我从昨天起就一直在谷歌上搜索我怎么做但却无济于事。我可以使用php和mysql问题完美地实现关系排名并返回特定学生的排名。但我有一个问题。问题是,如果一个学生在一个科目中失败(通过标记被给出33)在表格中按主题(可用的不同科目)分组,我想跳过使用mysql计算他的排名。
$query = "SELECT id, Names,
TOTALSCORE, Rank
FROM(
(SELECT t.*, IF(@p = TOTALSCORE, @n, @n := @n + 1)
AS Rank, @p := TOTALSCORE
FROM(
(SELECT id, Names,
SUM(score) TOTALSCORE
FROM exam, (SELECT @n := 0, @p := 0) n
GROUP BY id
ORDER BY TOTALSCORE DESC
) t
) r";
$myrank = mysql_query($query, $dbconnect) or die(mysql_error());
$i = 0;
$j = 0;
$data = array();
while($row_myrank = mysql_fetch_assoc($myrank)){
$data[$i] = $row_myrank;
if(isset($data[$i - 1]) && $data[$i - 1]['TOTALSCORE'] == $data[$i]['TOTALSCORE']){
$data[$i]['Rank'] = $j;
}else{
$data[$i]['Rank'] = ++$j;
}
$i++;
}
foreach($data as $key => $value){
if($value['id'] == $id){
if($value['Rank']>0){
return $value['Rank'];
} else{
return "<font color=red>Not get Ranking.</font>";
}
}
}
这是我想要实现的样本镜头。在这个镜头中,我只有三个不同的学生,在应用程序中我可能有超过300个不同的学生。学生失败的主题标记为红色。
答案 0 :(得分:2)
我不太确定“pass mark is given 33
”是什么意思。
我假设它是Fullmark
的33%,
例如,如果Fullmark
= 75,则通过分数为75 = 24,75分的33%,
这意味着学生必须获得至少24,75分才能通过考试。
在这种情况下,使用反连接进行简单检查可以提供帮助:
WHERE NOT EXISTS(
SELECT null FROM exam e
WHERE e.id = r.id
AND e.score/e.Fullmark < 0.33
);
以上意思是:“只给我这些记录,其中不存在任何带有过去标记的考试<33%”
完整查询如下所示,此处有演示 - &gt; http://www.sqlfiddle.com/#!2/50ef8/2
SELECT id, Names, TOTALSCORE, Rank
FROM
(
SELECT t.*,
IF(@p = TOTALSCORE, @n, @n := @n + 1) AS Rank,
@p := TOTALSCORE
FROM(
SELECT id, Names,SUM(score) TOTALSCORE
FROM exam, (SELECT @n := 0, @p := 0) n
GROUP BY id
ORDER BY TOTALSCORE DESC
) t
) r
WHERE NOT EXISTS(
SELECT null FROM exam e
WHERE e.id = r.id
AND e.score/e.Fullmark < 0.33
);
- 编辑 -
上面的查询计算错误的排名值,这里是改进版本:
SELECT id, Names, TOTALSCORE, Rank
FROM
(
SELECT t.*,
IF(@p = TOTALSCORE, @n, @n := @n + 1) AS Rank,
@p := TOTALSCORE
FROM(
SELECT id, Names,SUM(score) TOTALSCORE
FROM exam e1, (SELECT @n := 0, @p := 0) n
WHERE NOT EXISTS(
SELECT null FROM exam e2
WHERE e1.id = e2.id
AND e2.score/e2.Fullmark < 0.33
)
GROUP BY id
ORDER BY TOTALSCORE DESC
) t
) r
;
和演示: - &gt; http://www.sqlfiddle.com/#!2/50ef8/4