Twitter Bootstrap Question posted in

Inserting Data mysqli not working – Twitter bootstrap

Here is my HTML with form section… I cant understand where is my mistake

<!DOCTYPE html>
<html>

<head>
  <title></title>
  <link rel="stylesheet" type="text/css" href="loginStyle.css">
  <link href="Bootstrap/css/bootstrap.min.css" rel="stylesheet">
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1 user-scalable=no">
  <link href='https://fonts.googleapis.com/css?family=Jura:600' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" type="text/css" href="Font Awseome Glyphicons/css/font-awesome.min.css">
  <link href='https://fonts.googleapis.com/css?family=Lora:700' rel='stylesheet' type='text/css'>
  <link href='https://fonts.googleapis.com/css?family=Lato' rel='stylesheet' type='text/css'>
  <link rel="stylesheet" href="Owl Carousel/owl-carousel/owl.carousel.css">
  <link rel="stylesheet" href="Owl Carousel/owl-carousel/owl.theme.css">
</head>

<body>


  <div class="container">
    <div class="row">
      <div class="col-md-12" style="margin-top:8%">
        <h2 class="text-center">Be a Part of World</h2>
      </div>
    </div>
    <div class="row">
      <div class="col-md-6 col-md-offset-3">
        <form role="form" id="loginForm" action="regCheck.php" method="POST">
          <div class="form-group">
            <label for="regName">
              <h4>Name</h4>
            </label>
            <input type="text" id="regName" class="form-control" placeholder="Your name here..." name="name">
            <div class="alert"></div>
          </div>
          <div class="form-group">
            <label for="regLogin">
              <h4>Login</h4>
            </label>
            <input type="text" placeholder="Login Here..." id="regLogin" class="form-control" name="login">
            <div class="alert"></div>
          </div>
          <div class="form-group">
            <label for="regPass">
              <h4>Password</h4>
            </label>
            <input type="password" placeholder="Password Here..." id="regPass" class="form-control" name="pass">
            <div class="alert"></div>
          </div>
          <div class="form-group">
            <input type="submit" value="Sign Up" class="btn btn-danger btn-lg" id="regSubmit" name="submit">
          </div>
        </form>
      </div>
    </div>
    <div class="row">
      <div class="col-md-12">
        <h3 class="text-center"><strong>...0r Join Us With</strong></h3>
      </div>
    </div>
    <div class="row">
      <div class="col-md-6 col-md-offset-4">
        <ul class="list-inline">
          <li><a href="facebook.com"><span class="fa fa-facebook-official " aria-hidden="true"></span> Facebook</a>
          </li>
          <li><a href="twitter.com"><span class="fa fa-twitter" aria-hidden="true"></span> Twitter</a>
          </li>
          <li><a href="youtube.com"><span class="fa fa-youtube " aria-hidden="true"></span> Youtube</a>
          </li>
          <li><a href="instagram.com"><span class="fa fa-instagram " aria-hidden="true"></span> Instagram</a>
          </li>
        </ul>
      </div>
    </div>
  </div>

  <script type="text/javascript" src="JQuery/jquery-1.11.3.min.js"></script>
  <script src="Owl Carousel/owl-carousel/owl.carousel.js"></script>
  <script src="Bootstrap/js/bootstrap.min.js"></script>
  <script type="text/javascript" src="script.js"></script>
</body>

</html>

Here is section with checking registration

<?php
$con = mysqli_connect("localhost", "root", "");
if (!$con) {
    die("Failed Databse Connection");
}

$name = $_POST["name"];
$login = $_POST["login"];
$pass = $_POST["pass"];
$query = "INSERT INTO db_users(name, pass,login) VALUES ('$name', '$pass', '$login')";

if (isset($_POST["submit"])) {
    mysqli_query($con, $query);
}
?>
Tags:

5

Answers


  1. 0

    Your mysqli syntax is mixing with simple mysql

    You have to give database name as last argument in mysqli_connect

    $con = mysqli_connect("localhost","root","","databasename");
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  2. 0

    You forgot to select the database

    $con = mysqli_connect("localhost","root","","NameOfYourDatabase");

    PHP manual for mysqli_connect()

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  3. 0

    This is the structure it should be:

    <?php

    $con = mysqli_connect("host","username","password","database name");
    if (!$con) {
        die("Failed Databse Connection");
    }

    $name = $_POST["name"];
    $login = $_POST["login"];
    $pass =  $_POST["pass"];
    $query = "INSERT INTO db_users(name, pass,login) VALUES ('$name', '$pass', '$login')";

        if (isset($_POST["submit"])) {
            mysqli_query($con,$query);
        }

?>

    But with that you would get an error as it will look for non posted keys in the POST array so try this:

    <?php

    $con = mysqli_connect("host","username","password","database name");
    if (!$con) {
        die("Failed Databse Connection");
    }





     if (isset($_POST["submit"])) {
        $name = $_POST["name"];
        $login = $_POST["login"];
        $pass =  $_POST["pass"];
        $query = "INSERT INTO db_users(name, pass,login) VALUES ('$name', '$pass', '$login')";
        mysqli_query($con,$query);
        }

?>
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  4. 0

    Please mention DB name in mysqli_connect function.

    mysqli_connect("localhost", "root", "", "dbName");
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  5. 0

    You can use PHP database mysqli syntax and mention database name also in mysqli_select_db function as given below:

    $connect = mysqli_connect(“database_hostname”,”database_username”,”database_password”);
    mysqli_select_db($connect,”database_name”);

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