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();
拉风的咖菲猫