如何通过 PHP 更新 MySQL 表上的数据

Question

I can't seem to trigger if(isset($_POST['schedule'])) condition via form input button click. How can I update the $schedule_one variable using the following code?

    if(isset($_POST['login'])){
    $password=$_POST['password'];
    $email=$_POST['email'];
    $ret= mysqli_query($con,"SELECT * FROM maid_marketplace_users WHERE email='$email' and password='$password'");
    $num=mysqli_fetch_array($ret);
    $nomee = $num['full_name'];
    $schedOne = $num['schedule_one'];
    echo "<p align=center>You are entitled to <b>%10 off</b> for the next <b>12 months.</b></p> ";
?>       
     <div class="uk-child-width-expand@s uk-text-center" uk-grid>
           <center> <div class="uk-card uk-card-default uk-card-body">Welcome <b>
            <?php
            echo "<br>";
            if (is_null($schedule_one)){
                echo "You don't have any maid services scheduled";
                if(isset($_POST['schedule']))
                {
                    $id = $_SESSION['id'];
                    $schedule_one = $_POST['schedule_one'];
                    $sql=mysqli_query($con,"UPDATE maid_marketplace_users Set schedule_one='$schedule_one' WHERE id='$id'");
                    echo "<script>alert('$id Succes! your next schedule is on $schedule_one');</script>";

                    
                }
            }
            else
            {
                $schedule_one =  $_POST['schedule_one'];
                echo "Your next schedule is at $schedule_one";
            }
            ?>
            <!--Click here to <a href="logout.php" tite="Logout">Logout. -->
                <form name="schedule" action="" method="post">
                    <div class="uk-width-1-1@s">
                        <div class="uk-inline">
                            <input class="uk-input" name="schedule_one" type="datetime-local">
                        </div>
                    </div>
                    <div class="uk-width-1-1@s">
                        <input type="submit" name="schedule" value="Schedule" >
                    </div>
                </form>                
            </div></center>
     </div>
    <?php
        die();
        }

NOTE: I know you aren't supposed to add password directly into Mysql database, I will encrypt it once it's done.

Answer 1

You can't have the if(isset($_POST['schedule'])) inside of another if(isset($_POST['login'])) conditional statement. You have to move it out of the first condition like this;

if(isset($_POST['login']))
{
 ....
}

and you add the other conditional if statement

if(isset($_POST['schedule']))

{
 ....
}

One has to finish before you start the other if statement.

Answer 2

It's good that you know you're creating security vulnerabilities.

One thing you would need to do is to not make your SQL update logically unreachable.

In this:

                echo "You don't have any maid services scheduled";
                die();
                if(isset($_POST['schedule']))
                {
                    $id = $_SESSION['id'];
                    $schedule_one = $_POST['schedule_one'];
                    $sql=mysqli_query($con,"UPDATE maid_marketplace_users Set schedule_one='$schedule_one' WHERE id='$id'");
                    echo "<script>alert('$id Succes! your next schedule is on $schedule_one');</script>";

                    
                }

the die() will always interrupt your code before it reaches the SQL update.