Question

这是我的代码

$sql="SET @rank=0; SELECT * FROM (SELECT *, @rank:=@rank+1 AS Rank FROM scoregame where userid=33 order by score DESC) AS t";
$query=mysql_query($sql);
if(mysql_num_rows($query) != "")
{
  $stt=1;
  while($row=mysql_fetch_array($query))
  {
    $stt++;
    echo $row['score'];
  }
}

但有些不对劲：

 Warning: mysql_num_rows() expects parameter 1 to be resource

非常感谢你！

Answer 1

您尝试一次运行2个查询。这不适用于这个PHP函数。但您可以将其缩减为一个查询

SELECT *, @rank:=@rank+1 AS Rank 
FROM scoregame 
cross join (select @rank := 0) r
where userid=33 
order by score DESC

我的代码出了什么问题？排名得分mysql，php

1 个答案: