Javascript 在express Handlebar中配置文件扩展名
我有一个把手配置Javascript 在express Handlebar中配置文件扩展名,javascript,express,express-handlebars,Javascript,Express,Express Handlebars,我有一个把手配置 const path = require('path') const exphbs = require('express-handlebars') const morgan = require('morgan'); const multer = require('multer'); const express = require('express'); const erroHandler = require('errorhandler') const routes = req
const path = require('path')
const exphbs = require('express-handlebars')
const morgan = require('morgan');
const multer = require('multer');
const express = require('express');
const erroHandler = require('errorhandler')
const routes = require('../routes/index');
module.exports = app => {
app.set('port', process.env.PORT || 80)
app.set('views', path.join(__dirname, '../views'));
app.engine('.hbs', exphbs({
defaultLayout: 'main',
partialsDir: path.join(app.get('views'), 'partials' ),
layoutDir: path.join(app.get('views'), 'layouts'),
helpers: require('../helpers')
}))
app.set('view engine', '.hbs');
// middlewares
app.use(morgan('dev'));
app.use(multer({dest: path.join(__dirname, '../public/upload/temp')}).single('image'))
app.use(express.urlencoded({extended: false}));
app.use(express.json());
//routes
routes(app)
// static files
app.use('/public', express.static(path.join(__dirname, '../public')));
// errorhandlers
if ('development' === app.get('env')) {
app.use(erroHandler)
}
return app;
}
但我得到了这个错误:
错误:eNote:没有这样的文件或目录,请打开“C:\Users\Diesan”
Romero\Desktop\redsocial\views\layouts\main.Handlebar'
这是我的目录,怎么了 由于您的把手文件使用
.hbs
作为扩展名,因此在实例化把手时需要正确定义,如下所示:
exphbs.create({
extname: '.hbs',
// rest of options
})
或者修改您的代码:
app.engine('.hbs', exphbs({
extName: '.hbs',
defaultLayout: 'main',
partialsDir: path.join(app.get('views'), 'partials' ),
layoutDir: path.join(app.get('views'), 'layouts'),
helpers: require('../helpers')
}))
或者只需将您的Handlebar文件扩展名更改为
filename.handlebar
我需要将创建函数放在哪里?引擎内部?@DiesanRomero已编辑。@DiesanRomero您的views
文件夹位于src
文件夹中。同时修复您的文件路径。我正在尝试使用这个:path.join(_dirname,../src/views)),但是我得到了src文件夹两次,当我删除它时,我没有得到该文件夹。那太令人困惑了me@DiesanRomero如何:path.join(uu dirname,'../../src/views')
?