如何使用uglifyjs將多個文件縮減爲一個文件？

Question

我試圖用uglifyjs將javascript文件縮小爲一個文件，但它不起作用。我使用$ node uglifyjs.js來運行該文件。下面是uglify.js代碼

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."); 
} 
}); 

當我運行這段代碼，我得到了以下錯誤消息：

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) 

Answer 1

如here所示，uglify.minify要求您在string或列表傳string s。

您的代碼：

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

試圖一次讀取多個文件，但fs.readFileSync只接受一個文件。

所以你的代碼改成這樣：

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."); 
    } 
});