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Rayj00

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Everything posted by Rayj00

  1. Appreciate your help. I only got more confused with your code...llike how does it call the php without a URL? I'll just keep digging I guess.
  2. 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?
  3. So hold on for a bit. I found something weird going on with my index page.
  4. 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(); ?>
  5. 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!'); });
  6. Sorry...I misunderstood. The html/javascript and $.ajax are all in the index.php file. I have a separate file for connecting and updating the database called connUpdate.php
  7. So, I said everything works WITHOUT the ajax code. How does that equate to my javascript being the culprit?
  8. So, are you saying the code I posted looks ok? Like I said. Everything works as coded...WITHOUT the ajax code. My only problem is with the ajax....which I have been trouble shooting for a few days now.
  9. Note that the above mentioned javascript is included, along with my html in a file called index.php.
  10. Sorry. I originally posted what I tried at that time. What I posted above is what I am trying to get to work right now. I am not getting any messages back. And the database is not updated.
  11. 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.
  12. For now I will only have the one record so id is not used. How do I update the streamId using ajax?
  13. 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.
  14. Right now I have only two records....Id, and streamID I want to update the steamId. So this is only one issue. Worse than this is I cannot seem to get a $.ajax to work! No matter what I try, I cannot see the ajax being executed?
  15. I already had the javascript code for this so I used it.
  16. 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?
  17. 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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