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Question

In my code below I am displaying a javascript validation. But when I click on the moduleSubmit button, it does not display anything after if (isset($_POST['moduleSubmit'])) { . My question is that when I click on the button, and the javascript validation has passed, then how do I display the form after the submit?

JavaScript

function validation() {

    var isDataValid = true;

    var courseTextO = document.getElementById("coursesDrop");
    var moduleTextO = document.getElementById("modulesDrop");

    var errModuleMsgO = document.getElementById("moduleAlert");

    if (courseTextO.value == "") {
        $('#targetdiv').hide();
        $('#assessmentForm').hide();
        $('#updateForm').hide();
        $('#submitupdatebtn').hide();
        errModuleMsgO.innerHTML = "Please Select a Course";
        isDataValid = false;
    } else {
        errModuleMsgO.innerHTML = "";
    }

    if(isDataValid){
        $("#myForm").submit();
    }

}​

PHP / HTML

       <?php

// connect to the database
include('connect.php');


/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    die();
}


$sql = "SELECT CourseId, CourseNo, CourseName FROM Course ORDER BY CourseId"; 

$sqlstmt=$mysqli->prepare($sql);

$sqlstmt->execute(); 

$sqlstmt->bind_result($dbCourseId, $dbCourseNo, $dbCourseName);

$courses = array(); // easier if you don't use generic names for data 

$courseHTML = "";  
$courseHTML .= '<select name="courses" id="coursesDrop">' . PHP_EOL; 
$courseHTML .= '<option value="">Please Select</option>' . PHP_EOL;  

$outputcourse = "";

while($sqlstmt->fetch()) 
{

    $course = $dbCourseId;
    $courseno = $dbCourseNo;
    $coursename = $dbCourseName; 

    $courseHTML .= "<option value='" . $course . "'>" . $courseno . " - " . $coursename . "</option>" . PHP_EOL;  

    if (isset($_POST['courses']) && ($_POST['courses'] == $course)) {
        $outputcourse = "<p><strong>Course:</strong> " . $courseno .  " - "  . $coursename . "</p>";
    }

} 

$courseHTML .= '</select>'; 

?>

<form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" onsubmit="return validation();">
    <table>
        <tr>
            <th>Course: <?php echo $courseHTML; ?></th>
        </tr>
    </table>
    <p>
        <input id="moduleSubmit" type="submit" value="Submit Course and Module" name="moduleSubmit" />
    </p>
    <div id="moduleAlert"></div>
    <div id="targetdiv"></div>
</form>


<?php

if (isset($_POST['moduleSubmit'])) {    

    $sessionquery = "
    SELECT SessionId, SessionName, SessionDate, SessionTime, CourseId, SessionActive
    FROM Session
    WHERE (CourseId = ? AND SessionActive = ?)
    ORDER BY SessionName 
    ";

    $active = 1;

    $sessionqrystmt=$mysqli->prepare($sessionquery);
    // You only need to call bind_param once
    $sessionqrystmt->bind_param("si",$course, $active);
    // get result and assign variables (prefix with db)

    $sessionqrystmt->execute(); 

    $sessionqrystmt->bind_result($dbSessionId,$dbSessionName,$dbSessionDate,$dbSessionTime, $dbCourseId, $dbSessionActive);

    $sessionqrystmt->store_result();

    $sessionnum = $sessionqrystmt->num_rows();   

    if($sessionnum == 0) {
        echo "<p><span style='color: red'>Sorry, You have No Assessments under this Module</span></p>";
    } 
    else 
    { 
        echo "";
    }

    ...

}

Answer 1

you must use redirect action in html form .That will redirect you to whereever you want to.. like,

if $valid=$form->validate();
{
  $this->redirect('form url')
}

Answer 2

I am afraid the value of input type=button won't be posted in form. You can use firebug plugin of Firefox or Developer tools in chrome to check the posted values in http request. Or you can also print_r($_POST) to see what values are posted in php side.

Try use a hidden input <input type="hidden" value="1" name="issubmit"/> to check if it's a post.