无法访问if语句中的变量

Question

I'm trying to use this code to be able to stop double bookings in my booking system. At the moment when you enter the same time twice then it comes up with this error:

Email is validthis time is already booked
Notice: Undefined variable: sql in C:\xampp\htdocs\book.php on line 44

Warning: mysqli_query(): Empty query in C:\xampp\htdocs\book.php on line 44
Error:

this is my code

<?php

//$error = ""; // Initialize error as blank
$con=mysqli_connect("localhost","","","");


if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

if ($_SERVER["REQUEST_METHOD"] == "POST") 
{
    $email = $_POST['email'];
    $time = $_POST["time"];
    $name = $_POST["name"];
    $surname = $_POST["surname"];
    $date = $_POST["date"];
    $adl1 = $_POST["adl1"];
    $adl2 = $_POST["adl2"];
    $postcode = $_POST["postcode"]; 

    if(!filter_var(($email), FILTER_VALIDATE_EMAIL))
    {
        echo "E-mail is not valid";
    } 
    else 
    {
        echo "Email is valid";

        $result = mysqli_query($con, "SELECT time FROM tbl_booking WHERE time = '$time'") or trigger_error("Query Failed! SQL: $result - Error: ".mysqli_error($con), E_USER_ERROR);
        if(mysqli_num_rows($result) == 0) 
        {
            $sql="INSERT INTO tbl_booking (name, surname, email, date, time, adl1, adl2, postcode) VALUES ('$name','$surname','$email','$date','$time','$adl1','$adl2','$postcode')";

        } 
        else 
        {
            echo("this time is already booked");
        }

        if (!mysqli_query($con, $sql))
        {
            die('Error: ' . mysqli_error($con));
        }

        mysqli_close($con);
    }
}

Basically I think it's trying to access the $sql inside the if statement but I have no idea why it can't. Unless i'm being stupid.

Answer 1

You can rewrite your code as, since in else condition there is no query to the mysqli_query(), that why you got the error,

if(mysqli_num_rows($result) == 0){
     $sql="INSERT INTO tbl_booking (name, surname, email, date, time, adl1, adl2, postcode) VALUES ('$name','$surname','$email','$date','$time','$adl1','$adl2','$postcode')";
     mysqli_query($con, $sql) or die('Error: ' . mysqli_error($con));
 } else {
     echo("this time is already booked");
  }

Answer 2

Set $sql as "global" -> you just need to create it before the if-statement.

Answer 3

Try to declare $sql value on top of your php code your code not execute

 if(mysqli_num_rows($result) == 0) 
        {
            $sql="INSERT INTO tbl_booking (name, surname, email, date, time, adl1, adl2, postcode) VALUES ('$name','$surname','$email','$date','$time','$adl1','$adl2','$postcode')";

        } 

this condition its actually executing your else body that why it generate exception that $sql variable not defined