在不声明列名的情况下回显 mysqli 结果中的行？

Question

如何在不声明每个列名的情况下回显查询结果中的所有行（作为 JSON）？即没有写作'location_id' => $row['location_id']等等，就像我在下面所做的那样。

<?php

require_once("./config.php"); //database configuration file

require_once("./database.php");//database class file

$location_id = isset($_GET["location_id"]) ? $_GET["location_id"] : '';

$db = new Database();

if (isset($_GET["location_id"])){

    $sql = "SELECT * FROM location WHERE location_id = $location_id";

} else {

    $sql = "SELECT * FROM location";

}

$results = $db->conn->query($sql);

if($results->num_rows > 0){

    $data = array();

    while($row = $results->fetch_assoc()) {

        $data[] = array(

        'location_id' => $row['location_id'],

        'customer_id' => $row['customer_id'],

        'location_id' => $row['location_id'],

        'location_name' => $row['location_name'],

        'payment_interval' => $row['payment_interval'],

        'location_length' => $row['location_length'],

        'location_start_date' => $row['location_start_date'],

        'location_end_date' => $row['location_end_date'],

        'location_status' => $row['location_status'],

        'sign_sides' => $row['sign_sides'],

        'variable_annual_price' => $row['variable_annual_price'],

        'fixed_annual_price' => $row['fixed_annual_price'],

        'location_file' => $row['location_file']);

    }

header("Content-Type: application/json; charset=UTF-8");

echo json_encode(array('success' => 1, 'result' => $data));

} else {

    echo "Records not found.";

}

?>

更新代码。现在使用@Dharman 推荐的参数化准备语句（谢谢！）。我Parse error: syntax error, unexpected '->' (T_OBJECT_OPERATOR)在第 17 行。我正在运行 PHP 7.3 版。怎么了？我应该如何回显 $data 以便它像以前一样是 JSON 对象？

<?php

header("Content-Type: application/json; charset=UTF-8");

//include required files in the script

require_once("./config.php"); //database configuration file

require_once("./database.php");//database class file

$object_contract_id = isset($_POST["object_contract_id"]) ? $_POST["object_contract_id"] : '';

//create the database connection

$db = new Database();

Answer 1

fetch_assoc()&nbsp;已经将您的数据作为关联数组返回，因此您不需要重新进行关联。您可以简单地将结果分配给您的数据。=>$data[] = $row有关如何fetch_assoc()工作的详细说明。这是文档。