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

i trying make simple booking form. 1st showing available cities , packages in page database when user clicks on book button in page, page redirect action page selected form/package should added booking table automatically , send confirmation mail.

the following code generates available packages database:

include 'inc/db_connect.php';

//working connection db

$sql="select pkg, fcity, dcity, price, details tour_package fcity='".$fcity_search."'  order 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 '<a href="#" class="booking-btn" name="bb'.$book_btn_id++.'"> book </a>';         echo '<input type="submit" value="book" class="booking-btn" name="bb'.$book_btn_id++.'">';         echo '<br><br/><hr/>';                                   $i++;     echo '</form>'; } 

i've set dynamic name book button.

book_tour_action.php:

echo $num_var_to_db; $sql = "insert booking (num_var_to_db) values ('$num_var_to_db')"; 

but says undefined variable when book button clicked.

so means data not passed book_tour_action.php guess. how can make work?

you need keep $num_var_to_db value in hidden input field in form value. try 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 '<a href="#" class="booking-btn" name="bb'.$book_btn_id++.'"> book </a>';      echo '<input type="submit" value="book" class="booking-btn" name="bb'.$book_btn_id++.'">';       echo '<br><br/><hr/>';                                  $i++;  echo '</form>'; 

and in php page can following value:

$num_var_to_db = $_request['num_var_to_db'];