2013-08-29 37 views
2

我做了下面的事情,和node.js一起玩。 zipfiles文件夹中的文件会相应地压缩,并且所有内容都可以正常工作。 但是我在cmd上发现错误,我不知道它来自哪里或者如何解决它。nodejs扔呃; //未处理的'错误'事件

events.js:72 
     throw er; // Unhandled 'error' event 
      ^
Error: write after end 
    at writeAfterEnd (_stream_writable.js:130:12) 
    at Gzip.Writable.write (_stream_writable.js:178:5) 
    at write (_stream_readable.js:583:24) 
    at flow (_stream_readable.js:592:7) 
    at ReadStream.pipeOnReadable (_stream_readable.js:624:5) 
    at ReadStream.EventEmitter.emit (events.js:92:17) 
    at emitReadable_ (_stream_readable.js:408:10) 
    at emitReadable (_stream_readable.js:404:5) 
    at readableAddChunk (_stream_readable.js:165:9) 
    at ReadStream.Readable.push (_stream_readable.js:127:10) 

这里是我的脚本:

var zlib = require('zlib'); 
var gzip = zlib.createGzip(); 
var fs = require('fs'); 
var zip = { 
    zipAll: function(dir){ 
     //files to zip 
     fs.readdir(dir, function(err, data){ 
      if(err) throw(err); 
      var arrayValue = data.toString().split(','); 

      //files with .gz at the end, needs to be excluded 
       for(var i=0; i<arrayValue.length; i+=1){ 
        console.log("Zipping following files: " + arrayValue[i]); 
        var input = fs.createReadStream('zipfiles/' + arrayValue[i]); 
        var output = fs.createWriteStream('zipfiles/input'+[i]+'.txt'+'.gz'); 
        input.pipe(gzip).pipe(output);     
       } 
     }); 
    } 
}; 
zip.zipAll('zipfiles'); 

感谢

回答

2

gzip的对象是有点靠不住的(据我所知无证)重用多个文件。解决问题的最简单方法是简单地使用每个文件的单独对象gzip进行压缩,例如;

for(var i=0; i<arrayValue.length; i+=1){ 
    console.log("Zipping following files: " + arrayValue[i]); 
    var input = fs.createReadStream('zipfiles/' + arrayValue[i]); 
    var output = fs.createWriteStream('zipfiles/input'+[i]+'.txt'+'.gz'); 
    input.pipe(zlib.createGzip()).pipe(output);     
} 
相关问题