目前,我有以下所有其他路线:
app.get('*', function(req, res){
console.log('404ing');
res.render('404');
});
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根据日志,即使路线在上面匹配,也会被解雇.如何在没有任何匹配的情况下才能让它发射?
Cha*_*les 53
你只需要把它放在所有路线的尽头.
看看传递路径控制的第二个例子:
var express = require('express')
, app = express.createServer();
var users = [{ name: 'tj' }];
app.all('/user/:id/:op?', function(req, res, next){
req.user = users[req.params.id];
if (req.user) {
next();
} else {
next(new Error('cannot find user ' + req.params.id));
}
});
app.get('/user/:id', function(req, res){
res.send('viewing ' + req.user.name);
});
app.get('/user/:id/edit', function(req, res){
res.send('editing ' + req.user.name);
});
app.put('/user/:id', function(req, res){
res.send('updating ' + req.user.name);
});
app.get('*', function(req, res){
res.send('what???', 404);
});
app.listen(3000);
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或者,您无能为力,因为所有不匹配的路由都将生成404.然后您可以使用此代码显示正确的模板:
app.error(function(err, req, res, next){
if (err instanceof NotFound) {
res.render('404.jade');
} else {
next(err);
}
});
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它在错误处理中有记录.
您可以在所有路线的末尾处执行此操作,
const express = require('express');
const app = express();
const port = 8080;
// All your routes and middleware here.....
app.use((req, res, next) => {
res.status(404).json({
message: 'Ohh you are lost, read the API documentation to find your way back home :)'
})
})
// Init the server here,
app.listen( port, () => {
console.log('Sever is up')
})Run Code Online (Sandbox Code Playgroud)
希望有帮助,我在路线底部使用了这段代码
router.use((req, res, next) => {
next({
status: 404,
message: 'Not Found',
});
});
router.use((err, req, res, next) => {
if (err.status === 404) {
return res.status(400).render('404');
}
if (err.status === 500) {
return res.status(500).render('500');
}
next();
});
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