以下是我的示例结果：
+------+-----------------+--------+
| rank | nick | amount |
+------+-----------------+--------+
| 1 | 038112125147518 | 0.2000 |
| 2 | 0tosan | 0.2000 |
| 3 | 13unnyeater | 0.2000 |
+------+-----------------+--------+
SET / SELECT语句返回的内容：
SET @rank=0; SELECT @rank:=@rank+1 AS rank, nick, amount FROM points;
返回行匹配的最有效方法是什么：
nick='13unnyeater'
...'rank'栏后用上面的表格列表？我期望的结果是：
+------+-----------------+--------+
| rank | nick | amount |
+------+-----------------+--------+
| 3 | 13unnyeater | 0.2000 |
+------+-----------------+--------+