Button in bootstrap does not function - php

Question

Submit button does not functioning when I click. All of the code seem to have no errors, but it just does not inserted to my database. I am currently using bootstrap. I don't know what is the error I am having.
index.html
<div class="container">
<div class="col-md-5">
<div class="form-area">
<form role="form">
<br style="clear:both">
<h3 style="margin-bottom: 25px; text-align: center;">Schedule Form</h3>
<form name="form2" method="post" action="scheduleform.php">
<div class="form-group">
<input type="text" class="form-control" id="tajuk" name="tajuk" placeholder="Tajuk" required>
</div>
<div class="form-group">
<input type="text" class="form-control" id="tarikh" name="tarikh" placeholder="Tarikh" required>
</div>
<div class="form-group">
<input type="text" class="form-control" id="mobile" name="mobile" placeholder="Mobile Number" required>
</div>
<div class="form-group">
<textarea class="form-control" type="textarea" id="maklumat" name="maklumat" placeholder="Maklumat" maxlength="140" rows="7"></textarea>
<span class="help-block"><p id="characterLeft" class="help-block ">You have reached the limit</p></span>
</div>
<button><input type="submit" name="submit" class="btn btn-primary pull-right">Submit Form</button>
</form></form>
</div>
</div>
</div>
<!-- /#page-content-wrapper -->
scheduleform.php
<?php
$server = "localhost";
$user = "root";
$pass = "";
$dbname = "kajangdb";
//Creating connection for mysqli
$conn = new mysqli($server, $user, $pass, $dbname);
//Checking connection
if($conn->connect_error){
die("Connection failed:" . $conn->connect_error);
}
$tajuk = mysqli_real_escape_string($conn, $_POST['tajuk']);
$tarikh = mysqli_real_escape_string($conn, $_POST['tarikh']);
$mobile = mysqli_real_escape_string($conn, $_POST['mobile']);
$maklumat = mysqli_real_escape_string($conn, $_POST['maklumat']);
$sql = "INSERT INTO schedule (tajuk, tarikh, mobile, maklumat) VALUES ('$tajuk', '$tarikh', '$mobile', '$maklumat')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>

Answer 1

Issues i found with your code.
1. You don't need to wrap form element with another form element
2.
$sql = "INSERT INTO schedule (tajuk, tarikh, mobile, maklumat) VALUES ('$tajuk', '$tarikh', '$mobile', '$maklumat')";
Here you added extra space after "schedule" and "VALUES" which may created problem. change your code as below.
$sql = "INSERT INTO schedule(tajuk, tarikh, mobile, maklumat) VALUES('$tajuk', '$tarikh', '$mobile', '$maklumat')";

Answer 2

1) Nested forms are definitely error-prone. Use more forms in a page, if
you wish, but never nest them. Here you are using form2 inside
another one. Don't do that, delete the outer one.
2) Your submit button syntax is wrong. Use input OR button.
Instead of:
<button><input type="submit" name="submit" class="btn btn-primary pull-right">Submit Form</button>
use:
<input type="submit" name="submit" value="Submit Form" class="btn btn-primary pull-right" />
or:
<button type="submit" name="submit" class="btn btn-primary pull-right">Submit Form</button>
3) Provide validation on $_POST values. Like:
if (
!isset($_POST['tajuk']) ||
!isset($_POST['tarikh']) ||
!isset($_POST['mobile']) ||
!isset($_POST['maklumat'])
) {
echo 'Error: not all values are valid. Please provide valid values.';
}
$conn = new mysqli($server, $user, $pass, $dbname);
//...
I recommend you to always prepare your sql statements for execution, in order to avoid the SQL injection risks. Use prepare() before querying. See: mysqli::prepare.
Also use exception handling. See please The mysqli_sql_exception class
EDIT 1:
It is not allowed to use paragraphs (<p>) inside spans (<span>). Use other container types, like <div> instead of <span>. So, replace
<span class="help-block">
<p id="characterLeft" class="help-block ">
You have reached the limit
</p>
</span>
with
<div class="help-block">
<p id="characterLeft" class="help-block ">
You have reached the limit
</p>
</div>