如何通过php从mysql获取行/数组/关联结果中的特定值中检索行号？

Question

假设我在 PHP 中有这个 sql 结果（$myresults = mysql_fetch_… assoc/row/array）：

SELECT table1.name as name, sum(table2.numbers) as numbers FROM table2 INNER JOIN
 table1 ON table2.fk_id = table1.id GROUP BY name ORDER BY numbers DESC

---------------
| John | 800  |
---------------
| Mark | 500  |
---------------
| Bill | 300  |
---------------

所以我被记录为 Mark ($_SESSION['name'] == "Mark") 并且我想知道值 'Mark' 位于哪一行（在本例中为第 1 行，考虑到第一行是 0)。

我如何通过 PHP 获得它？

谢谢……

编辑：把它想象成一个高分或排行榜表，我不需要用户 ID，但用户现在所在的行...

Answer 1

您应该以这种方式使用用户定义的变量：

SELECT table1.name as name, sum(table2.numbers) as numbers,
  @rank := @rank + 1 rank
FROM table2
CROSS JOIN (SELECT @rank := 0) init
JOIN table1 ON table2.fk_id = table1.id
GROUP BY name
ORDER BY numbers DESC

再想一想，group by 可能会给您计算 UDV 带来一些麻烦。这是另一种替代方法，但性能不如以前的方法。

SELECT *, @rank := @rank + 1 rank FROM (
    SELECT table1.name as name, sum(table2.numbers) as numbers
    FROM table2
    JOIN table1 ON table2.fk_id = table1.id
    GROUP BY name
) s
CROSS JOIN (SELECT @rank := 0) init
ORDER BY numbers DESC

无论如何，我建议直接在 PHP 中计数。这将更加灵活和高效。

Answer 2

修改您的 SQL 以选择主 ID 以及其他数据：

SELECT 
    table1.id as id,
    table1.name as name, 
    sum(table2.numbers) as numbers 
FROM 
    table2 
INNER JOIN
    table1 ON table2.fk_id = table1.id 
GROUP BY 
    name 
ORDER BY 
    numbers DESC