如何仅使用uglifyjs将多个文件缩小为一个文件?

时间:2017-05-26 01:04:56

标签: javascript node.js angular express uglifyjs

我尝试使用uglifyjs将javascript文件缩小为一个文件,但它无效。我使用$ node uglifyjs.js来运行该文件。以下是uglify.js中的代码

  var fs = require('fs');

  var uglifyjs = require('uglify-js');

  var files = ['app.js', 'geolocation.service.js'];

  var result = uglifyjs.minify(fs.readFileSync(files, 'utf8'));

 console.log(result.code);

 fs.writeFile("output.min.js", result.code, function(err){
 if(err){
  console.log(err);
  } else {
  console.log("File was successfully saved.");
 }
 });

当我运行此代码时,我收到以下错误消息:

  fs.js:549
  return binding.open(pathModule._makeLong(path), stringToFlags(flags), 
  mode);
             ^

   TypeError: path must be a string
   at TypeError (native)
   at Object.fs.openSync (fs.js:549:18)
   at Object.fs.readFileSync (fs.js:397:15)
   at Object.<anonymous> (C:\Users\HP\Desktop\uglify\uglify.js:7:33)
   at Module._compile (module.js:435:26)
   at Object.Module._extensions..js (module.js:442:10)
   at Module.load (module.js:356:32)
   at Function.Module._load (module.js:311:12)
   at Function.Module.runMain (module.js:467:10)
   at startup (node.js:134:18)

1 个答案:

答案 0 :(得分:4)

here所示,uglify.minify要求您传递stringstring个列表。

您的代码:

var result = uglifyjs.minify(fs.readFileSync(files, 'utf8'));

尝试一次读取多个文件,但fs.readFileSync只接受一个文件。

所以将代码更改为:

var filesContents = ['app.js', 'geolocation.service.js'].map(function (file) {
    return fs.readFileSync(file, 'utf8');
})

var result = uglifyjs.minify(filesContents);

if (result.error) {
    console.error("Error minifying: " + result.error);
}

fs.writeFile("output.min.js", result.code, function (err) {
    if (err) {
        console.error(err);
    } else {
        console.log("File was successfully saved.");
    }
});