PHP:警告:count():参数必须是数组或实现 Countable 的对象

Question

我想选择最佳的查找结果。我收到一个错误:这里我在下面的文件中显示了 HTML 表单操作和 DB 连接，请检查它。并且错误消息也提到了这一部分。我使用的是 php 7.2 而不是一些问题

Warning: count(): Parameter must be an array or an object that implements
Countable in D:\xammp\htdocs\search\index.php on line 64

警告:在第 37 行的 D:\xammp\htdocs\search\index.php 中使用未定义的常量 city_id - 假定为“city_id”(这将在 PHP 的 future 版本中引发错误)

帮帮我...错误消息。

PHP:

<?php
$i = 1;
if (count($searchdata) > 0) {
    foreach ($searchdata as $places) {
        echo '<tr>';
        echo '<th>' . $i . '</th>';
        echo '<td>' . $places['city'] . '</td>';
        echo '<td>' . $places['visiting_place'] . '</td>';
        echo '<td>' . $places['history'] . '</td>';
        echo '</tr>';
        $i++;
    }
} else {
    echo '<td colspan="4">No Search Result Found.</td>';
}
?>

我正在使用 PHP，例如:

<?php
$searchdata = [];
$keyword = '';
if (isset($_POST['search'])) {
    $city = $_POST['city'];
    $keyword = $_POST['keyword'];
    $searchdata = $model->getVisitinPlaceData($city, $keyword);
}
?>

数据库连接和表格数据获取

<?php 
class Db {
    private $hostname = 'localhost';
    private $username = 'root';
    private $password = '';
    private $database = 'test';
    private $conn;
    public function __construct() { 
        $this->conn = mysqli_connect($this->hostname, $this->username, $this->password, $this->database); 
        if(!$this->conn) {
            echo 'Database not connected';
        }
    }
    public function getTouristCity(){
        $query = "SELECT * FROM tourist_city WHERE is_enabled = '1'";
        $result = mysqli_query($this->conn, $query);
        return $result;
    }
    public function getVisitingPlaces(){
        $query = "SELECT * FROM visiting_places WHERE is_enabled = '1'";
        $result = mysqli_query($this->conn, $query);
        return $result;
    }
    public function getVisitinPlaceData($cityid, $keyword){
        $sWhere = '';
        $where = array();
        if($cityid > 0) {
            $where[] = 'V.city_id = '.$cityid.' AND V.is_enabled = "1"';
        }

        if($keyword != '') {
            $keyword = trim($keyword);
            $where[] = "( V.visiting_place LIKE '%$keyword%' OR  V.history LIKE '%$keyword%'  OR  C.city LIKE '%$keyword%' )";
        }
        $sWhere     = implode(' AND ', $where);
        if($sWhere) {
            $sWhere = 'WHERE '.$sWhere;
        } 
        if(($cityid > 0) || ($keyword != '')) {
            $query = "SELECT * FROM visiting_places AS V JOIN tourist_city AS C ON C.city_id = V.city_id $sWhere ";
            $result = mysqli_query($this->conn, $query);
            return $result;
        }
    }
}
?>

html 表单操作

<form action="" method="post" > 

            <div class="col-sm-3"> 

                <select name="city" class="form-control">

                <option value="0">Select City</option>
                <?php foreach($turistCity as $city) {
                    $checked = ($_POST['city'] == $city[city_id])? 'selected' : '';
                    echo '<option value="'.$city[city_id].'" '.$checked.'>'.$city[city].'</option>';
                }
                ?>
                </select>
            </div>
            <div class="col-sm-3"> 
             <input type="text" name="keyword" placeholder="Keword" value="<?php echo $keyword;?>" class="form-control"> 
            </div>

            <button name="search" type="submit" class="btn btn-primary">Search</button>
        </form>

Answer 1

只需将它包装在一个检查它是否可数的条件中，所以..

// PHP >= 7.3
if(is_countable($searchdata)) {
    // Do something
}

// PHP >= 7.1
if(is_iterable($searchdata)) {
    // Do something
}

注意:除非你明确地定义它，否则你永远不应该假设任何事情都是真的。 count($something) 暗示它确实是一个数组或你可以计算的东西。如果情况并非总是如此，那么您在继续之前请仔细检查。

EDIT:在这种情况下，将 is_iterable() 用于 PHP >= 7.1 或 is_countable() 用于 >= 7.3。上面的片段更新了