NodeJS trying to use a URL inside of a fs.createfilestream
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0
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I'm trying to do a post request onto my api, the api works perfectly ( I am able to post files, but not through a url), but now I'm trying to post through an url.
this is the code I have now, I removed some lines that aren't relevant to the question or were for testing.
request({
url: url + "gettoken"
, json: true
}, function (error, response, body) {
user = body;
var rs = fs.createReadStream(up.url);
var ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
ws.on('drain', function () {
rs.resume();
});
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
})
Can anyone please help me.
javascript node.js
asked Nov 22 at 11:38
Gido Selten
176
add a comment |
up vote
0
down vote
favorite
I'm trying to do a post request onto my api, the api works perfectly ( I am able to post files, but not through a url), but now I'm trying to post through an url.
this is the code I have now, I removed some lines that aren't relevant to the question or were for testing.
request({
url: url + "gettoken"
, json: true
}, function (error, response, body) {
user = body;
var rs = fs.createReadStream(up.url);
var ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
ws.on('drain', function () {
rs.resume();
});
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
})
Can anyone please help me.
javascript node.js
asked Nov 22 at 11:38
Gido Selten
176
What isup.url?fs.createReadStreamonly works for (local) files.
– robertklep
Nov 22 at 14:11
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
fs.createReadStreamis meant to read files, not to make POST requests, that's what therequest.post()is for. Do you mean that you want to download a file fromup.urland subsequently upload that file tourl +"upload?..."?
– robertklep
Nov 22 at 14:27
yes, that's what I want
– Gido Selten
Nov 22 at 14:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to do a post request onto my api, the api works perfectly ( I am able to post files, but not through a url), but now I'm trying to post through an url.
this is the code I have now, I removed some lines that aren't relevant to the question or were for testing.
request({
url: url + "gettoken"
, json: true
}, function (error, response, body) {
user = body;
var rs = fs.createReadStream(up.url);
var ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
ws.on('drain', function () {
rs.resume();
});
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
})
Can anyone please help me.
javascript node.js
asked Nov 22 at 11:38
Gido Selten
176
I'm trying to do a post request onto my api, the api works perfectly ( I am able to post files, but not through a url), but now I'm trying to post through an url.
this is the code I have now, I removed some lines that aren't relevant to the question or were for testing.
request({
url: url + "gettoken"
, json: true
}, function (error, response, body) {
user = body;
var rs = fs.createReadStream(up.url);
var ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
ws.on('drain', function () {
rs.resume();
});
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
})
Can anyone please help me.
javascript node.js
javascript node.js
asked Nov 22 at 11:38
Gido Selten
176
asked Nov 22 at 11:38
Gido Selten
176
asked Nov 22 at 11:38
Gido Selten
176
asked Nov 22 at 11:38
Gido Selten
176
asked Nov 22 at 11:38
Gido Selten
176
176
What isup.url?fs.createReadStreamonly works for (local) files.
– robertklep
Nov 22 at 14:11
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
fs.createReadStreamis meant to read files, not to make POST requests, that's what therequest.post()is for. Do you mean that you want to download a file fromup.urland subsequently upload that file tourl +"upload?..."?
– robertklep
Nov 22 at 14:27
yes, that's what I want
– Gido Selten
Nov 22 at 14:49
add a comment |
What isup.url?fs.createReadStreamonly works for (local) files.
– robertklep
Nov 22 at 14:11
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
fs.createReadStreamis meant to read files, not to make POST requests, that's what therequest.post()is for. Do you mean that you want to download a file fromup.urland subsequently upload that file tourl +"upload?..."?
– robertklep
Nov 22 at 14:27
yes, that's what I want
– Gido Selten
Nov 22 at 14:49
What is
up.url? fs.createReadStream only works for (local) files.– robertklep
Nov 22 at 14:11
What is
up.url? fs.createReadStream only works for (local) files.– robertklep
Nov 22 at 14:11
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
fs.createReadStream is meant to read files, not to make POST requests, that's what the request.post() is for. Do you mean that you want to download a file from up.url and subsequently upload that file to url +"upload?..."?– robertklep
Nov 22 at 14:27
fs.createReadStream is meant to read files, not to make POST requests, that's what the request.post() is for. Do you mean that you want to download a file from up.url and subsequently upload that file to url +"upload?..."?– robertklep
Nov 22 at 14:27
yes, that's what I want
– Gido Selten
Nov 22 at 14:49
yes, that's what I want
– Gido Selten
Nov 22 at 14:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
So the idea is to upload a file that's located at up.url to another server at url + "upload?...".
Since fs.createReadStream is meant to read local files, and not URL's, you need something that can create a stream from a URL (or rather, retrieve that URL and stream the response).
You can also use request for that:
request({
url: url + "gettoken",
json: true
}, function (error, response, body) {
const user = body;
const rs = request.get(up.url);
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
});
Typically, file uploads work through multipart/form-data, but your code doesn't suggest that being used here. If it is, the code would become something like this:
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`, {
formData : {
the_file : rs
}
});
// no `rs.pipe(ws)`
answered Nov 22 at 15:04
robertklep
134k17231240
that works thank you!
– Gido Selten
Nov 22 at 15:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
So the idea is to upload a file that's located at up.url to another server at url + "upload?...".
Since fs.createReadStream is meant to read local files, and not URL's, you need something that can create a stream from a URL (or rather, retrieve that URL and stream the response).
You can also use request for that:
request({
url: url + "gettoken",
json: true
}, function (error, response, body) {
const user = body;
const rs = request.get(up.url);
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
});
Typically, file uploads work through multipart/form-data, but your code doesn't suggest that being used here. If it is, the code would become something like this:
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`, {
formData : {
the_file : rs
}
});
// no `rs.pipe(ws)`
answered Nov 22 at 15:04
robertklep
134k17231240
that works thank you!
– Gido Selten
Nov 22 at 15:12
add a comment |
up vote
1
down vote
accepted
So the idea is to upload a file that's located at up.url to another server at url + "upload?...".
Since fs.createReadStream is meant to read local files, and not URL's, you need something that can create a stream from a URL (or rather, retrieve that URL and stream the response).
You can also use request for that:
request({
url: url + "gettoken",
json: true
}, function (error, response, body) {
const user = body;
const rs = request.get(up.url);
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
});
Typically, file uploads work through multipart/form-data, but your code doesn't suggest that being used here. If it is, the code would become something like this:
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`, {
formData : {
the_file : rs
}
});
// no `rs.pipe(ws)`
answered Nov 22 at 15:04
robertklep
134k17231240
that works thank you!
– Gido Selten
Nov 22 at 15:12
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
So the idea is to upload a file that's located at up.url to another server at url + "upload?...".
Since fs.createReadStream is meant to read local files, and not URL's, you need something that can create a stream from a URL (or rather, retrieve that URL and stream the response).
You can also use request for that:
request({
url: url + "gettoken",
json: true
}, function (error, response, body) {
const user = body;
const rs = request.get(up.url);
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
});
Typically, file uploads work through multipart/form-data, but your code doesn't suggest that being used here. If it is, the code would become something like this:
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`, {
formData : {
the_file : rs
}
});
// no `rs.pipe(ws)`
answered Nov 22 at 15:04
robertklep
134k17231240
So the idea is to upload a file that's located at up.url to another server at url + "upload?...".
Since fs.createReadStream is meant to read local files, and not URL's, you need something that can create a stream from a URL (or rather, retrieve that URL and stream the response).
You can also use request for that:
request({
url: url + "gettoken",
json: true
}, function (error, response, body) {
const user = body;
const rs = request.get(up.url);
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`);
rs.on('end', function () {
console.log(filename);
});
ws.on('error', function (err) {
console.error('cannot send file ' + err);
});
rs.pipe(ws);
});
Typically, file uploads work through multipart/form-data, but your code doesn't suggest that being used here. If it is, the code would become something like this:
const ws = request.post(url + "upload?token=" + `${user.token}&key=${user.key}&filename=${filename}`, {
formData : {
the_file : rs
}
});
// no `rs.pipe(ws)`
answered Nov 22 at 15:04
robertklep
134k17231240
answered Nov 22 at 15:04
robertklep
134k17231240
answered Nov 22 at 15:04
robertklep
134k17231240
answered Nov 22 at 15:04
robertklep
134k17231240
134k17231240
that works thank you!
– Gido Selten
Nov 22 at 15:12
add a comment |
that works thank you!
– Gido Selten
Nov 22 at 15:12
that works thank you!
– Gido Selten
Nov 22 at 15:12
that works thank you!
– Gido Selten
Nov 22 at 15:12
add a comment |
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What is
up.url?fs.createReadStreamonly works for (local) files.– robertklep
Nov 22 at 14:11
up.url is the url I'm using it is returned somewhere, the fs.createReadstream is only used as that is the way I found out how to do post requests, but if there is an other way, that would be possible too, but I have no idea what I could do
– Gido Selten
Nov 22 at 14:25
fs.createReadStreamis meant to read files, not to make POST requests, that's what therequest.post()is for. Do you mean that you want to download a file fromup.urland subsequently upload that file tourl +"upload?..."?– robertklep
Nov 22 at 14:27
yes, that's what I want
– Gido Selten
Nov 22 at 14:49