要在 MySQL 中查询相同记录并仅保留时间最晚的那一条，你可以使用以下几种方法：

方法一：使用子查询和 GROUP BY

假设你的表名为 your_table，时间字段为 create_time，其他用于判断记录相同的字段为 field1, field2 等：

SELECT t1.*
FROM your_table t1
INNER JOIN (SELECT field1, field2, MAX(create_time) AS max_timeFROM your_tableGROUP BY field1, field2
) t2 ON t1.field1 = t2.field1 AND t1.field2 = t2.field2 AND t1.create_time = t2.max_time;

方法二：使用窗口函数（MySQL 8.0+）

如果你使用的是 MySQL 8.0 或更高版本，可以使用窗口函数更简洁地实现：

WITH ranked_data AS (SELECT *,ROW_NUMBER() OVER (PARTITION BY field1, field2 ORDER BY create_time DESC) AS rnFROM your_table
)
SELECT * FROM ranked_data WHERE rn = 1;

方法三：使用 LEFT JOIN 排除法

SELECT t1.*
FROM your_table t1
LEFT JOIN your_table t2 ON t1.field1 = t2.field1 AND t1.field2 = t2.field2 AND t1.create_time < t2.create_time
WHERE t2.id IS NULL;

实际示例

假设有一个 orders 表，包含 order_id, customer_id, product_id, order_time 字段，我们想保留每个客户对每个产品的最新订单：

-- 方法一示例
SELECT o1.*
FROM orders o1
INNER JOIN (SELECT customer_id, product_id, MAX(order_time) AS latest_timeFROM ordersGROUP BY customer_id, product_id
) o2 ON o1.customer_id = o2.customer_id AND o1.product_id = o2.product_id AND o1.order_time = o2.latest_time;-- 方法二示例（MySQL 8.0+）
WITH latest_orders AS (SELECT *,ROW_NUMBER() OVER (PARTITION BY customer_id, product_id ORDER BY order_time DESC) AS rnFROM orders
)
SELECT * FROM latest_orders WHERE rn = 1;

注意事项