PHP Drop Down Menu Populated from sql query that generates a table

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Hi,

I have a drop down list called courses. When the user chooses a course, I should show him information about the teacher that gives this course. So, on change I need to get the value selected, and show him the results generated from an sql query.

This is my php code:

$sql= "SELECT id, course_period_id from schedule WHERE STUDENT_ID='$_SESSION[student_id]'";
$result=mysql_query($sql);

$options="";

while ($row=mysql_fetch_array($result)) {

$id=$row["id"];
$course_period_id=$row["course_period_id"];
$course= DBGet(DBQuery("SELECT title FROM course_periods WHERE course_period_id='$course_period_id'"));
$options.="<OPTION VALUE=\"$course[1]['TITLE']\">".$course[1]['TITLE'].'</option>';
}

echo '';

echo "<SELECT>
<OPTION VALUE=0>Choose
$options
</SELECT>";

echo '';

I created a php file "teachers_info.php" with the following code:

if(!empty($_GET['Course']))
{
$sql="SELECT teacher_id FROM course_periods where title= '$course'";
$teacher_id= DBGet(DBQuery($sql));

$result= DBGet(DBQyery(" SELECT first_name, last_name, phone, email FROM staff WHERE staff_id = '$teacher_id[1]['teacher_id']'"));

echo "
";

echo "";
echo "";
echo "";
echo "";
echo "";
echo "";

echo "

Firstname	Lastname	Phone	E-mail
" . $result[1]['first_name'] . "	" . $result[1]['last_name'] . "	" . $result[1]['phone'] . "	" . $result[1]['email'] . "

";
}

?>

I know I need to use jquery and ajax but I have no idea how to do so. I've watched so many tutorials but still I dont understand how can I apply this to my code. Note that the project I'm working on supports jquery and ajax.

XML

I've asked this question and got the following answer:

$(function(){

$("select").on("change", function(){

var value = $(this).value(); //get the selected id
$.get("requesturl.php", {course: value}, function(){
// do something with the response
}, "json");

})

})

The requesturl.php file would look as follows:

$course = $_GET["course"]
if($course){
//execute Database query and store it in $result
echo json_encode($result);
}

So,should I add this to my code:

echo "<script type = 'text/javascript'>function(){

$("select").on("change", function(){

var value = $(this).value(); //get the selected id
$.get("requesturl.php", {course: value}, function(){
// do something with the response
}, "json");

})

})

</script>

I tried this:

echo "<script type = 'text/javascript'>function populate(course)
    {
    alert(test);
    }</script>";

but nothing changed and in debug it didn't even pass through this.

I think I should add a submit button that gets the value of the chosen option (course) and goes to the Teachers_Info.php file to generat the tables cz I really can't understand jquery right now I'm so confused. Any ideas please??

Thanks :)

Posted 15-Oct-14 0:09am

Member 11154955

Updated 15-Oct-14 0:24am

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Prava-MFS · Answer 1 · 2014-10-17T00:22:00

Hello,

Let's try the below workout for fixing your issue.

1) First, you are getting all the course details, the student acquired.

$sql = "SELECT S.id, S.course_period_id, CP.title
        FROM schedule AS S 
        LEFT JOIN course_periods AS CP
        ON S.course_period_id = CP.course_period_id
        WHERE STUDENT_ID = '$_SESSION[student_id]'";
$result = mysql_query($sql);

You got the resultant array now.

2) Need to loop into it, to get the data

echo '<select onchange="javascript: showTeacherInfo();" id="sel_courseList">';
echo '<option value="0">Choose Course</option>';

while ($row = mysql_fetch_array($result)) {
    echo '<option value="' . $row['title'] . '">' . $row['title'] . '</option>';
}

echo '</select>';

Drop down is ready now.

3) Now, time to write javascript method, where we will do the AJAX call to get teacher's info.

function showTeacherInfo() {
    var courseTitle = $('#sel_courseList :selected').val();
    var url = 'teachers_info.php';

    $.post(url,
        {
            courseTitle: courseTitle
        },
        function(data) {
            // success message
        }
    );
}

We are done with the AJAX call.

4) Let's go to teachers_info.php file.

if(isset($_POST['courseTitle']) && !empty($_POST['courseTitle'])) {
    $courseTitle = $_POST['courseTitle'];
    $sql = "SELECT S.first_name, S.last_name, S.phone, S.email
            FROM course_periods AS CP
            INNER JOIN staff AS S
            ON CP.teacher_id = S.staff_id
            WHERE CP.title= '$courseTitle'";
    $result = mysql_query($sql);
    $teacher_details = mysql_fetch_row($result);

    // Now, all set and we can get the details of teacher.
    $teacher_first_name = $teacher_details[0];
    $teacher_last_name = $teacher_details[1];
    $teacher_phone = $teacher_details[2];
    $teacher_email = $teacher_details[3];
}

Hopefully, the above code will work for you. Though we have many ways and can say better way to fix your problem, I have chosen the simplest way.

Please let me know, if you are still finding any issue. Thanks and all the best :)