php stop form from posting - php

Question

Basically i have a form where a studentID is inputted, i then want to check id the inputted studentID is in the database, if it is post the form to the next page. If not then display an error on the page where you input studentID
Don't really know where to start
Cheers

Answer 1

is this what you want?
<form id = "form" action = "./?page=markandfeedback" method = "post">
<br>
Mark for:
<INPUT id="stud" onkeypress="return isNumberKey(event)" type="text" name="stud" value="Enter Student Number">
<input type="submit" value = 'Continue'>
<?
$studID = $_POST['stud'];
$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 $_SESSION['module'];
if ($result == NULL) { // nothing found
echo "the student id you entered is not in the database";
}
else {
$_SESSION['student'] = $studID;
Header("Location: http://www.whereever.com/"); // send the browser where you want
exit();
}
?>
EDIT:
I went over the other answers. I assume you check for mysql injection properly. I recommend implementing AJAX AFTER everything works and is secure. The idea behind my solution was to solve the problem as simple as possible. If you want to make something fancy out of it you could:
generate the whole form via php and tell the user in the input field, that the id wasn't found
tell your Javascript to present the information in some fancy way
Use AJAX. Everybody loves forms with AJAX.
You could, as suggested, assume that the user entered a valid id. You would check on the "whereever" page wether the id is actually valid. If it weren't, you would simply send the user back to the form and tell the php to output an error message (maybe via get). This possibility is not usual, I am not sure if it has any advantages.
the mysql_num_rows hint is nice, too, if you don't want any data from the user. I thought you wanted to do something with the data because of the SELECT *.

Answer 2

Make a seperate controller that does the checking of the username.
Use ajax to check if user input is valid or not.
So you'll have something like this:
<input id="stud" onchange="checkStudentId(this)" />
<script>
function checkStudentId(inputElement) {
var id = inputElement.value();
$.ajax({
url: "test.html",
context: {id:id}
}).done(function() {
// Check the return result
});
}
</script>
Here is a reference to jquery ajax
http://api.jquery.com/jQuery.ajax/

Answer 3

You actually have to connect to the server in some fashion to figure out of the student exists. What you'd normally do in this situation is submit the form to the server and do validation server-side. If the student exists, you return the "next" page. If the student doesn't exist, then you return (or redirect to using a Location header) the same form again with an error message.
Another popular method would be to use an AJAX request to check asynchronously (which I see many other people are recommending). I'd only recommend this way if you're actually doing validation right as they've finished entering the student id and are showing an error message in real-time, effectively. In this way, AJAX is a nice-to-have to provide quick user feedback, but not a real solution. Keep in mind that regardless of this, you need to check for and handle this when the form is submitted anyway, or at the least, consider what will happen when the form is submitted with an invalid id.
People can bypass this check (EVERY request from the client side is considered hostile, you can't implicitly trust anything)
Another user may have deleted the student ID between the time the check was done and the form was submitted
There could be an error in your code that causes validation to falsely pass or not to recognize a negative response
Doing AJAX onsubmit makes no sense, because effectively you're doubling the amount of work by making the server handle two separate requests in a row. It's simply the wrong answer to the problem.
The biggest trouble with this implementation is the PHP code can quickly get quite hairy and hard to follow as you have everything mixed together.
This is where you probably start to tip over using PHP like a templating language (mixed php code and html markup) and start getting into using a framework where your views (the HTML) are decoupled from your PHP code (if you're using the very-populate MVC pattern, this code is called your controller -- precisely because it controls how the server responds). This is how any professional developer will work. Kohana, CakePHP, and Zend are all examples of fairly popular MVC frameworks, all of which are used professionally.

Answer 4

You can do this in two different ways
AJAX - make ajax call to your server and check the ID if its exist display the error else go to the next page
PHP - put a hidden input in your form and make the action of the form to the same page and check everything their and keep the values of the input fields is the $_POST['field_name'];
And you can make the action into another page and return back variable or make a session to hold the error message

Answer 5

Try this:
<?
if(isset($_POST['stud'])){
$studID = $_POST['stud'];
$module2 = $_SESSION['module'];
$ex = $_POST['exer'];
$studerr = array();
$host="hostname";//your db host
$user="user";//your db user
$pass="pass";//your db pass
$conn=mysql_connect($host,$user,$pass);
$sql = 'SELECT * FROM `student`, `modules` WHERE `studentID` = '.$studID.' AND `moduleCode` = '.$_SESSION['module'];
$result = mysql_query ($sql,$conn);
if(mysql_num_rows($result)>0){//the id was found in the DB, do whatever here...
echo $_SESSION['module'];
$_SESSION['student'] = $studID;
Header("Location: http://www.whereever.com/");//redirect to wherever
$error=false;
}
else{//id was not found
$error=true;}
}//end of isset
?>
<? if($error===true){?> <div> The id was not found.... </div> <?}?>
<form id = "form" action = "<? echo $_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI']; ?>" method = "post">
<br>
Mark for:
<INPUT id="stud" onkeypress="return isNumberKey(event)" type="text" name="stud" value="Enter Student Number">
<input type="submit" value = 'Continue'>
So what this does is: When the user hits submit, conects to the DB, and checks if the ID exists...if it does, then it redirects it to wherever.com (see comments) and if it don't an error messege will show up. Be sure to change the db variable values to your own ($host, $user, $pass).