How would I retrieve the data on a different page?

Question

I'm Sending this Video Blob to the uploadVideo.php file how would i retrieve it on the uploadvideo.php page?

function postVideoToServer(videoblob) {
        var data = {};
        data.video = videoblob;
        data.metadata = 'test metadata';
        data.action = "upload_video";
        jQuery.post("http://mysite.com/uploadvideo.php", data, onUploadSuccess);
    }

More Details

function postVideoToServer(videoblob) {
    var data = {};
    data.video = videoblob;
    data.metadata = 'test metadata';
    data.action = "upload_video";
    jQuery.post("http://mysite.com/uploadvideo.php", data, onUploadSuccess);
}

And On the uploadvideo.php i have this:

    <?php
    require("connect.php");

    $video = $_POST["video"];
    $up = mysql_query("INSERT INTO video VALUES ('$video')");
    ?>

But it still does not seem to work? I Think i am doing something wrong?

`$_POST['video']`, `$_POST['metadata']`, `$_POST['action']` ? — Brewal, Jun 25 '13 at 09:05
try `$video = addslashes($_POST["video"]);`, as special characters like `'` can be in `$video` — kajacx, Jun 27 '13 at 13:10

score 2 · Answer 1 · answered Jun 25 '13 at 09:07

2

$video = $_POST['video'];
$metadata = $_POST['metadata'];
$action = $_POST['action'];

answered Jun 25 '13 at 09:07

Kautil

1,321
9
13

score 1 · Answer 2 · edited May 23 '17 at 11:50

1

You can use either $_REQUEST["key"]

$video = $_REQUEST["video"];
$metadata = $_REQUEST["metadata"]; //test metadata
$action = $_REQUEST["action"]; //upload_video

or $_POST["key"]

$video = $_POST["video"];
$metadata = $_POST["metadata"]; //test metadata
$action = $_POST["action"]; //upload_video

to retrieve the data sent to the PHP script.

To read a little bit more about different data retrieval techniques and their comparison in PHP refer link.

edited May 23 '17 at 11:50

Community

1
1

answered Jun 25 '13 at 09:14

Faisal Sayed

783
4
12

I Put this into the uploadvideo.php but it does not seen to work maybe its the script for the getusermedia? require("connect.php"); $video = $_POST["video"]; $up = mysql_query("INSERT INTO video VALUES ('$video')"); – Alexander Nazarkin Jun 25 '13 at 10:59

score 0 · Answer 3 · answered Jun 25 '13 at 09:08

0

uploadvideo.php:

$video = $_POST["video"];
$metadata = $_POST["metadata"]; //test metadata
$action = $_POST["action"]; //upload_video
...

now you can do whatever you want with these.

answered Jun 25 '13 at 09:08

kajacx

12,361
5
43
70

Hi, I Added A bit more details still does's not seem to work? – Alexander Nazarkin Jun 25 '13 at 23:58

How would I retrieve the data on a different page?

3 Answers3