PHP と MySQL を使用して JSON データを検索およびフィルターするにはどうすればよいですか?

CREATE TABLE json_data(
    id INT PRIMARY KEY AUTO_INCREMENT,
    data JSON
);

INSERT INTO json_data (data) VALUES 
    ('{"name":"John","age":25,"city":"New York"}'),
    ('{"name":"Alice","age":30,"city":"London"}'),
    ('{"name":"Bob","age":35,"city":"Paris"}'),
    ('{"name":"Emma","age":28,"city":"Dublin"}');

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";

// 创建连接
$conn = new mysqli($servername, $username, $password, $dbname);

// 检测连接
if ($conn->connect_error) {
    die("连接失败: " . $conn->connect_error);
}
?>

<?php
$sql = "SELECT data FROM json_data";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // 输出数据
    while($row = $result->fetch_assoc()) {
        $json = json_decode($row['data'], true); // 将JSON数据解码为关联数组

        // 搜索条件
        $searchTerm = "John";

        // 如果姓名匹配搜索条件，则打印出结果
        if($json['name'] == $searchTerm) {
            echo "姓名：" . $json['name'] . ", 年龄：" . $json['age'] . ", 城市：" . $json['city'] . "<br>";
        }
    }
} else {
    echo "0 结果";
}
$conn->close();
?>

<?php
$sql = "SELECT data FROM json_data";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // 输出数据
    while($row = $result->fetch_assoc()) {
        $json = json_decode($row['data'], true);

        // 过滤条件
        $ageFilter = 30;
        $cityFilter = "London";

        // 如果满足过滤条件，则打印出结果
        if($json['age'] > $ageFilter && $json['city'] == $cityFilter) {
            echo "姓名：" . $json['name'] . ", 年龄：" . $json['age'] . ", 城市：" . $json['city'] . "<br>";
        }
    }
} else {
    echo "0 结果";
}

$conn->close();
?>