Why this ajax in registration form won't read php

Question

I have issue with ajax registration form. I've tried many options but I cannot find where I made a mistake. The problem is with email. Ajax cannot read if email exist (check-email.php). Everything else works. I used test@ok.com to test it.

table `users`

| id | email       |
|  1 | test@ok.com |

HTML

<script src="https://static.zrcdn.net/js/fw/jquery/1.7.2/jquery-1.7.2.min.js"></script>
<script>
  function checkStepOne() {
    jQuery("#owner_email").parents("li").find("label.error").hide();
    jQuery("#owner_password").parents("li").find("label.error").hide();
    jQuery("#owner_email").parents("li").find("span.check-1-r").removeClass('chkValid');
    jQuery("#owner_password").parents("li").find("span.check-2-r").removeClass('chkValid');

    var owner_email = jQuery("#owner_email").val();
    var owner_password = jQuery("#owner_password").val();

    var chkemail = 1;
    var chkpass = 1;
    var reg = /^([A-Za-z0-9_\-\.])+\@([A-Za-z0-9_\-\.])+\.([A-Za-z]{2,4})$/;

    $.ajax({
      type: "POST",
      url: "check-email.php",
      data: {
        owner_email,
      },
      success: function(responseEmail) {
        if (responseEmail == 0) {
          jQuery("#owner_email").parents("li").find("#e_exist").show();
          chkemail = 0;
        } else {
          chkemail = 1;
        }
      }
    });

    if (owner_email == "" || reg.test(owner_email) == false) {
      chkemail = 0;
    }

    if (chkemail == 1) {
      jQuery("#owner_email").parents("li").find("span.check-1-r").addClass('chkValid');
    } else {
      jQuery("#owner_email").parents("li").find("#e_avaib").show();
    }

    if (owner_password.length < 5 || owner_password == "") {
      chkpass = 0;
    }

    if (chkpass == 1) {
      jQuery("#owner_password").parents("li").find("span.check-2-r").addClass('chkValid');
    } else {
      jQuery("#owner_password").parents("li").find("label.error").show();
    }

    if (chkpass == 1 && chkemail == 1) {
      jQuery(".step-1").hide();
      jQuery(".step-2").addClass('chkValid').show();
    }
    return false;
  }
</script>


<form method="post" id="frmEmailSignup" onsubmit="return submitResform();">
  <ul>
    <li>
      <label class="side-label" for="owner[email]">Email</label>
      <input type="email" class="email required" id="owner_email" name="owner[email]" placeholder="Email" tabindex="0" onkeyup="return forceLower(this);" style="margin-bottom: 6px;"/>
      <label class="error" id="e_exist" style="margin-top: -9px;">Email exist.</label>
      <label class="error" id="e_avaib" style="margin-top: -9px;">Wrong email.</label>
      <span class="check-1-r"></span>
    </li>
    <li class="password-row">
      <label class="side-label" for="owner[password]">Password</label>
      <input type="password" class="password required" id="owner_password" name="owner[password]" placeholder="Password" tabindex="0"/>
      <label class="error">Wrong password.</label>
      <span class="check-2-r"></span>
    </li>
    <li class="full">
      <button type="button" class="button cta btn-disabled next next-one" tabindex="0" data-plan="register" onclick="checkStepOne();">Sign up</button>
    </li>
  </ul>
</form>

check-email.php

<?php
$email = $_POST['owner[email]'];
$query = mysql_query("SELECT email FROM users WHERE email = '$email'");
if (mysql_num_rows($query) == 1) {
  echo '0';
} else {
  echo '1';
}
?>

Because you're passing owner_email and checking owner[email] — Saad Suri, Jul 17 '18 at 10:52
`$_POST['owner[email]'];` is never set, should be `$_POST['owner_email']` — brombeer, Jul 17 '18 at 10:53
Also please note that `mysql_*` functions are deprecated. Use `mysqli` or `PDO` instead — brombeer, Jul 17 '18 at 10:54

score 1 · Accepted Answer · answered Jul 17 '18 at 10:58

1

You should pass data to AJAX as

data: {
    owner_email: owner_email
}

and in PHP get it as

$_POST['owner_email']

answered Jul 17 '18 at 10:58

Daniel Vítek

185
12

You're right. Thanks! – gdfhfghfg Jul 17 '18 at 11:08

score 0 · Answer 2 · answered Jul 17 '18 at 11:03

0

you have pass data in ajax its wrong. you have used this.

data: {owner_email}

it should be like this

data :{email:owner_email}

in check-email.php file get email like this

$email = $_POST['email'];

answered Jul 17 '18 at 11:03

ashish kumar

172
7

Why this ajax in registration form won't read php

2 Answers2