我有一个工作api,用于在server.js中的路由中提供图像,并希望将其抽象到一个单独的模块中。
之前:
app.get('/api/image/:filename', function(req, res){
var resourcePath = 'uploads/public/projectnumber/issues/small/' + req.params.filename + '.png';
console.log(resourcePath)
if(fs.existsSync(resourcePath)) {
var file = fs.readFileSync(resourcePath);
res.writeHead(200, 'Content-Type:application/pdf:image/png');
res.end(file,'binary');
}
else {
res.send(400, 'No image found');
}
})
我想要这样的东西:
var ImageRouter = require('./routes/imageRouter');
app.use('/api/image/:filename', ImageRouter);
我试着在我的imageRouter.js文件中这样写:
var express = require('express');
var fs = require('fs');
var router = express.Router();
router.use(function(req, res, next) {
var resourcePath = 'public/images/' + req.params.filename + '.png';
if(fs.existsSync(resourcePath)) {
var file = fs.readFileSync(resourcePath);
res.writeHead(200, 'Content-Type:application/pdf:image/png');
res.end(file,'binary');
}
else {
res.send(400, 'No image found');
}
next();
});
module.exports = router;
但是req.params.filename未定义。我哪里错了?
谢谢!
您应该在imageRouter.js
路由器上使用get()
,并在主应用程序上加前缀。
use()
用于中间件。
这是imageRouter.js
:
var router = require('express').Router();
var fs = require('fs');
router.get('/:filename', function(req, res) {
var resourcePath = 'public/images/' + req.params.filename + '.png';
if(fs.existsSync(resourcePath)) {
var file = fs.readFileSync(resourcePath);
res.writeHead(200, 'Content-Type:application/pdf:image/png');
res.end(file,'binary');
}
else {
res.send(400, 'No image found');
}
});
module.exports = router;
和你的服务器.js:
var express = require('express');
var app = express();
var ImageRouter = require('./routes/imageRouter');
app.use('/api/image', ImageRouter);