在express中root后通过可选参数传递路由控制?

发布于 2024-11-25 16:53:51 字数 1392 浏览 0 评论 0原文

我正在开发一个简单的网址缩短应用程序,并具有以下快速路线:

app.get('/', function(req, res){
  res.render('index', {
    link: null
  });
});

app.post('/', function(req, res){
  function makeRandom(){
    var text = "";
    var possible = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";

    for( var i=0; i < 3 /*y u looking at me <33??*/; i++ )
      text += possible.charAt(Math.floor(Math.random() * possible.length));
    return text;
  }
  var url = req.body.user.url;
  var key = makeRandom();
  client.set(key, url);
  var link = 'http://50.22.248.74/l/' + key;
  res.render('index', {
    link: link
  });
  console.log(url);
  console.log(key);
});

app.get('/l/:key', function(req, res){
  client.get(req.params.key, function(err, reply){
    if(client.get(reply)){
      res.redirect(reply);
    }
    else{
      res.render('index', {
        link: null
      });
    }
  });
});

我想从我的路线中删除 /l/ (以使我的网址更短)并设置 :key 参数选修的。这是执行此操作的正确方法吗:

app.get('/:key?', function(req, res, next){
  client.get(req.params.key, function(err, reply){
    if(client.get(reply)){
      res.redirect(reply);
    }
    else{
      next();
    }
  });
});

app.get('/', function(req, res){
  res.render('index, {
    link: null
  });
});

不确定我是否需要指定我的 / 路线是“下一个”路线。但由于我唯一的其他路线是更新后的 / 路线,我想它会工作得很好。

I'm working on a simple url-shortening app and have the following express routes:

app.get('/', function(req, res){
  res.render('index', {
    link: null
  });
});

app.post('/', function(req, res){
  function makeRandom(){
    var text = "";
    var possible = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";

    for( var i=0; i < 3 /*y u looking at me <33??*/; i++ )
      text += possible.charAt(Math.floor(Math.random() * possible.length));
    return text;
  }
  var url = req.body.user.url;
  var key = makeRandom();
  client.set(key, url);
  var link = 'http://50.22.248.74/l/' + key;
  res.render('index', {
    link: link
  });
  console.log(url);
  console.log(key);
});

app.get('/l/:key', function(req, res){
  client.get(req.params.key, function(err, reply){
    if(client.get(reply)){
      res.redirect(reply);
    }
    else{
      res.render('index', {
        link: null
      });
    }
  });
});

I would like to remove the /l/ from my route (to make my url's shorter) and make the :key parameter optional. Would this be the correct way to do this:

app.get('/:key?', function(req, res, next){
  client.get(req.params.key, function(err, reply){
    if(client.get(reply)){
      res.redirect(reply);
    }
    else{
      next();
    }
  });
});

app.get('/', function(req, res){
  res.render('index, {
    link: null
  });
});

Not sure if I need to specify that my / route is the one to be "nexted" to. But since my only other route would be my updated post / route, I would imagine it would work fine.

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评论(2

江南月 2024-12-02 16:53:51

这将取决于 client.get 在将 undefined 作为其第一个参数传递时执行的操作。

像这样的事情会更安全:

app.get('/:key?', function(req, res, next) {
    var key = req.params.key;
    if (!key) {
        next();
        return;
    }
    client.get(key, function(err, reply) {
        if(client.get(reply)) {
            res.redirect(reply);
        }
        else {
            res.render('index', {
                link: null
            });
        }
    });
});

在回调中调用 next() 没有问题。

根据 this,处理程序按照添加的顺序调用,因此只要你的下一个路由是 app.get('/', ...) 如果没有键,它就会被调用。

That would work depending on what client.get does when passed undefined as its first parameter.

Something like this would be safer:

app.get('/:key?', function(req, res, next) {
    var key = req.params.key;
    if (!key) {
        next();
        return;
    }
    client.get(key, function(err, reply) {
        if(client.get(reply)) {
            res.redirect(reply);
        }
        else {
            res.render('index', {
                link: null
            });
        }
    });
});

There's no problem in calling next() inside the callback.

According to this, handlers are invoked in the order that they are added, so as long as your next route is app.get('/', ...) it will be called if there is no key.

难理解 2024-12-02 16:53:51

Express版本:

"dependencies": {
    "body-parser": "^1.19.0",
    "express": "^4.17.1"
  }

可选参数非常方便,您可以使用express轻松声明和使用它们:

app.get('/api/v1/tours/:cId/:pId/:batchNo?', (req, res)=>{
    console.log("category Id: "+req.params.cId);
    console.log("product ID: "+req.params.pId);
    if (req.params.batchNo){
        console.log("Batch No: "+req.params.batchNo);
    }
});

在上面的代码中,batchNo是可选的。 Express 会将其视为可选,因为在 URL 构造之后,我给出了一个“?” batchNo '/:batchNo?' 之后的符号

现在我可以仅使用categoryId 和productId 进行调用,或者使用所有三个参数进行调用。

http://127.0.0.1:3000/api/v1/tours/5/10
//or
http://127.0.0.1:3000/api/v1/tours/5/10/8987

输入图像描述这里
输入图像描述这里

Express version:

"dependencies": {
    "body-parser": "^1.19.0",
    "express": "^4.17.1"
  }

Optional parameter are very much handy, you can declare and use them easily using express:

app.get('/api/v1/tours/:cId/:pId/:batchNo?', (req, res)=>{
    console.log("category Id: "+req.params.cId);
    console.log("product ID: "+req.params.pId);
    if (req.params.batchNo){
        console.log("Batch No: "+req.params.batchNo);
    }
});

In the above code batchNo is optional. Express will count it optional because after in URL construction, I gave a '?' symbol after batchNo '/:batchNo?'

Now I can call with only categoryId and productId or with all three-parameter.

http://127.0.0.1:3000/api/v1/tours/5/10
//or
http://127.0.0.1:3000/api/v1/tours/5/10/8987

enter image description here
enter image description here

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