Question

我正在努力让搜索工作，我已经回应并使用了var dump，我在浏览器中得到的是：

0没有搜索结果

object(mysqli_result)#2 (5) { ["current_field"]=> int(0) 
                              ["field_count"]=> int(4) 
                              ["lengths"]=> NULL 
                              ["num_rows"]=> int(0) 
                              ["type"]=> int(0) 
                            }

但是我输入的是在我的数据库中的巴黎，所以它不应该回应没有搜索。

以下是我的代码。不知道我哪里出错了，任何帮助都会很有意义。

require_once('config1.php');
error_reporting(E_ALL);

$output = '';
if(isset($_POST['search'])){
$searchq = $_POST['search'];

$sql = "SELECT a.attraction_name, a.lat, a.long, a.cost FROM zz_attractions a 
INNER JOIN zz_city c ON a.city_id = c.city_id WHERE c.city_name LIKE 
'%searchq%'";

$result = mysqli_query($conn, $sql);
$count = mysqli_num_rows($result);

if ($count == 0) { 
    $output = 'there was no search results';

    echo $count;
    echo $output;
    var_dump($result);
} else {
    while ($row = mysql_fetch_array($sql)) { 
    $attraction_name = $row['attraction_name'];
    $lat = $row['lat'];
    $long = $row['long'];
    $cost = $row['cost'];

    $output .= '<div>'.$attraction_name.' '.$lat.' '.$long.' '.$cost.'</div>';

    }
}
}


city_id  attraction_id  attraction_name  lat  long  cost
1           1             Eiffel Tower   49    2     25

Answer 1

您错过了传递给SQL语句的变量的$ off。尝试

$sql = "SELECT a.attraction_name, a.lat, a.long, a.cost FROM zz_attractions a 
INNER JOIN zz_city c ON a.city_id = c.city_id WHERE c.city_name LIKE 
'%$searchq%'";

另外，看一下sql语句的绑定参数，而不是直接传递它们，因为你当前的代码让你很容易注入 https://www.w3schools.com/sql/sql_injection.asp | http://php.net/manual/en/mysqli-stmt.bind-param.php

即使数据库中有数据，也会回显“无搜索结果”

1 个答案: