NodeJS trying to use a URL inside of a fs.createfilestream

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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.

asked Nov 22 at 11:38

Gido Selten

176

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

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

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.

asked Nov 22 at 11:38

Gido Selten

176

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

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

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.

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

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

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

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

add a comment |

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

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

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

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

add a comment |

1 Answer
1

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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

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

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

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

add a comment |

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