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  1. Hello Guys, I have been going over the book and wanted to do small exercise..., where I have form with three radio buttons and when clicking each one I am alerted (alert() method) with the value of the radio button. HTML markup looks like: <form action="#" method="post" id="theForm"> <fieldset><legend>Select Radio Button</legend> <div> <p> <input type="radio" name="options" id="a1" value="a1"> a1 <input type="radio" name="options" id="b2" value="b2"> b2 <input type="radio" name="options" id="c3" value="c3"> c3 </p> </div> <div> <p id="res"></p> </div> <input type="submit" value="Submit" id="submit"> </fieldset> </form> JavaScript Code: window.onload = function () { 'use strict'; document.getElementsByName('options').onchange = displaySelection; //document.getElementById('a1').onclick = displaySelection; //document.getElementById('b2').onclick = displaySelection; //document.getElementById('c3').onclick = displaySelection; }; function displaySelection() { 'use strict'; var selection = document.getElementsByName('options'); var result; for (var i = 0, count = selection.length; i < count; i++) { if (selection[i].checked) { result = selection[i].value; alert(result); break; } } } It works perfectly when using onclick event with reference to the id (three lines commented at the beginning) but it does not work when I refere to the element by name… with event onchange or onclick. Can you guys explain why is that? Thanks, Tom
  2. Hello, I'm new to this forum thanks to a stupid problem I've been struggling with in the past couple of days. Now please keep in mind that I'm a newbie PHP dev, as in I'm working on a school project and sadly there's nobody who could give me a hand there. I need to warn you that my code uses the mysql functions, not mysqli or whatever else, since I'm using XAMPP and I don't really need to upgrade to the newest version. So, the problem is this: for some reason I can't seem to be able to validate my radio button group or my select field(drop down list). If I don't validate, will my data be sent to the db anyway because that would be the better option. If not, could you please help me out? Here's my code: <?php $query_fill_table = "SELECT `movie_title`, `play_time_1`, `play_time_2`, `play_time_3` FROM `on_screen`"; if($query_fill_table_run = mysql_query($query_fill_table)) { echo '<table border="1"> <tr> <th>Movie Title</th> <th>Show Times</th> <th>Number of Seats</th> </tr>'; while($query_rows = mysql_fetch_assoc($query_fill_table_run)) { echo '<tr>'; echo '<td>'.$query_rows['movie_title'].'</td>'; echo '<td>'; ?> <form action="make_reservation.php" method="POST"> <input type = "radio" name = "showtimes" value = "1"><?php echo $query_rows['play_time_1'];?> <input type = "radio" name = "showtimes" value = "2"><?php echo $query_rows['play_time_2'];?> <input type = "radio" name = "showtimes" value = "3"><?php echo $query_rows['play_time_3'];?> </form> <?php echo '</td>'; echo '<td>'; ?> <form action="make_reservation.php" method="POST"> <select name = "seats"> <option value="0"></option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> </form> <?php echo '</td>'; } echo '</table>'; if(isset($_POST['showtimes'], $_POST['seats'])) { $showtimes = $_POST['showtimes']; $seats = $_POST['seats']; //this is just a check if($showtimes==1) { echo 'ok'; } } else { echo 'not ok.'; } } ?> <form action="make_reservation.php" method="POST"> <input type="submit" value="Submit reservation"> </form> This isn't the complete code, but it is the validation section which is what seems to not be working. I will greatly appreciate any opinions. Thanks! ~Vanya D.
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