在 Mysqli 准备语句上获取 json_encode() 时出现问题

Question

我在使用 MySQLi 导出准备语句的结果时遇到一些问题。

$age = "young";
$rows = array();
$mysqli = new mysqli($db_hostname, $db_username, $db_password, $db_database);

        $stmt = $mysqli->prepare("SELECT fname, lname FROM users WHERE age=?");
        $stmt->bind_param('s', $age);
        $stmt->execute();
        $stmt->bind_result($thefName,$thelName);
                while($stmt->fetch()) 
                  {
                    array_push($rows, $thefName,$thelName);
                  }
            $stmt->close();

$mysqli->close();
echo json_encode($rows);

如您所见，我正在尝试将 json_encode($rows); 每一行回显为 JSON 数组，如下所示：

[
    {"firstName":"John", "lastName":"Doe"}, 
    {"firstName":"Anna", "lastName":"Smith"}, 
    {"firstName":"Peter","lastName":"Jones"}
]

但我得到的是

["John", "Doe", "Anna", "Smith", "Peter", "Jones"]

能否请您告诉我为什么会发生这种情况以及我该如何解决

Answer 1

推送正确的数组 -

array_push($rows, array('firstName' => $thefName, 'lastName' => $thelName));

在 Mysqli 准备语句上获取 json_encode() 时出现问题

Having Issue on Getting json_encode() On Mysqli Prepared Statement

php

mysqli

prepared-statement