可选的获取参数

The following:

以下几点:

app.get('/foo/start/:start/end/:end', blah.someFunc);

matches

匹配

/foo/start/1/end/4

but I want it to also match an optional parameter

但我希望它也匹配一个可选参数

/foo/start/1/end/4/optional/7

I've tried this:

我已经试过这个:

app.get('/foo/start/:start/end/:end(/optional/:value)?', blah.someFunc);

but it doesn't match either of the above two examples. I think it's because I'm trying to give it a RegExp when it's expecting something else?

但这两个例子都不匹配。我想是因为我试图给它一个RegExp当它期望别的东西时?

Thanks.

谢谢。

4 个解决方案

#1

Why don't you add another rule before the one you have, like this

你为什么不在已有的规则之前再加一条，像这样

app.get('/foo/start/:start/end/:end/optional/:value', blah.someFunc);
app.get('/foo/start/:start/end/:end', blah.someFunc);

It will be used before the one without the optional value.

它将在没有可选值的值之前使用。

If you want to use just one line try this:

如果你只想用一行的话，试试这个:

app.get('/foo/start/:start/end/:end/optional?', blah.someFunc)

see the docs for an example.

请参阅文档中的示例。

#2

If you are using Express 4.x Then I think its better to use array format for route. For example I have route /service which gives all service list and same route when used with id /service/id/:id gives single service with id in the param.

如果你使用的是Express 4。然后我认为最好使用数组格式来进行路由。例如，在使用id/ service/id/ id时，我有提供所有服务列表和相同路由的路由/服务:id在param中提供id。

app.get(['/service', '/service/id/:id'], function(req, res) {});

#3

In this example, if the url is /hello or /hello/world it works. The ? makes the parameter become optional (express 4).

在本例中，如果url是/hello或/hello/world，那么它就可以工作。的吗?使参数成为可选的(express 4)。

// app.js
var index = require('/routes/index');
app.use('/hello', index);

// routes/index.js
router.get('hello/:name?', function(req, res){
  var name = req.params.name;
  var data = {
    name: name
  };
  res.json(data);
});

#4

You can also use regular expressions in routes. Perhaps something like:

您还可以在路由中使用正则表达式。也许类似:

app.get(/^\/foo\/start\/:start\/end\/:end(\/optional\/:value)?/, function (req, res, next) {

#1