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下面是我的node.js文件的一個片段 我想從LESS編譯代碼到CSS。 我缺少什麼LESS文件沒有得到編譯
app.use(lessMiddleware({
src: path.join(__dirname, '/public/stylesheets/styles', 'less'),
dest: path.join(__dirname, '/public/stylesheets/css'),
prefix : '/stylesheets',
}));
當我刷新我的index.html,它拋出我下面的錯誤
TypeError: Arguments to path.join must be strings
at f (path.js:204:15)
at Object.filter (native)
at Object.exports.join (path.js:209:40)
at C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\less-middleware\lib\mi
ddleware.js:161:27
at Layer.handle [as handle_request] (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\n
ode_modules\express\lib\router\layer.js:95:5)
at trim_prefix (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\l
ib\router\index.js:312:13)
at C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\lib\router\ind
ex.js:280:7
at Function.process_params (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modul
es\express\lib\router\index.js:330:12)
at next (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\lib\rout
er\index.js:271:10)
at expressInit (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\l
ib\middleware\init.js:33:5)
TypeError: Arguments to path.join must be strings
at f (path.js:204:15)
at Object.filter (native)
at Object.exports.join (path.js:209:40)
at C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\less-middleware\lib\mi
ddleware.js:161:27
at Layer.handle [as handle_request] (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\n
ode_modules\express\lib\router\layer.js:95:5)
at trim_prefix (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\l
ib\router\index.js:312:13)
at C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\lib\router\ind
ex.js:280:7
at Function.process_params (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modul
es\express\lib\router\index.js:330:12)
at next (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\lib\rout
er\index.js:271:10)
at expressInit (C:\SrkOwnRepo\Hotaaal\HotelUI10892515\node_modules\express\l
ib\middleware\init.js:33:5)
請提供一個解決辦法或任何其他中間件。