Php form data won't insert data on my PhpMyAdmin database - php

Question

I've been trying to resolve this but i can't see the error. please help me. This is my code so far. What should i do? i checked every detail but cant find it. when i click the button, there's no error and it will return in the reservation.php page. I also have a "res_id" in my database which is auto increment. how should i include it in the insert statement?
Php code:
<?
$conn = mysqli_connect("localhost","root","") or die(mysql_error());
mysqli_select_db("hoteldb",$hoteldb) or die("no db found");
if(isset($_POST['confirm'])){
$fname= $_POST['fname'];
$lname= $_POST['lname'];
$contact= $_POST['contact'];
$username= $_POST['username'];
$password= $_POST['password'];
$email= $_POST['email'];
$address= $_POST['address'];
$checkin= $_POST['checkin'];
$checkout= $_POST['checkout'];
$query="INSERT INTO reservation(fname,lname,contact,username,password,email,address,checkin,checkout)
VALUES ('$fname','$lname','$contact','$username','$password','$email','$address','$checkin','$checkout')";
mysql_query($query,$conn);
mysqli_close($conn);
}
?>
Html code:
<form action="reservation.php" method="post">
<div>
<label>Title:</label>
<select name="title">
<option value="ms">Ms.</option>
<option value="mr">Mr.</option>
<option value="mrs">Mrs.</option>
</select>
</div>
<div >
<label>First Name:</label>
<input type="text" name="fname">
</div>
<div>
<label>Last Name:</label>
<input type="text" name="lname">
</div>
<div>
<label>Phone Number:</label>
<input type="text" name="contact">
</div>
<div>
<label>Username:</label>
<input type="text" name="username" >
</div>
<div >
<label>Password:</label>
<input type="password" name="password" >
</div>
<div >
<label>Email Address:</label>
<input type="text" name="email" >
</div>
<hr>
<h5>Address</h5>
<div >
<label>Full Address:</label>
<input type="text" name="address" >
</div>
<hr>
<h5>Date</h5>
<div >
<legend>Check-in</legend>
<input type="date" name="checkin">
<legend>Checkout </legend>
<input type="date" name="checkout">
</div>
<input type="submit" name="confirm">
</form>
or is it because of the date picker? if so, can you please teach me how to use the date picker? its my first time using it. Thank you! :)

Answer 1

try to get error description: (add after $query's definition)
if (!mysqli_query($conn, $query)) {
echo("Error description: " . mysqli_error($conn));
}

Answer 2

Try turning on PHP errors to not get just an empty page.
http://php.net/manual/en/mysqli.query.php
mysql_query($query, $conn);
this should be
mysqli_query($conn, $query);
I also have a "res_id" in my database which is auto increment. how should i include it in the insert statement?
Since it's set to "auto increment", you dont need to do anything with it
Also, try looking at PDO, it's safer and takes less code to use.

Answer 3

When you connect to the database you use mysql, after this you are trying to insert data using mysqli_query, it is impossible. You have to replace
mysql_query($query,$conn);
with
mysqli_query($conn, $query);
so your php will look like this
<?
$conn = mysqli_connect("localhost","root","") or die(mysql_error());
mysqli_select_db("hoteldb",$hoteldb) or die("no db found");
if(isset($_POST['confirm'])){
$fname= $_POST['fname'];
$lname= $_POST['lname'];
$contact= $_POST['contact'];
$username= $_POST['username'];
$password= $_POST['password'];
$email= $_POST['email'];
$address= $_POST['address'];
$checkin= $_POST['checkin'];
$checkout= $_POST['checkout'];
$query="INSERT INTO reservation(fname,lname,contact,username,password,email,address,checkin,checkout)
VALUES ('$fname','$lname','$contact','$username','$password','$email','$address','$checkin','$checkout')";
mysqli_query($conn, $query);
mysqli_close($conn);
}
?>
This is main problem.
You can use this templates any time u need insert something in your database
https://www.w3schools.com/php/php_mysql_insert.asp
If u want to use mysqli go to
Example (MySQLi Procedural) part
But I would recommend to use PDO;

Answer 4

The error in Mysqli_select_db statement.
You need correct your statement.
mysqli_select_db("hoteldb",$hoteldb) or die("no db found");
insted of this
mysqli_select_db($conn,"hoteldb") or die("no db found");