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so I have a site that can record a video stream from the users webcam and convert it to a blob using getUserMedia()
. I now want to send this blob to the server so the video can be saved and processed. I am having trobule sending a blob file via ajax, I have tried using both the formData + XMLHttpRequest() method as well as pure ajax. It is imperetive for my application that users can record video and immediately send this video to my server.
Any help would be greatly appreciated.
so HTML:
<div class="demo">
<video id="preview" autoplay width="400" height="300"></video>
<video id="recording" width="400" height="300" style="display:none;" controls></video>
<div class="progress">
<div class="progress-bar"></div>
<span>01:00</span>
</div>
<button class="record">Record</button>
<button class="upload">Upload</button>
</div>
Relevant JS functions:
function captureVideo () {
const preview = document.querySelector('video#preview');
const recording = document.querySelector('video#recording');
navigator.mediaDevices.getUserMedia({video: true}).then((stream) => {
preview.srcObject = stream;
preview.captureStream = preview.captureStream || preview.mozCaptureStream;
return new Promise(resolve => preview.onplaying = resolve);
}).then(() => {
let recorder = new MediaRecorder(preview.captureStream());
let data = [];
recorder.ondataavailable = event => data.push(event.data);
recorder.start();
log(recorder.state + " for " + (60000/1000) + " seconds...");
let stopped = new Promise((resolve, reject) => {
recorder.onstop = resolve;
recorder.onerror = event => reject(event.name);
});
$('button.stop').click(function () {
recorder.stop();
});
return Promise.all([ stopped ]).then(() => data);
}).then ((recordedChunks) => {
let recordedBlob = new Blob(recordedChunks, {
type: "video/webm"
});
recording.src = URL.createObjectURL(recordedBlob);
$('#preview').hide();
$('#recording').show();
log("Successfully recorded " + recordedBlob.size + " bytes of " +
recordedBlob.type + " media.");
$('button.upload').click(function() {
sendVideoToAPI(recordedBlob);
});
}).catch(log);
}
function sendVideoToAPI (blob) {
let fd = new FormData();
let file = new File([blob], 'recording');
fd.append('data', file);
console.log(fd); // test to see if appending form data would work, it didn't this is pletely empty.
let form = new FormData();
let request = new XMLHttpRequest();
form.append("file",file);
request.open("POST", "/demo/upload", true);
request.send(form); // hits the route but doesn't send the file
console.log(request.response) // returns nothing
// I have also tried this method which hits the route and gets a response however the file is not present in the request when it hits the server.
// $.ajax({
// url: Routing.generate('upload'),
// data: file,
// contentType: false,
// processData: false,
// error: function (res) {
// console.log(res);
// },
// success: function(res) {
// console.log(res);
// }
// });
}
so I have a site that can record a video stream from the users webcam and convert it to a blob using getUserMedia()
. I now want to send this blob to the server so the video can be saved and processed. I am having trobule sending a blob file via ajax, I have tried using both the formData + XMLHttpRequest() method as well as pure ajax. It is imperetive for my application that users can record video and immediately send this video to my server.
Any help would be greatly appreciated.
so HTML:
<div class="demo">
<video id="preview" autoplay width="400" height="300"></video>
<video id="recording" width="400" height="300" style="display:none;" controls></video>
<div class="progress">
<div class="progress-bar"></div>
<span>01:00</span>
</div>
<button class="record">Record</button>
<button class="upload">Upload</button>
</div>
Relevant JS functions:
function captureVideo () {
const preview = document.querySelector('video#preview');
const recording = document.querySelector('video#recording');
navigator.mediaDevices.getUserMedia({video: true}).then((stream) => {
preview.srcObject = stream;
preview.captureStream = preview.captureStream || preview.mozCaptureStream;
return new Promise(resolve => preview.onplaying = resolve);
}).then(() => {
let recorder = new MediaRecorder(preview.captureStream());
let data = [];
recorder.ondataavailable = event => data.push(event.data);
recorder.start();
log(recorder.state + " for " + (60000/1000) + " seconds...");
let stopped = new Promise((resolve, reject) => {
recorder.onstop = resolve;
recorder.onerror = event => reject(event.name);
});
$('button.stop').click(function () {
recorder.stop();
});
return Promise.all([ stopped ]).then(() => data);
}).then ((recordedChunks) => {
let recordedBlob = new Blob(recordedChunks, {
type: "video/webm"
});
recording.src = URL.createObjectURL(recordedBlob);
$('#preview').hide();
$('#recording').show();
log("Successfully recorded " + recordedBlob.size + " bytes of " +
recordedBlob.type + " media.");
$('button.upload').click(function() {
sendVideoToAPI(recordedBlob);
});
}).catch(log);
}
function sendVideoToAPI (blob) {
let fd = new FormData();
let file = new File([blob], 'recording');
fd.append('data', file);
console.log(fd); // test to see if appending form data would work, it didn't this is pletely empty.
let form = new FormData();
let request = new XMLHttpRequest();
form.append("file",file);
request.open("POST", "/demo/upload", true);
request.send(form); // hits the route but doesn't send the file
console.log(request.response) // returns nothing
// I have also tried this method which hits the route and gets a response however the file is not present in the request when it hits the server.
// $.ajax({
// url: Routing.generate('upload'),
// data: file,
// contentType: false,
// processData: false,
// error: function (res) {
// console.log(res);
// },
// success: function(res) {
// console.log(res);
// }
// });
}
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asked Oct 12, 2018 at 15:01
UUake UpUUake Up
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- 1 I +1 this because of a clear explanation of how this works with FormData and posting capture files. Thanks. – Antony Commented Apr 7, 2020 at 16:01
1 Answer
Reset to default 4You are sending an ajax request you won't have the response immediately after the send() mand. you can access the response value using the request onload event
request.onload = function () {
if (request.readyState === request.DONE) {
if (request.status === 200) {
console.log(request.response);
}
}
};
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