Question

据说我在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'Mark'所在的行＃（在这种情况下，行号1，考虑到第一行是0）。

如何通过PHP获取？

...谢谢

编辑：将其视为高分或排行榜表，我不需要用户ID，而是用户所在的行...

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

经过一番思考后，小组可能会给你带来一些UDV计数的麻烦。这是另一种选择，但效果不如前一种方法。

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，我建议直接用PHP计算。这将更加灵活和高效。

Answer 2

修改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

如何通过php从mysql fetch row / array / assoc结果中检索特定值的行号？

2 个答案: