Why is my ajax code not actually sending?

Question

I have my ajax code that I'm trying to post from the values of textbox's. I want to send the info to my php page to where it does a query.

    function addUser() {
        myOutput = document.getElementById('add_user_result');
        var member_username = $('#username_name').val();
        var member_email    = $('#email_name').val();
        var member_cpukey   = $('#cpukey_name').val();
    
        if(member_username != "" & member_email != "" & member_cpukey != "") {
            $.ajax({
                type: "POST",
                url: 'includes/ajax_data_add_member.php',
                data: { username: member_username, email: member_email, cpukey: member_cpukey },
                success: function(response) {
                    $("#add_user_result").show();
                    $('#add_user_result').fadeOut(3000).html(response);
                    header('Location: admin_members.php');
                },
                error: function() {
                    $("#add_user_result").show();
                    $('#add_user_result').fadeOut(3000).html(response);
                    header('Location: admin_members.php');
                }
            });
        } else {
            $("#add_user_result").show();
            $('#add_user_result').fadeOut(3000);
            myOutput.innerHTML = "<font style='color: red;'>You must fill in all the blanks.</font>";
        }
        return false;
    }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="row">
  <div class="col-lg-12">
    <div class="p-20">
      <div class="card-box" style="background: #eeeeee; padding: 10px; border: 1px solid #E3E3E3; border-radius: 0px;">
        <div class="form-inline">
          <div class="form-group" style="padding-left: 4%;">
            <div class="form-group m-r-12">
              <label class="col-sm-12 control-label">Add User</label>
            </div>
            <input type="text" name="username_name" id="username_name" placeholder="Username" class="form-control" required/>
            <input type="text" name="email_name" id="email_name" placeholder="Email" class="form-control" required/>
            <input type="text" name="cpukey_name" id="cpukey_name" placeholder="CPUKey" class="form-control" required/>
            <button onclick="addUser()" class="btn btn-success"><i class="fa fa-user-plus"></i> Add User</button>
            <select class="form-control" id="selection_value" name="selection_value">
              <option value="account_credits">Account Credits</option>
              <option value="free_gifted_credits">Free Gifted Credits</option>
              <option value="time">Member Server Days</option>
            </select>
            <input type="text" name="add_to_all_value" id="add_to_all_value" onkeypress="return isNumberKey(event)" placeholder="Value to add to current" class="form-control" required/>
            <button id="button1" onclick="add_to_all()" class="btn btn-primary"><i class="fa fa-user-plus"></i> Add To All Users</button>
          </div>
        </div>
      </div>
      <div id="add_user_result" class="add_user_result"></div>
    </div>
  </div>
</div>

My PHP code

if(!empty($_POST) && isset($_POST)) {
$username           = mysqli_real_escape_string($con, $_POST['username']);
$email              = mysqli_real_escape_string($con, $_POST['email']);
$cpukey             = mysqli_real_escape_string($con, $_POST['cpukey']);
$default_password   = grab_default_user_password();

$insert_query = mysqli_query($con, "INSERT INTO users SET username = '$username', password = '$default_password',  email = '$email', cpukey='$cpukey'");
if($insert_query) {
    echo '<font style="color: green;">Successfully Inserted user <b>'.$username.'</b></font>';
    echo '<META HTTP-EQUIV="Refresh" CONTENT="3; URL=admin_members.php">';
}
else {
    echo '<font style="color: red;">Failed to insert user <b>'.$username.'</b></font>';
    echo '<META HTTP-EQUIV="Refresh" CONTENT="3; URL=admin_members.php">';
}
}

Does anyone know why when I click on and type in each of the textbox's, it says " You must fill in all the blanks. " even though none of them are equal to '' (blank). And then, when I refresh, it sends me back to the index.php (signs my session out).

@PhiterFernandes yes I did mean &&, I changed my javascript but it still gives me the message of all fields need to be entered. — BenZa Music, Dec 14 '16 at 16:21

score 1 · Answer 1 · edited May 23 '17 at 12:09

1

Using && instead of & will cause the conditional to work properly.

See this answer for more details on why.

Also, you didn't define response in your ajax's error function, it will throw errors.

edited May 23 '17 at 12:09

Community

1
1

answered Dec 13 '16 at 23:55

Phiter

14,570
14
50
84

while && vs & is something that the OP should be doing, it seems the "logic" would actually be correct – Jaromanda X Dec 13 '16 at 23:57
Yes, the logic is correct. I debugged his code and the reason why they didn't work became clear. – Phiter Dec 13 '16 at 23:58
@PhiterFernandes I'm now using this : ` if(member_username != "" && member_email != "" && member_cpukey != "")` but it's still saying the message "You must fill in all the blanks." – BenZa Music Dec 14 '16 at 16:18

score 0 · Answer 2 · answered Dec 14 '16 at 00:05

0

first of all , you need to put double "&" to mean the "and" in JavaScript so

if(member_username != "" & member_email != "" & member_cpukey != "") {

so let's suppose now that the request had been well sent to the server side , the request that you're setting there will not work cause you are having a big syntax error , you can not say in SQL INSERT INTO users SET usernam the correct form is to write it as this :

"INSERT INTO users VALUES('$username','$default_password','$email','$cpukey')"

and another remark is that "header('Location: admin_members.php');" is a PHP code so it will not be interpreted on your JS code so for that i suggest you to use the window.location.href = "admin_members.php"; i hope it helped.

answered Dec 14 '16 at 00:05

PacMan

1,358
2
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In mysql, you can use SET for INSERTS. It's an mysql extension. [Check this](http://stackoverflow.com/questions/861722/mysql-insert-into-table-values-vs-insert-into-table-set). – Phiter Dec 14 '16 at 10:04
`"INSERT INTO users VALUES('$username','$default_password','$email','$cpukey')" ` how is this specifying the column to enter it into? – BenZa Music Dec 14 '16 at 16:22
it will do the insert asynchronously to the order of your fields in the database – PacMan Dec 14 '16 at 20:53

score 0 · Answer 3 · answered Dec 14 '16 at 20:04

I found out the reason it was not working correctly.

I was calling my my database file too late in my php file. My new working code is :

<?php
session_start();
include_once('../configuration/db.php');

function user_session_timeout_manual() {
global $con;
    $grab_user_timeout_sql = mysqli_query($con, "SELECT user_session_timeout FROM sitesettings");
    $row = mysqli_fetch_array($grab_user_timeout_sql);
    return $row['user_session_timeout'];
}

function grab_default_user_password() {
global $con;
    $grab_default_user_password_sql = mysqli_query($con, "SELECT default_user_insert_password FROM sitesettings");
    $row = mysqli_fetch_array($grab_default_user_password_sql);
    return $row['default_user_insert_password'];
}

if(!isset($_SESSION['username']) || time() - $_SESSION['login_time'] > user_session_timeout_manual()) {
    unset($_SESSION['login']);
    unset($_SESSION['username']);
    session_destroy();
    header("Location: index.php");
}

else {
    $_SESSION['login_time'] = time();

if(!isset($_SESSION['login']) || $_SESSION['login'] !== true) {
    unset($_SESSION['login']);
    unset($_SESSION['username']);
    session_destroy();
    header('Location: index.php');
}
else {

if(!empty($_POST) && isset($_POST)) {
    $username_grab          = mysqli_real_escape_string($con, $_POST['username']);
    $email_grab             = mysqli_real_escape_string($con, $_POST['email']);
    $cpukey_grab            = mysqli_real_escape_string($con, $_POST['cpukey']);
    $default_password_grab  = grab_default_user_password();

    $insert_query = mysqli_query($con, "INSERT INTO users SET username = '$username_grab', password = '$default_password_grab',  email = '$email_grab', cpukey='$cpukey_grab'");
    if($insert_query) {
        echo '<font style="color: green;">Successfully Inserted user <b>'.$username_grab.'</b></font>';
        echo '<META HTTP-EQUIV="Refresh" CONTENT="3; URL=admin_members.php">';
    }
    else {
        echo '<font style="color: red;">Failed to insert user <b>'.$username_grab.'</b></font>';
        echo '<META HTTP-EQUIV="Refresh" CONTENT="3; URL=admin_members.php">';
    }
}

}}

?>

Why is my ajax code not actually sending?

3 Answers3