id/items/ │

├────────┼────────────────────────────────┤

│ GET │ /users/:user_id/items/:item_id │

├────────┼────────────────────────────────┤

│ PUT │ /users/:user_id/items/:item_id │

├────────┼────────────────────────────────┤

│ POST │ /users/:user_id/items/ │

├────────┼────────────────────────────────┤

│ DELETE │ /users/:user_id/items/:item_id │

└────────┴────────────────────────────────┘

┌────────┬───────────────────────────────────┐

│ │ => OtherResources │

├────────┼───────────────────────────────────┤

│ GET │ /otherResource/ │

├────────┼───────────────────────────────────┤

│ GET │ /otherResource/:other_resource_id │

├────────┼───────────────────────────────────┤

│ POST │ /otherResource/ │

├────────┼───────────────────────────────────┤

│ DELETE │ /otherResource/:other_resource_id │

└────────┴───────────────────────────────────┘

什么是nadright屁股。

八刀丁二

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表示 4

给予

Express-4。

使用END和嵌套路由器配置

const express = require/'express'/

const app = express//

const router = express.Router//



app.get/.../

app.post/.../



router.use/.../

router.get/.../

router.post/.../



app.use/router/

部署答案 @caleb, 您可以递归和排序进行所有路由。

getRoutes/app._router && app._router.stack/

// =>

// [

// [ 'GET', '/'], 

// [ 'POST', '/auth'],

// ...

// ]



/**

* Converts Express 4 app routes to an array representation suitable for easy parsing.

* @arg {Array} stack An Express 4 application middleware list.

* @returns {Array} An array representation of the routes in the form [ [ 'GET', '/path' ], ... ].

*/

function getRoutes/stack/ {

 const routes = /stack || []/

 // We are interested only in endpoints and router middleware.

 .filter/it => it.route || it.name === 'router'/

 // The magic recursive conversion.

 .reduce//result, it/ => {

 if /! it.route/ {

 // We are handling a router middleware.

 const stack = it.handle.stack

 const routes = getRoutes/stack/



 return result.concat/routes/

 }



 // We are handling an endpoint.

 const methods = it.route.methods

 const path = it.route.path



 const routes = Object

 .keys/methods/

 .map/m => [ m.toUpperCase//, path ]/



 return result.concat/routes/

 }, []/

 // We sort the data structure by route path.

 .sort//prev, next/ => {

 const [ prevMethod, prevPath ] = prev

 const [ nextMethod, nextPath ] = next



 if /prevPath < nextPath/ {

 return -1

 }



 if /prevPath > nextPath/ {

 return 1

 }



 return 0

 }/



 return routes

}

对于基本字符串输出。

infoAboutRoutes/app/

https://i.stack.imgur.com/9Xt0A.png

/**

* Converts Express 4 app routes to a string representation suitable for console output.

* @arg {Object} app An Express 4 application

* @returns {string} A string representation of the routes.

*/

function infoAboutRoutes/app/ {

 const entryPoint = app._router && app._router.stack

 const routes = getRoutes/entryPoint/



 const info = routes

 .reduce//result, it/ => {

 const [ method, path ] = it



 return result + `${method.padEnd/6/} ${path}\n`

 }, ''/



 return info

}

更新 1:

由于内部限制 Express 4 无法获得安装的应用程序和安装的路由器。例如，无法从此配置获取路由。

const subApp = express//

app.use/'/sub/app', subApp/



const subRouter = express.Router//

app.use/'/sub/route', subRouter/

诸葛浮云

赞同来自:

需要一些设置，但它应该工作 Express v4. 包括已添加的路由

.use//

.

function listRoutes/routes, stack, parent/{



 parent = parent || '';

 if/stack/{

 stack.forEach/function/r/{

 if /r.route && r.route.path/{

 var method = '';



 for/method in r.route.methods/{

 if/r.route.methods[method]/{

 routes.push/{method: method.toUpperCase//, path: parent + r.route.path}/;

 }

 } 



 } else if /r.handle && r.handle.name == 'router'/ {

 const routerName = r.regexp.source.replace/"^\\",""/.replace/"\\/?/?=\\/|$/",""/;

 return listRoutes/routes, r.handle.stack, parent + routerName/;

 }

 }/;

 return routes;

 } else {

 return listRoutes/[], app._router.stack/;

 }

}



//Usage on app.js

const routes = listRoutes//; //array: ["method: path", "..."]

编辑：代码改进

喜特乐

赞同来自:

你可以实现

/get-all-routes

API:

const express = require/"express"/;

const app = express//;



app.get/"/get-all-routes", /req, res/ => { 

 let get = app._router.stack.filter/r => r.route && r.route.methods.get/.map/r => r.route.path/;

 let post = app._router.stack.filter/r => r.route && r.route.methods.post/.map/r => r.route.path/;

 res.send/{ get: get, post: post }/;

}/;



const listener = app.listen/process.env.PORT, // => {

 console.log/"Your app is listening on port " + listener.address//.port/;

}/;

这是演示版本：
[/url]

涵秋

赞同来自:

稍微更新和更新的答案方法 @prranay's:

const routes = app._router.stack

 .filter//middleware/ => middleware.route/

 .map//middleware/ => `${Object.keys/middleware.route.methods/.join/', '/} -> ${middleware.route.path}`/



console.log/JSON.stringify/routes, null, 4//;

二哥

赞同来自:

Express Router的初始化

let router = require/'express'/.Router//;

router.get/'/', function /req, res/ {

 res.json/{

 status: `API Its Working`,

 route: router.stack.filter/r => r.route/

 .map/r=> { return {"path":r.route.path, 

 "methods":r.route.methods}}/,

 message: 'Welcome to my crafted with love!',

 }/;

 }/;

导入用户控制器

var userController = require/'./controller/userController'/;

用户路线

router.route/'/users'/

 .get/userController.index/

 .post/userController.new/;

router.route/'/users/:user_id'/

 .get/userController.view/

 .patch/userController.update/

 .put/userController.update/

 .delete/userController.delete/;

Export API 路线

module.exports = router;

出口

{"status":"API Its Working, APP Route","route": 

[{"path":"/","methods":{"get":true}}, 

{"path":"/users","methods":{"get":true,"post":true}}, 

{"path":"/users/:user_id","methods": ....}

冰洋

赞同来自:

它为我工作了

let routes = []

app._router.stack.forEach/function /middleware/ {

 if/middleware.route/ {

 routes.push/Object.keys/middleware.route.methods/ + " -> " + middleware.route.path/;

 }

}/;



console.log/JSON.stringify/routes, null, 4//;

O/P:

[

 "get -> /posts/:id",

 "post -> /posts",

 "patch -> /posts"

]

江南孤鹜

赞同来自:

在 Express 3.5.x 我在运行应用程序之前添加它以进行打印路由 terminal :

var routes = app.routes;

for /var verb in routes/{

 if /routes.hasOwnProperty/verb// {

 routes[verb].forEach/function/route/{

 console.log/verb + " : "+route['path']/;

 }/;

 }

}

也许它会有所帮助......

冰洋

赞同来自:

详细的路由信息包含一条路线列表 "express": "4.x.x",

import {

 Router

} from 'express';

var router = Router//;



router.get/"/routes", /req, res, next/ => {

 var routes = [];

 var i = 0;

 router.stack.forEach/function /r/ {

 if /r.route && r.route.path/ {

 r.route.stack.forEach/function /type/ {

 var method = type.method.toUpperCase//;

 routes[i++] = {

 no:i,

 method: method.toUpperCase//,

 path: r.route.path

 };

 }/

 }

 }/



 res.send/'<h1>List of routes.</h1>' + JSON.stringify/routes//;

}/;

简单的代码输出

List of routes.



[

{"no":1,"method":"POST","path":"/admin"},

{"no":2,"method":"GET","path":"/"},

{"no":3,"method":"GET","path":"/routes"},

{"no":4,"method":"POST","path":"/student/:studentId/course/:courseId/topic/:topicId/task/:taskId/item"},

{"no":5,"method":"GET","path":"/student/:studentId/course/:courseId/topic/:topicId/task/:taskId/item"},

{"no":6,"method":"PUT","path":"/student/:studentId/course/:courseId/topic/:topicId/task/:taskId/item/:itemId"},

{"no":7,"method":"DELETE","path":"/student/:studentId/course/:courseId/topic/:topicId/task/:taskId/item/:itemId"}

]

冰洋

赞同来自:

所以我正在寻找所有问题的答案......我不喜欢最多......我拿了一些东西 ... 这样做：

const resolveRoutes = /stack/ => {

 return stack.map/function /layer/ {

 if /layer.route && layer.route.path.isString/// {

 let methods = Object.keys/layer.route.methods/;

 if /methods.length > 20/

 methods = ["ALL"];



 return {methods: methods, path: layer.route.path};

 }



 if /layer.name === 'router'/ // router middleware

 return resolveRoutes/layer.handle.stack/;



 }/.filter/route => route/;

};



const routes = resolveRoutes/express._router.stack/;

const printRoute = /route/ => {

 if /Array.isArray/route//

 return route.forEach/route => printRoute/route//;



 console.log/JSON.stringify/route.methods/ + " " + route.path/;

};



printRoute/routes/;

不是最美丽的......但筑巢，并做他的工作

还有注意 20 那里......我只是假设没有正常路线 20 方法..所以我得出结论是所有的..

石油百科

赞同来自:

只需使用此包裹。 npm, 他将提供一个Web结论以及结论 terminal 以良好的格式化表格形式。

https://i.stack.imgur.com/iNnKn.png
https://www.npmjs.com/package/ ... logue

江南孤鹜

赞同来自:

在表达中 4.*

//Obtiene las rutas declaradas de la API

 let listPathRoutes: any[] = [];

 let rutasRouter = _.filter/application._router.stack, rutaTmp => rutaTmp.name === 'router'/;

 rutasRouter.forEach//pathRoute: any/ => {

 let pathPrincipal = pathRoute.regexp.toString//;

 pathPrincipal = pathPrincipal.replace/'/^\\',''/;

 pathPrincipal = pathPrincipal.replace/'?/?=\\/|$//i',''/;

 pathPrincipal = pathPrincipal.replace//\\\//g,'/'/;

 let routesTemp = _.filter/pathRoute.handle.stack, rutasTmp => rutasTmp.route !== undefined/;

 routesTemp.forEach//route: any/ => {

 let pathRuta = `${pathPrincipal.replace//\/\//g,''/}${route.route.path}`;

 let ruta = {

 path: pathRuta.replace/'//','/'/,

 methods: route.route.methods

 }

 listPathRoutes.push/ruta/;

 }/;

 }/;console.log/listPathRoutes/

二哥

赞同来自:

我发布了一个打印所有中间软件的包，以及路由，在尝试审核快速应用程序时非常有用。您使用包作为中间软件，因此甚至打印：

https://github.com/ErisDS/middleware-stack-printer
他打印出像树上的东西：

- middleware 1

- middleware 2

- Route /thing/

- - middleware 3

- - controller /HTTP VERB/

龙天

赞同来自:

以下是Express中的美丽打印路由的单线功能

app

:

const getAppRoutes = /app/ => app._router.stack.reduce/

 /acc, val/ => acc.concat/

 val.route ? [val.route.path] :

 val.name === "router" ? val.handle.stack.filter/

 x => x.route/.map/

 x => val.regexp.toString//.match//\/[a-z]+//[0] + /

 x.route.path === '/' ? '' : x.route.path// : []/ , []/.sort//;

如何在表达中获取所有注册路由？

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如何在表达中获取所有注册路由？

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