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Rayj00

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Posts posted by Rayj00

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    So one thing I forgot....I already had a submit button.  It's used to initialize the broadcaster stream and start the broadcast.

    <button onclick="startPublishing()" class="btn btn-info" disabled
                                            id="start_publish_button">Start Broadcasting</button>

    So how would your <button id="btnsend">Send Stream ID</button>  integrate with what I have?

    Or can I just insert your code after the startPublishing() function?

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    And this is connUpdate.php:

    <?php
    error_reporting(E_ALL);

    ini_set('display_errors', '1');

    $servername = "localhost";
    $username = "root";
    $password = "password";
    $dbname = "WebRTCApp";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $stream1 = $_POST['streamId'];

    $sql = "UPDATE broadcast SET streamID='stream1' WHERE id=1";

    if ($conn->query($sql) === TRUE) {
        echo "Record updated successfully";
    } else {
        echo "Error updating record: " . $conn->error;
    }

    $conn->close();
    ?>

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    This is the $.ajax:

     $.ajax({
                    type: 'POST',
                    url: 'connUpdate.php',
                    data: {streamId : streamId}
                    )}.done(function(){
                    alert('AJAX was successfull!');
                    alert('Data to the server');
                    }).fail(function(){
                    alert('There was some error performing the AJAX call!');
                    });
     

     

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    Here is my connUpdate.php: 

    <?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "WebRTCApp";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$stream1 = $_POST['streamId'];

$sql = "UPDATE broadcast SET streamID='stream1' WHERE id=1";

if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully";
} else {
    echo "Error updating record: " . $conn->error;
}

$conn->close();
?>

    And my javascript: 

     $.ajax({
                type: 'POST',
                url: 'connUpdate.php',
                data: {streamId : streamId}
                )}.done(function(){
                alert('AJAX was successfull!');
                alert('Data to the db was successfull!);
                }).fail(function(){
                alert('There was some error performing the AJAX call!');
                });

    One thing that is really bugging me is I can never see or find any error logging.  And yes, my php.ini file is configured for logging errors with E_ALL.

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    So you have to bear with me as I am not a programmer or developer.  I'm kinda learning as I go.

    But here is the database:

    mysql> SELECT * from broadcast;
    +----+----------+
    | id | streamID |
    +----+----------+
    |  1 | 23456    |
    +----+----------+
    1 row in set (0.00 sec)

    I need to update the streamID using, I assume,  ajax.

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    17 hours ago, benanamen said:

    OP, What is the point of adding another layer (Javascript) just to get a random value? Why not just generate in Php?

    And rather than tell us about your attempted solution to the real problem, how about tell us what the real problem is.

    I already had the javascript code for this so I used it.

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    17 hours ago, gizmola said:

    A few things you may or may not understand that are important:

    HTTP is stateless.  There is a request/response protocol built into HTTP that you need to understand for clarity on many issues involving web applications.

    This is important to you for these reasons:

    1. User A makes HTTP GET request to your page, returns HTML.  Closes connection.
    2. User A's browser now runs your javascript code, generates random var.  
    3. User A (I assume but you did not state this explicitly) makes HTTP GET request to other page.  However, server has no way of knowing that this is still user A.

    FIrst to the cut and dried answer:

    To pass your javascript variable to the database, you need to use AJAX.  The typical way of doing this would be to have it use the POST method.  Your PHP script that will be the target of the Ajax call, will read the value from the $_POST superglobal, so you can write that first.

    Again the problem is that, if this is a multi user application, how do you identify one user from another?  You need some sort of id to tell them apart, and this has to be a candidate key in the database, or you will have no idea in Page 2 etc which row you should be reading back.

    In general, PHP sessions solve many problems for this application, and in an unauthenticated scenario, might remove the need entirely for a database.  You could simply save the javascript variable as a session variable and read that in Page 2 etc.

    Like all things, the devil is in the details.  Oftentimes people are coy about the nature of the requirement, and don't understand that there are caveats and use cases for many features.  Very few of the people who answer the vast majority of questions here have the patience to exhaustively cover every possible scenario and solution in the hopes that one of them will match the problem.   

     

    Here is the ajax in my index.html file: 

    var test = randomvalue;
$.ajax({
                              url: "index.php",
                              method: "POST",
                              data: {"test": test }
                      })

     

    ...and here is the code in the php file: 

    $stream1 = $_POST[test];

$sql = "UPDATE broadcast SET streamID='$test'";
if ($conn->query($sql) === TRUE) {
    echo "Record updated successfully";

    Other code connects to the database where I have created the database/table/field = streamID

    So,  does this look right, functionally?

    Newbee Needs Help With Javascript/PHP/Mysql

    in PHP Coding Help

    Posted

    Ok, I consider myself a dabbler in programming so bear with me.

    What I need to do sounds fairly simple, but I cannot figure it out.

    I have an html page.  Inside is some basic html and some Javascript. 

    The javascript basically generates a random value consisting of numbers and characters in:  var = randomToken.

    The length of the random value generated is 11 characters on chrome and 19 on FF? I have no idea why the discrepancy between browsers, but that's not a problem.  I'll look into that later.

    What I'm having trouble doing is right now,  I need to take that random value  that was generated by a javascript function and save it out to a database, so on the other end, when someone browses

    to another page, I need to retrieve the saved value from the database.   I've looked at many examples but they always include a bunch of stuff I don't need, like tables, multiple stuff in the database, etc.

    All I need to do is be able to store a variable and later read it out of a database.  The database just needs a field to store the random value.  Nothing else is needed in the database (for now anyway).

    I know I need php, but i am super green at using it. 

    I have a mysql database/table built with field:  randomVar

    So in a nutshell: 

    The starting html page needs to:

    • Generate random value in Javascript. Save it in a variable.  Done.
    • Pass it to php Not Done
    • Update the database randomVar with the random value Not Done
    • Zero out the randomVar field in the database at some point. (when originator hits stop button or closes the page.)  Not Done

    Another html page needs to:

    • PHP Read the  randomVar from the database;  if > 0 use it Not Done
    • If  = 0,  echo back some text to the html page....like "Try Later...  Not Done
    • Pass it to javascript Not Done

    I am not looking for the exact code unless you want to supply it?  :)

    Thanks,

    Ray

     

     

     

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