tommie20 Posted November 18, 2011 Share Posted November 18, 2011 Hi guys, I have a login page where the user can login, the input is being send to php where it checks if the input is in the mysql database. If so i would like to send back the username that is typed in to autofill it in a form completed later on. In code: $(document).ready(function(){ $("#username").focus(); $("#submit").click(function(){ $.ajax({ url: "http://serverip/session.php", cache: false, type: "POST", data:{ username: $("#username").val(), password: $("#password").val() }, success: function(data){ if (data == '1') window.location.replace("home.html"); else{ alert("error"); } } }); }); }); PHP: <?php session_start(); if(!isset($_SESSION['username'])){ if(isset($_POST['username'])){ $db = new mysqli("localhost", "root", "", "gegevens"); $stmt = $db->prepare("SELECT username FROM members WHERE username = ? AND password = ?"); $stmt->bind_param('ss', htmlentities($_POST['username'], ENT_QUOTES), htmlentities($_POST['password'], ENT_QUOTES)); $stmt->execute(); $stmt->store_result(); if($stmt->num_rows() == 1){ $stmt->bind_result($username); $stmt->fetch(); $_SESSION['username'] = $username; echo "1"; exit; } }else{ echo "0"; } } ?> I read in other posts that i could use datatype json to send multiple variables back. But i couldn't get it to work properly. I would like to send back the username typed in but the '1' has to be send back too. Thanks, Tom Link to comment https://forums.phpfreaks.com/topic/251374-sending-back-variable-to-ajax/ Share on other sites More sharing options...
trq Posted November 18, 2011 Share Posted November 18, 2011 I read in other posts that i could use datatype json to send multiple variables back. But i couldn't get it to work properly. Why not post the code that isn't working then? It's really as simple as: echo json_encode(array('username' => $username, 'status' => true)); In your php, then something like.... success: function(data) { if (data.status == 1) { console.log('You have logged in as ' + data.username); } else { console.log('Login failed'); } } in you Javascript. Link to comment https://forums.phpfreaks.com/topic/251374-sending-back-variable-to-ajax/#findComment-1289293 Share on other sites More sharing options...
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