您好我必须在一个如下所示的架构中配置一个快速应用程序:
![my architecture][1]
Build
Core/pulse.core/assets/styles/global.css
app.js
Express_app
views/home.jade
controller/routes.js
node_modules
我的app.js看起来像这样:
var express = require('express'),
http = require('http'),
fs = require('fs'),
path = require('path');
var app = express();
html_templates = __dirname + '/Express_app';
app.set('views', html_templates + '/views');
app.set('view engine', 'jade');
app.use("/Core", express.static(__dirname + 'Core/Pulse.Core/Assets/Styles/'));
app.listen(3000, function () {
console.log("express has started on port 3000");
});
require('./Express_app/controller/routes.js')(app);
在我的home.jade文件中,我有:
link(rel="stylesheet", href=levelarbo+"../../Core/Pulse.Core/Assets/Styles/global.css")
我的html正常加载,但是,我的css文件得到404,请帮忙。我被困在这几个小时:(
答案 0 :(得分:1)
app.use("/Core", express.static(__dirname + 'Core/Pulse.Core/Assets/Styles/'));
应该是(注意/
的位置)......
app.use("/Core", express.static(__dirname + '/Core'));
URL中的大小写应与文件系统中的大小写匹配。如果实际路径为Core/pulse.core/assets/styles/global.css
,则玉文件应为...
link(rel="stylesheet", href=levelarbo+"../../Core/pulse.core/assets/styles/global.css")