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&nbsp;($row&nbsp;=&nbsp;mysqli_fetch_array($result))&nbsp;{ }