问题
I'm having a node app which needs to get some zip file from client Postman and extract it to a folder in my fileSystem,Im using express I did the following which doesnt work,
what am I missing here?
I've created sample node app to simulate the issue.
var express = require('express');
var upload = require('multer')({ dest: 'uploads/' });
var admZip = require('adm-zip');
var app = express();
app.post('/',upload.single('file'),function(req,res){
debugger;
var zip = new admZip(req.file);
zip.extractAllTo("C://TestFolder//TestPathtoExtract", true);
res.send("unzip");
});
var server = app.listen(3001,function(){
var host = server.address().address;
var port = server.address().port;
console.log('Example app listening at http://%s:%s',host,port);
})
This is how I use it im postman
If there is other way to do it with different open source this can be great! I use https://github.com/cthackers/adm-zip
which can be change to any other library
I've also find this lib but not sure how to use it with express https://www.npmjs.com/package/decompress-zip
Thanks!
回答1:
This is the set up I did for Postman
, first this is my form-data
body
Now in the header I left in blank after trying to set multipart/form-data
manually and utterly failed, so no header here.
Here I did a pair of console.log
, one of req.headers
to be sure of Postman
sending the right multipart/form-data
and another of req.file
And well the output seems to be fine
Edit: the code.
var express = require('express');
var upload = require('multer')({
dest: 'uploads/'
});
var admZip = require('adm-zip');
var app = express();
app.post('/', upload.single('file'), function(req, res) {
console.log('%c > req.headers test.js [9] <=================================', 'color:blue;', req.headers);
debugger;
console.log('%c > req.file test.js [10] <=================================', 'color:blue;', req.file);
//instead of just req.file I use req.file.path as admzip needs the actual file path
var zip = new admZip(req.file.path);
zip.extractAllTo("/Users/myuser/Desktop/ext", true);
res.send("unzip");
});
var server = app.listen(3001, function() {
var host = server.address().address;
var port = server.address().port;
console.log('Example app listening at http://%s:%s', host, port);
});
回答2:
You need to pass filename as argument.
Use req.file.path
var zip = new admZip(req.file.path);
来源:https://stackoverflow.com/questions/34043569/how-to-extract-zip-from-client-in-node