将考试成绩分为等级

发布于 2024-10-31 02:22:54 字数 1405 浏览 0 评论 0原文

大家好,我想制作一个以学术为导向的匹配网站,让男孩和女孩通过学习成绩和兴趣爱好找到潜在的伴侣。

我只是在考虑转换考试成绩的数量,例如。 A、B、C级并转换为S级、A级等联赛。首先,我想将考试成绩 A 转换为 5 的整数,将考试成绩 B 转换为 4 的整数,依此类推。然后我总结它们并将它们分类到各自的类别中。

原始且累人的代码...我不确定它是否会起作用。

$obtaindata = mysql_fetch_assoc(mysql_query('SELECT * FROM userinfo WHERE primaryemel="' . $_COOKIE['smkdtuser'] . '"'));

$pmrresults = json_decode($obtaindata['pmr']);
$spmresults = json_decode($obtaindata['spm']);
$upsresults = json_decode($obtaindata['upsr']);

function calculateClassForPMR ($pmrresults) {
$aquality = (int)$pmrresults['a'] * 5;
$bquality = (int)$pmrresults['b'] * 4;
$cquality = (int)$pmrresults['c'] * 3;
$dquality = (int)$pmrresults['d'] * 2;
$gquality = (int)$pmrresults['g'] * 1;

$additup = $aquality + $bquality + $cquality + $dquality + $gquality;

//Classify sum of scores to their respective class
if ($additup => 35) {$classified = "s";}
elseif ($additup >= 29 && $additup <= 34) {$classified = "a";}
elseif ($additup >= 23 && $additup <= 28) {$classified = "b";}
elseif ($additup >= 17 && $additup <= 22) {$classified = "c";}
elseif ($additup >= 11 && $additup <= 16) {$classified = "d";}
elseif ($additup >= 0 && $additup <= 10) {$classified = "e";}
else {$classified = "wtf";};

return $classified; }

不过,不要怪我,我刚刚开始学习 php 并尝试做一些奇怪的事情......

任何回复或评论都非常感谢。

Hey guys I want to make an academic-oriented matching website that allows boys and girls to find their potential partners through academic performance and hobbies.

I was just thinking about convert quantity of exam grades eg. A, B, C and convert them into leagues like S-class, A-class. First I want to convert exam grade A into integer of 5, exam grade B into integer of 4 and so on. Then I sum them up and classify them into their respective classes.

The original and tiring code... I'm not sure whether it's going to work.

$obtaindata = mysql_fetch_assoc(mysql_query('SELECT * FROM userinfo WHERE primaryemel="' . $_COOKIE['smkdtuser'] . '"'));

$pmrresults = json_decode($obtaindata['pmr']);
$spmresults = json_decode($obtaindata['spm']);
$upsresults = json_decode($obtaindata['upsr']);

function calculateClassForPMR ($pmrresults) {
$aquality = (int)$pmrresults['a'] * 5;
$bquality = (int)$pmrresults['b'] * 4;
$cquality = (int)$pmrresults['c'] * 3;
$dquality = (int)$pmrresults['d'] * 2;
$gquality = (int)$pmrresults['g'] * 1;

$additup = $aquality + $bquality + $cquality + $dquality + $gquality;

//Classify sum of scores to their respective class
if ($additup => 35) {$classified = "s";}
elseif ($additup >= 29 && $additup <= 34) {$classified = "a";}
elseif ($additup >= 23 && $additup <= 28) {$classified = "b";}
elseif ($additup >= 17 && $additup <= 22) {$classified = "c";}
elseif ($additup >= 11 && $additup <= 16) {$classified = "d";}
elseif ($additup >= 0 && $additup <= 10) {$classified = "e";}
else {$classified = "wtf";};

return $classified; }

Don't blame me though ,I was just started learning php and try to do something weird...

Any replies or comments are highly appreciated.

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欢你一世 2024-11-07 02:22:54

使用 json_decode 时,我假设您有 {"a":1,"b":2,"c":3,"d":4,"e":5} 类似这样的内容在该表格行中。尝试使用:

$pmrresults = var_dump(json_decode($obtaindata['pmr'], true)) 然后你可以使用 $pmrresults['a'];

你有当您将值存储到数据库时,使用 json_encode()

我不想租另一台服务器只是为了测试它......它的成本很高。 ...用户 XAMPP 它是免费的用于测试。

when using json_decode i assume that you have {"a":1,"b":2,"c":3,"d":4,"e":5} something like this containt in that tabel row. Try to use :

$pmrresults = var_dump(json_decode($obtaindata['pmr'], true)) then you can use $pmrresults['a'];

You have to json_encode() when you store values to DB

I don't want to rent another server just to test it... it's costly. ... user XAMPP it's free for testing.

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