sending a form to different pages depending on values - php

Question

I was wandering how you would go about sending a form to different pages.
If there is an error in the form, stay on the page, showing an error and if there is no error go onto another page
here is my code
<form id = "form" action = "./?page=markandfeedback" method = "post">
<br>
Mark for:
<INPUT id="txtChar" onkeypress="return isNumberKey(event)" type="text" name="txtChar" value="Enter Student Number" style="min-width:165px;">
<input type="button" value = 'Continue' onclick="validate()">
<?
$studID = $_POST['txtChar'];
$module2 = $_SESSION['module'];
$ex = $_POST['exer'];
$studerr = array();
$sql = 'SELECT * FROM `student`, `modules` WHERE `studentID` = '.$studID.' AND `moduleCode` = '.$_SESSION['module'];
$result = mysql_query ($sql);
echo $_SESSION['module'];
if(isset($studID)){
if($result == null){
$studerr[] = "No Student with that ";
print_r($studerr);
}
$_SESSION['student'] = $studID;
}
echo' <SCRIPT TYPE="text/javascript" LANGUAGE="javascript">
function validate() {
if('.$result.' == null)
{
return false
}
else
{
return true;
}
}
</script>';
?>
</form>
cheers

Answer 1

Very briefly (without knowing other details):
if (!empty($errorMessage)) {
echo $errorMessage;
} else {
header("Location: http://domain.com/success.php");
}

Answer 2

Have you searched something about that ???
YOu can use normal php validation for error checking and
if(there is no error )
{
header('location: redirect2page.php')
}
else
{
show errors
}

Answer 3

you have to do simple javascript validation in form submit if error occures it will return false so, thats remain on same page
like
function validate()
{
if(document.getElementById('elementid').value=='')
{
return false;
}
else
{
return true;
}
just call this function on submit button click