How to implement form inside a tab panel? - php

Question

I have written a php code for a form which is inside a tab panel. But the values are not getting inside database. I have written the php code on the top of the php file.After submitting the form submit it goes to the second tab when the php code is written on top of the file before tag and the values obtained from the form is not getting inserted into the database. Why is it like that ? I dont know how to correct this. Can anyone help me on this ?
php code :
<?php
include("database.php");
if(isset($_REQUEST['save'])){
if($_POST['start']!=''&&$_POST['end']!=''&&$_POST['firstname']!=''&&$_POST['lastname']!=''&&$_POST['age']!='')
{
$start = $_POST['start'];
$end = $_POST['end'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$age = $_POST['age'];
$cur = date("d.m.Y");
$j = 'CUS'.rand(1,1000000);
$sql = "insert into tbl_customer(id,customer_id,firstname,lastname,age,rentalfromdate,rentaltodate,date) values(' ','$j','$firstname','$lastname','$age','$start','$end','$cur')";
mysqli_query($con,$sql) or die(mysqli_error($con));
}
}
?>
tab :
<div class="tab-pane active" role="tabpanel" id="step1">
<div class="step1">
<div class="spec">
<div class="col-md-6">
</div>
<div class="col-md-6">
</div>
</div>
<div class="main_cont">
<div class="main-agileits">
<div class="register-box">
<form name="customer" id="customer" method="post" action="#step1">
<div id="flight-datepicker" class="input-daterange input-group">
<div class="col-md-6 s_pad">
<label>First day of rental</label><span class="fontawesome-calendar"></span>
<input type="text" id="start-date" name="start" placeholder="Select Date" data-date-format="DD, MM d" class="input-sm form-control" /><span class="date-text date-depart"></span>
</div>
<div class="col-md-6 s_pad">
<label>Last day of rental</label><span class="fontawesome-calendar"></span>
<input type="text" id="end-date" name="end" placeholder="Select Date" data-date-format="DD, MM d" class="input-sm form-control"/><span class="date-text date-return"></span>
</div>
</div>
<div>
<div class="form-group col-md-12 s_pad">
<label for="exampleInputEmail1">First Name</label>
<input type="text" class="form-control" id="exampleInputEmail1" name="firstname" placeholder="First Name" />
</div>
<div class="form-group col-md-12 s_pad">
<label for="exampleInputEmail1">Last Name</label>
<input type="text" class="form-control" id="exampleInputEmail1" name="lastname" placeholder="Last Name" />
</div>
<div class="form-group col-md-6 s_pad">
<label for="exampleInputEmail1" class="ctred">Age only if under 18</label>
<input type="text" class="form-control" name="age" id="exampleInputName" placeholder="Age">
</div>
</div>
<div>
<button type="submit" name="save" class="btn btn-primary mag_top2 next-step">Continue</button>
</div>
</form>
</div>
</div>
</div>
</div>
</div>

Answer 1

Do like this. Then you will get the error or success message:-
if (mysqli_query($con, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($con);
}

Answer 2

I hope id field is primary key so you dont require it in your sql query so try the following query.
$sql = "insert into tbl_customer(customer_id,firstname,lastname,age,rentalfromdate,rentaltodate,date) values('$j','$firstname','$lastname','$age','$start','$end','$cur')";