PHP从表1中选择并在表2中插入id return为0的表2

<?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, $dbname);
if ($conn->connect_error) {
    die("数据库连接失败: " . $conn->connect_error);
}

// 从表1中选择id为0的记录
$sql = "SELECT * FROM 表1 WHERE id = 0";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // 将选中的记录插入到表2中
    $row = $result->fetch_assoc();
    $id = $row["id"];
    $name = $row["name"];
    $age = $row["age"];

    $sql = "INSERT INTO 表2 (id, name, age) VALUES ('$id', '$name', '$age')";
    if ($conn->query($sql) === TRUE) {
        echo "成功插入记录到表2";
    } else {
        echo "插入记录到表2失败: " . $conn->error;
    }
} else {
    echo "在表1中找不到id为0的记录";
}

// 关闭数据库连接
$conn->close();
?>