如何正确使用if else语句？

Question

I want check, whether booking id is not in the database as well as it is greater than today. The database checking part working. But this part not going through if the condition is true. I think something wrong with the if else statements.

else if($checkindate > $today)
{
    $bidErr="This booking is not comming today. Please check again";
}

I have include full code here. booking Id bid coming from a form.

<?php
    $today=date("Y-n-j"); 
    echo "<h4>Today is <font color='red'>".$today."</font></h4><br><br>";
    // define variables and set to empty values
    $bidErr =  "";
    $bid =  "";

    if ($_SERVER["REQUEST_METHOD"] == "POST") {
        $flag = 1;
        if (empty($_POST["bid"])) {
            $bidErr = "Booking ID is required.";
            $flag=0;
        } else {
            $bid = test_input($_POST["bid"]);
            // check if name only contains letters and whitespace
            if (!preg_match("/^[0-9]*$/",$bid)) {
                $bidErr = "Only Numbers are allowed"; 
                $flag=0;
            }
        }
        include("connect.php");
        if($flag=="1"){
            $SQL="SELECT guestid,checkindate FROM bookings WHERE bookingid='$bid'";
            $run=mysql_query($SQL,$con) or die ("SQL error");
            $rec=mysql_fetch_array($run);
            $row=mysql_num_rows($run);

            $checkindate = $rec['checkindate'];

            echo $checkindate;

            if ($row < 1) {
                $bidErr="Invalid BookingID. Please check again";
            }   
            else if($checkindate > $today) {
                $bidErr="This booking is not comming today. Please check again";
            } else {
                $_SESSION["chinbid"] = $bid ;
                header("Location: checkinhandler.php");
                exit;
            }
        }
    }

    function test_input($data) {
        $data = trim($data);
        $data = stripslashes($data);
        $data = htmlspecialchars($data);
        return $data;
    }
?>

Answer 1

Try this:

else if(date("Y-n-j", strtotime($checkindate)) > $today) 

I'm guessing your $checkindate is not in the same format (Ynj) as your $today variable.

Answer 2

Check your $checkindate is same as $today date.

Try to check $checkindate format is like date("Ymd") (2016-09-16). IF so then $today will be like

$today=date("Y-m-d"); 

Check the php manual

Also i suggest to go for mysqli_* instead of using mysql_*. Because It has been deprecated since php 5.5 and completely removed in php 7.0 .

Answer 3

<?php
$row='0';
date_default_timezone_set('UTC');
$date=date_create("2016-09-15");
$today = date("Ymd"); 
$date= date_format($date, 'Ymd');
echo "Checkin date is " . $date. "<br>";
echo "Today is " . $today."<br>";
if ($row < '1'){
    echo "no records";
} elseif ($date == $today) {
    echo "dates are the same";
} else {
    echo "check in handler";
}
?>