unzip the file and parse it to js - node.js

Question

hi i had tried to unzip the file from my c drive and trying to parse to javascript object
here is the code
var AdmZip = require('adm-zip');
var fs = require('fs'), xml2js = require('xml2js');
var parser = new xml2js.Parser();
var paramdata = 'c:/sample/kusuma.zip';
console.log(paramdata);
var zip = new AdmZip(paramdata);
var zipEntries = zip.getEntries();
var obj = [];
var count = 0;
zipEntries.forEach(function(zipEntry) {
var len = zipEntries.length;
console.log(zipEntry.toString());
console.log(zipEntry.entryName);
fs.readFile("", function(err, data) {
console.log(data);
parser.parseString(data, function(err, result) {
count++;
console.log(count);
obj.push(result);
if (count === len) {
console.log(obj);
res.send(obj);
}
});
});
});
please check the code once and provide me some more examples

Answer 1

Well, fs.readFile() is for reading files that are themselves directly on disk, which these aren't.
However, adm-zip is already reading in the contents of the .zip, so you shouldn't need fs. Each zipEntry has getData() and getDataAsync() methods that can be used to retrieve contents.
zipEntries.forEach(function (zipEntry) {
zipEntry.getDataAsync(function (data) {
parser.parseString(data, function (err, result) {
console.log(result);
});
});
});
Also, as zipEntries is an Array, you can use .filter() to reduce it to only XML files.
var zipEntries = zip.getEntries().filter(function (zipEntry) {
return !zipEntry.isDirectory && /\.xml$/.test(zipEntry.entryName);
});
You'll also want to determine len once from the collection rather than from each entry. You can also test that against use obj.length rather than having to keep count separately:
var len = zipEntries.length;
var obj = [];
zipEntries.forEach(function (zipEntry) {
zipEntry.getDataAsync(function (data) {
parser.parseString(data, function (err, result) {
obj.push(result);
if (obj.length === len) {
res.send(obj);
}
});
});
});