为什么节点中间件不工作？ [英] Why node-sass-middleware is not working?
问题描述
我已经在我的快速应用程序中安装了 node-sass-middleware 模块，但是我并没有这样做，只是因为中间件读取的是不正确的源，当我调试控制台日志是：
GET / 200 558.983 ms - 4651
source：/ home / karim / snippets / my-financial / public / stylesheets / sass / stylesheets / main.sass
dest：/home/karim/Snippets/my-financial/public/stylesheets/stylesheets/main.css
阅读：/ home / karim / snippets / my-financial / public / stylesheets / stylesheets / main.css
目录是错误的，为什么中间件在源 / dest之间添加字符串（ stylesheets / .. public / stylesheets / sass / ）和 .sass文件 / .css文件（ main.sass 和 main.css ） ？
我的 app.js 中有这个配置：
var sassMiddleware =要求（节点萨斯中间件）;
...
...
var app = express（）;
app.use（sassMiddleware（{
src：path.join（__ dirname，'public / stylesheets / sass'），
dest：path.join（__ dirname，'public / stylesheets'），
debug：true，
indentedSyntax：true，
outputStyle：'compressed'
}））;
显然这并不是编译任何东西，因为目录是错误的。
在$ code> .. public / stylesheets / sass / 文件夹内我只有一个文件， main.sass 我想编译并将结果移动到 sass / 文件夹之外，我的意思是在 .. public / stylesheets /
这是因为 - 我很确定 - 在你的html文件上有这样的东西：
< head>
<！ - 所有的头部东西和 - >
< link rel =stylesheethref =/ stylesheets / main.css/>
< / head>
- 让我们把这个href称为你的爱好者一段时间。
当您的服务器收到任何 get 请求时，中间件将查找编译的 main.sass / home / karim / Snippets / my-financial / public / stylesheets （ dest yourAwesomeHref ，导致此路由：
/ Which那个文件显然根本不存在！
所以你必须添加前缀：/ stylesheets在你的中间件避免这个问题。
最终的代码是：
var sassMiddleware = require（'node-sass-middleware'） ;
...
...
var app = express（）;
app.use（sassMiddleware（{
src：path.join（__ dirname，'public / stylesheets / sass'），
dest：path.join（__ dirname，'public / stylesheets'），
debug：true，
indentedSyntax：true，
outputStyle：'compressed'，
前缀：'/ stylesheets'
}））;
I have installed the node-sass-middleware module on my express application, but i'm not getting that working, just because the middleware is reading an incorrect source, when i debug the console log is:
GET / 200 558.983 ms - 4651
source: /home/karim/Snippets/my-financial/public/stylesheets/sass/stylesheets/main.sass
dest: /home/karim/Snippets/my-financial/public/stylesheets/stylesheets/main.css
read: /home/karim/Snippets/my-financial/public/stylesheets/stylesheets/main.css
which both directories are wrong, why the middleware is adding the string stylesheets/ between the source/dest (..public/stylesheets/sass/) and the .sass file/.css file (main.sass and main.css)?
I have this configuration inside my app.js:
var sassMiddleware = require('node-sass-middleware');
...
...
var app = express();
app.use(sassMiddleware({
src: path.join(__dirname, 'public/stylesheets/sass'),
dest: path.join(__dirname, 'public/stylesheets'),
debug: true,
indentedSyntax: true,
outputStyle: 'compressed'
}));
Obviously this is not compiling anything, becuase the directories are wrong.
Inside the ..public/stylesheets/sass/ folder i just have one file, main.sass which i want to compile and move the result outside the sass/ folder, i mean at ..public/stylesheets/.
解决方案 That is because -- i am pretty sure -- on your html file there is something like that:
<head>
<!--All your head stuff and-->
<link rel="stylesheet" href="/stylesheets/main.css"/>
</head>
-- Lets call that href as yourAwesomeHref for a moment.
When your server receive any get request, the middleware will look for the compiled main.sass on /home/karim/Snippets/my-financial/public/stylesheets (dest option for the middleware) following by yourAwesomeHref, resulting this route:
/home/karim/Snippets/my-financial/public/stylesheets/stylesheets/main.css
Which that file obviously does not exist at all!
So you have to add prefix: "/stylesheets" on your middleware for avoid that problem.
The final code is:
var sassMiddleware = require('node-sass-middleware');
...
...
var app = express();
app.use(sassMiddleware({
src: path.join(__dirname, 'public/stylesheets/sass'),
dest: path.join(__dirname, 'public/stylesheets'),
debug: true,
indentedSyntax: true,
outputStyle: 'compressed',
prefix: '/stylesheets'
}));
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