使用ajax将文件数据提交到php文件

Question

I have to submit only two value using ajax call to a php file. The php file will handle data with $_POST[] function. Problem is my ajax code not sending data properly so i am not getting any of result in console.log(result) function. How can i fix that? Any idea?

Jquery:

<script type="text/javascript">
                $("#signup_submit").on( "click", function() {
                  //alert("foo");
                  var _username = "foo";
                  var _email = "test@gmail.com";


                  $.post("check.php", {username: _username, email: _email}, function(result){
                    console.log(result);
                  });



                });
            </script>

PHP:

$username = "";
$email = "";
if ($_POST["username"] && $_POST["email"]) {
    $username = $_POST["username"];
    $email = $_POST["email"];
}

if ($username != "" && $email != "") {
    if (!username_exists($username) && !signup_email($signup_email)) {
        json_encode("good");
    }else{
        json_encode("bad");
    }
}

Answer 1

I've commented you the possible answer but I'll write it here to simplify. Your javascript code seems to be fine but change PHP script to this and try:

$username = "";
$email = "";
$response = ""
if ($_POST["username"] && $_POST["email"]) {
    $username = $_POST["username"];
    $email = $_POST["email"];
}

if ($username != "" && $email != "") {
    if (!username_exists($username) && !signup_email($signup_email)) {
        $response = json_encode("good");
    }else{
        $response = json_encode("bad");
    }
}
echo $response;

echo $response cause if you don't add that line you won't get any result from your ajax request so console.log won't show anything.

Hope this helps you out!

Answer 2

i think this code will work for you perfect.

 $("#signup_submit").on( "click", function() {
  var _username = "foo";
  var _email = "test@gmail.com";

  $.ajax({
    type: 'POST',
    url: 'check.php',
    dataType: 'text',
    data: {
     username: _username,
     email: _email
   },
   success: function(response) {
    console.log(response);
   },
   error: function(err) {
      alert(err);
   }
  });
 });

And like user #molinet writed - in php file do with $response and then echo it.

$username = "";
$email = "";
if ($_POST["username"] && $_POST["email"]) {
 $username = $_POST["username"];
 $email = $_POST["email"];
}

if ($username != "" && $email != "") {
 if (!username_exists($username) && !signup_email($signup_email)) {
    $response = "good";
 }else{
   $response = "bad";
 }
}
echo $response;