how to pass a retried database data to another page? - php

Question

I am trying to make a simple booking form.
1st showing available cities and packages in a page from database
when the user clicks on the BOOK button in the page, the page will redirect to the action page where the selected form/package should be added to the booking table automatically and send a confirmation mail.
The following code generates the available packages from the database:
include 'inc/db_connect.php';
//working connection to the DB
$sql="SELECT pkg, fcity, dcity, price, details FROM tour_package WHERE fcity='".$fcity_search."' ORDER BY fcity ASC";
$sqlresult=mysqli_query($con,$sql);
$afct=mysqli_affected_rows($con);
while($pkg=mysqli_fetch_array($sqlresult,MYSQLI_ASSOC)){
$num[]=$pkg['pkg'];
$fcity[]=$pkg['fcity'];
$dcity[]=$pkg['dcity'];
$price[]=$pkg['price'];
$details[]=$pkg['details'];
$num++;
}
$i=0;
$book_btn_id = 1;
while($i < $afct)
{
echo '<form action="book_tour_action.php">';
$num_var_to_db = $num[$i];
echo '<span class="search-class-row"><div class="search-results-label">Package : </div>'.$num[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">From : </div>'. $fcity[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Destination : </div>' . $dcity[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Price : </div>' .$price[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Details :</div>' . $details[$i].'<br/></span>';
//echo ' BOOK ';
echo '<input type="submit" value="BOOK" class="booking-btn" name="bb'.$book_btn_id++.'">';
echo '<br><br/><hr/>';
$i++;
echo '</form>';
}
I've set a dynamic name to the BOOK button.
book_tour_action.php:
echo $num_var_to_db;
$sql = "INSERT INTO booking (num_var_to_db)
VALUES ('$num_var_to_db')";
But it says UNDEFINED VARIABLE when the BOOK button is clicked.
So that means the data is not passed to book_tour_action.php I guess.
How can I make it work?

Answer 1

You need to keep $num_var_to_db value in hidden input field in the form to get the value. Try do like below :
echo '<form action="book_tour_action.php">';
$num_var_to_db = $num[$i];
echo '<input type="hidden" name="num_var_to_db" value="'.$num_var_to_db.'">'
echo '<span class="search-class-row"><div class="search-results-label">Package : </div>'.$num[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">From : </div>'. $fcity[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Destination : </div>' . $dcity[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Price : </div>' .$price[$i].'<br/></span>';
echo '<span class="search-class-row"><div class="search-results-label">Details :</div>' . $details[$i].'<br/></span>';
//echo ' BOOK ';
echo '<input type="submit" value="BOOK" class="booking-btn" name="bb'.$book_btn_id++.'">';
echo '<br><br/><hr/>';
$i++;
echo '</form>';
and in the php page you can do following to get the value:
$num_var_to_db = $_REQUEST['num_var_to_db'];