show error message when the course is already register in php

Question

i want this code to show me a error message when student already register a course

 $reds = mysqli_query($db,"SELECT count(*) FROM `course` " );
    $row = mysqli_fetch_row($reds);
    $num = $row[0];

echo $num>0 ; die();
   var_dump($num) ;die();
    if ($num>0){
   // echo "record already exits";
}

else {


    $sd = "INSERT INTO `users_course`(users_id,course_id) VALUES ( ' {$_SESSION['users_id']}', '$course_id')";
    $r = mysqli_query($db, $sd);
        if ($r) {

        $m = ' register';
        header('location: registercourse.php'); //redirect to index page after inserting

please provide the proper question, enough explanation is required in order to get a proper answer. — Prateik Darji, Dec 05 '17 at 09:43

score 0 · Answer 1 · answered Dec 05 '17 at 09:42

0

You just need to pass users_id in your query to check if a particular student (user id) has registered for the course or not.

$reds = mysqli_query($db,"SELECT count(*) FROM `users_course` where users_id = `".$_SESSION['users_id']."` and course_id='".$course_id."'" );

Because with current SELECT query, you are checking only if there is any course present in the course table. So as there would be some rows, it is giving your error message that course is already there.

Hope this is the clear explaination for you.

answered Dec 05 '17 at 09:42

Himanshu Upadhyay

6,558
1
20
33

Would be better if you would use prepared statements and bind variables. – Nigel Ren Dec 05 '17 at 09:47
thank you very very much – dannycool Dec 05 '17 at 10:07
Happy to help @dannycool – Himanshu Upadhyay Dec 05 '17 at 10:18

score 0 · Answer 2 · answered Dec 05 '17 at 09:42

0

I don't know what is the structure of your database but i think you have to change this query 'SELECT count() FROM course ' to "SELECT count() FROM users_course WHERE course_id = '{$_SESSION['users_id']}' AND user_id = '$course_id'"

answered Dec 05 '17 at 09:42

Momen Shaker

141
1
9

Alive to die - Anant · Accepted Answer · 2017-12-05T10:20:54.003

Do like below:-

<?php
    session_start(); //start session to use session variable
    $user_id = $_SESSION['users_id'];
    $reds = mysqli_query($db,"SELECT * FROM `users_course` WHERE users_id = $user_id AND course_id= $course_id " );

    if (mysqli_num_rows($reds)>0){
        echo "record already exits";exit;
}else {
    $sd = "INSERT INTO `users_course`(users_id,course_id) VALUES ( $user_id, $course_id)";
    $r = mysqli_query($db, $sd);
    if ($r) {
        $m = ' register';
        header('location: registercourse.php'); //redirect to index page after inserting
    }else{
        echo "error:-".mysqli_error($db);
    }
}

Note:- May be above code worked for you currently, but it is wide open for SQL Injection, so try to read prepared statements and use them.

mysqli.prepare

So check this for an example:-

<?php
    session_start(); //start session to use session variable
    $user_id = $_SESSION['users_id'];
    $stmt = mysqli_prepare($db, "SELECT * FROM `users_course` WHERE users_id = ? AND course_id= ?");
    mysqli_stmt_bind_param($stmt, 'ii', $user_id, $course_id);

    /* execute query */
    mysqli_stmt_execute($stmt);

    /* store result */
    mysqli_stmt_store_result($stmt);

    if(mysqli_stmt_num_rows($stmt)>0){
        /* close statement */
        mysqli_stmt_close($stmt);
        echo "record already exits";exit;
    }else{
        $stmt = mysqli_prepare($db,"INSERT INTO `users_course`(users_id,course_id) VALUES (?, ?)";
        mysqli_stmt_bind_param($stmt, 'ii', $user_id, $course_id);
        if ( mysqli_stmt_execute($stmt)) {
            $m = ' register';
            header('location: registercourse.php'); //redirect to index page after inserting
        }else{
            echo "error:-".mysqli_error($db);
        }
    }

/* close connection */
mysqli_close($db);
?>

@dannycool glad to help you. Please read my note section carefully and try the second solution once.Thanks — Alive to die - Anant, Dec 05 '17 at 10:20
prevent your code from `SQL INJECTION`. search `SQL INJECTION` on google and read about it — Alive to die - Anant, Dec 05 '17 at 11:13
@dannycool please ask a new question with your code effort as well as error statement. You will get answer for-sure — Alive to die - Anant, Dec 05 '17 at 11:41

show error message when the course is already register in php

3 Answers3