如何将多个SQL查询合并为一个，以便在PHP代码中输出为JSON？

<?php
// 连接数据库
$servername = "数据库服务器地址";
$username = "数据库用户名";
$password = "数据库密码";
$dbname = "数据库名";

$conn = new mysqli($servername, $username, $password, $dbname);

// 检查连接是否成功
if ($conn->connect_error) {
    die("连接失败: " . $conn->connect_error);
}

// 执行多个SQL查询
$sql1 = "SELECT * FROM table1";
$sql2 = "SELECT * FROM table2";
$sql3 = "SELECT * FROM table3";

$result1 = $conn->query($sql1);
$result2 = $conn->query($sql2);
$result3 = $conn->query($sql3);

// 合并查询结果
$mergedResult = array();

if ($result1->num_rows > 0) {
    while ($row = $result1->fetch_assoc()) {
        $mergedResult[] = $row;
    }
}

if ($result2->num_rows > 0) {
    while ($row = $result2->fetch_assoc()) {
        $mergedResult[] = $row;
    }
}

if ($result3->num_rows > 0) {
    while ($row = $result3->fetch_assoc()) {
        $mergedResult[] = $row;
    }
}

// 将合并后的结果转换为JSON
$jsonResult = json_encode($mergedResult);

// 输出JSON
echo $jsonResult;

// 关闭数据库连接
$conn->close();
?>