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POJQuery with PHP, using $.ajax
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copied!<p>I've been struggling for the entire day, I can't seem to get it right to send through a JSON object to a PHP file using the $.ajax function. I eventually got it right to send it through but now it refuses to decode.</p> <p>//JQuery File</p> <pre><code> var user = { "email" : document.getElementById('email1').value, "password" : document.getElementById('pwd1').value, "fname" : document.getElementById('firstname').value, "lname" : document.getElementById('lastname').value, "gender" : Validator.getGender(), "dob" : document.getElementById('dob').value }; JSON.stringify(user); user = {json:user}; $.ajax({ type: "POST", url: "register.php", dataType: 'json', data: user, success: function(result) { alert("It worked :D"); alert(result); }, failure: function() { alert('whoops'); } }); </code></pre> <p>//PHP File</p> <pre><code><?php $json = $_REQUEST['json']; $json = stripslashes($json); $jsonobj = json_decode($json); $fname = $jsonobj->fname; $lname = $jsonobj->lname; $password = $jsonobj->password; $email = $jsonobj->email; $gender = $jsonobj->gender; $dob = $jsonobj->dob; echo $gender; </code></pre> <p>?></p> <p>Am I doing something completely silly ? I'm just echoing the gender variable back for now but am actually going to use this to enter it into a database.</p> <p>Thanks in advance :)</p> <p>EDIT:</p> <p>How would it would it work if my php file looked like this:</p> <pre><code> session_start(); $db = mysql_connect("localhost", "root"); if(!$db) { die("DB connection failed: " . mysql_error()); } $db_select = mysql_select_db("tinyspace", $db); if(!$db_select) { die("DB connection failed: " . mysql_error()); } $json_string = $_REQUEST["json"]; $jsonobj = json_decode($_REQUEST["json"]); $fname = $jsonobj -> fname; $lname = $jsonobj -> lname; $pwd = $jsonobj -> password; $email = $jsonobj -> email; $gender = $jsonobj -> gender; $dob = $jsonobj -> dob; $sql("INSERT INTO tinyspace.users (email, password, firstname, lastname, gender, dob) VALUES ('$email','$password','$fname', '$lname','$gender','$dob')"); if (!mysql_query($sql,$db)) { die('Error: ' . mysql_error()); } mysql_close($db); </code></pre>

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