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php - 如何改进这个php mysql代码?

转载 作者:行者123 更新时间:2023-11-29 00:28:21 25 4
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使用这些代码,我回显了 regd 等于 $regd 的学生的 Rank。事实上,这是一个工作代码。但是,有 friend 告诉我,在mysql语句中,DistinctGroup By不应该一起使用。但是作为一个新手,我无法弄清楚如果不使用 Distinct 我将如何实现它,因为它不会返回没有 Distinct 的行。谁能建议我如何改进这些代码?

<?php 
mysql_select_db($database_dbconnect, $dbconnect);
$query_myrank = "SELECT Distinct regd, Name_of_exam,
Name_of_Student, TOTALSCORE, Rank
FROM (SELECT *, IF(@marks = (@marks := TOTALSCORE),
@auto, @auto := @auto + 1) AS Rank
FROM (SELECT Name_of_Student, regd,
Name_of_exam, SUM(Mark_score) AS TOTALSCORE
FROM cixexam, (SELECT @auto := 0,
@marks := 0) AS init
GROUP BY regd
ORDER BY TOTALSCORE DESC) t) AS result
HAVING (Name_of_exam='First Terminal Exam' OR
Name_of_exam='First Term Test')";

$myrank = mysql_query($query_myrank, $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['regd'] == $regd)
{
echo $value['Rank'];
}
}
?>

最佳答案

DistinctGroup By 慢。您可以在不同时使用 Group ByDistinct 的情况下这样做,这可能是您想要实现的。

SELECT regd, Roll_no, Name_of_Student, Name_of_exam,
TOTALSCORE, Rank
FROM
(
SELECT t.*, IF(@p = TOTALSCORE, @n, @n := @n + 1) AS Rank, @p := TOTALSCORE
FROM
(
SELECT regd, Roll_no, Name_of_Student, Name_of_exam,
SUM(Mark_score) TOTALSCORE
FROM cixexam, (SELECT @n := 0, @p := 0) n
WHERE (Name_of_exam='First Terminal Exam' OR Name_of_exam='First Term Test')
GROUP BY regd
ORDER BY TOTALSCORE DESC
) t
) r

关于php - 如何改进这个php mysql代码?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17859234/

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