我如何将 mysqli_expects 参数 1 修复为给定的 mysqli_result 字符串

Question

当我尝试执行查询并获取行数时，会生成错误。我已经尝试解决这个问题好几天了，而且我是 php 新手，所以它可能甚至不是语法问题。  <?php  // error_reporting(0);  $flying_from = "";  $flying_to = "";  $departure_date = "";  $return_date = "";  $number_of_adults = "";  $number_of_children = "";  $flight_class = "";  // Connect to database  $conn = mysqli_connect('localhost', 'presh', '1234', 'saffron');  if (isset($_GET['submit'])) {    $flying_from = $_GET['flying_from'];    $flying_to = $_GET['flying_to'];    $departure_date = $_GET['departure_date'];    $return_date = $_GET['return_date'];    $number_of_adults = $_GET['number_of_adults'];    $number_of_children = $_GET['number_of_children'];    $flight_class = $_GET['flight_class'];    $query = "SELECT * FROM flights WHERE flying_from LIKE $flying_from AND flying_to LIKE $flying_to AND departure_date LIKE $departure_date AND return_date LIKE $return_date AND    number_of_adults LIKE $number_of_adults AND number_of_children LIKE $number_of_children AND flight_class LIKE $flight_class";    $result = mysqli_query($conn,$query);    $count = mysqli_num_rows($result);    if($count > 0){        $output = 'There was no search results';    }else{      while ($row = mysqli_fetch_array($query)) {        $fFrom = $row['flying_from'];        $fTo = $row['flying_to'];        $dDate = $row['departure_date'];        $rDate = $row['return_date'];        $nAdults = $row['number_of_adults'];        $nChildren = $row['number_of_children'];        $fClass = $row['flight_class'];        $output .= '<div>'.$fFrom.' '.$fTo. ' '.$dDate.' '.$rDate.' '.$nAdults.' '.$nChildren.' '.$fClass.'</div>';    }  }}?>

Answer 1

您需要传递的$result变量$query不在mysqli_fetch_array()

while ($row = mysqli_fetch_array($result)) {

}