PHP，如果SQL語句有條件

Question

我正在處理一種表格，其中要求我檢查是否存在一個或多個特定值的實例。 問題是在測試中，即使選擇的值不是，我仍然得到“日期已經預訂”。

這是我的代碼-

if ($_POST) {
   if (isset($_POST['proceedtopaypal'])){
    $apartment = $_POST['apartment'];
    $name = $_POST['name'];
    $surname = $_POST['surname'];
    $email = $_POST['email'];
    $address = $_POST['address'];
    $mobile = $_POST['mobile'];
    $pax = $_POST['pax'];
    $address = $_POST['address'];
    $remarks = $_POST['remarks'];
    $day_from = $_POST['day_from'];
    $month_from = $_POST['month_from'];
    $year_from = $_POST['year_from'];
    $booking_from = $year_from."-".$month_from."-".$day_from; (format- yyyy-mm-dd)
    $day_to = $_POST['day_to'];
    $month_to = $_POST['month_to'];
    $year_to = $_POST['year_to'];
    $booking_to = $year_to."-".$month_to."-".$day_to;
    $no_of_nights = abs(strtotime($booking_to) - strtotime($booking_from));     
    $days = floor($no_of_nights / (60*60*24));
    $validdate = false;
    $buttonpressed = false;

            function IsInjected($str)
    {
      $injections = array('(\n+)',
                  '(\r+)',
                  '(\t+)',
                  '(%0A+)',
                  '(%0D+)',
                  '(%08+)',
                  '(%09+)'
                  );
      $inject = join('|', $injections);
      $inject = "/$inject/i";
      if(preg_match($inject,$str))
        {
        return true;
      }
      else
        {
        return false;
      }
    }

if (!checkdate($month_from, $day_from, $year_from)) {
        echo "Check in date is invalid";
    }
    else if (!checkdate($month_to, $day_to, $year_to)) {
        echo "Check out date is invalid";
    }
    else if ($booking_from > $booking_to) {
             echo "Check in date is after check out date";
    }
    // check if all info is filled in 
    else if (($name == "Name") || ($surname == "surname") || ($email == "Email") || ($mobile == "mobile") || ($address == "Address")) {
        echo "Please fill in the missing information";
    }
    else if (IsInjected($email)) {
        echo "Not an email";
    } 
    else if ($validdate == false) {
        $final = true;
        include 'connect.php';
        $sql = "SELECT COUNT(*)  FROM room_nights WHERE apartmentID= '$apartment' AND dates >= '$booking_from' AND dates < '$booking_to'";
        $result = mysqli_query($conn, $sql);
        $data = mysqli_num_rows($result);
        $row=mysqli_fetch_row($result);
        echo $row[0];   

        if ($data > 0) {
            echo "Dates are already booked!";
        }
        else {
            echo "Proceed";
        }

        }

    }
}

什么是解決此問題的最佳解決方案，以便我繼續我的工作。

Answer 1

$data = mysqli_num_rows($result); 是沒用的。 在查詢中使用SELECT COUNT(*)時，不應使用mysqli_num_rows函數。 您可以這樣使用：

$sql = "SELECT COUNT(*)  FROM room_nights WHERE apartmentID= '$apartment' AND dates >= '$booking_from' AND dates < '$booking_to'";
    $result = mysqli_query($conn, $sql);
    $row=mysqli_fetch_row($result);
    echo $row[0];   

    if ($row[0] > 0) {
        echo "Dates are already booked!";
    }
    else {
        echo "Proceed";
    }