计算巨大数据库的百分位数

Question

我有庞大的数据库，其中包含有关测试代码的学生信息以及为这些测试代码实现的标记我需要为每个测试代码对应的学生重新计算百分位数。我有一系列测试代码的代码，但它无法正常工作。

function recompute_percentiles()
{
if($_REQUEST[testcode]=="CAT B1" or $_REQUEST[testcode]=="CAT B2" or $_REQUEST[testcode]=="CAT B3" or $_REQUEST[testcode]=="CAT B4")
{

echo "<br />Got testcode: ".$_REQUEST[testcode];


$getsortedq=mysql_query("SELECT username, section1right as m from kmarks where testcode='.$_REQUEST[testcode].' order by section1right DESC");

if(!$getsortedq) 
echo "Could not get the sorted query";
else 
echo "got the sorted query quick";



$totalcount=mysql_num_rows($getsortedq);

while($r=mysql_fetch_array($getsortedq))
{
$u=$r[username];
$m=$r[m];
$array[$u]=$m;
}



$array2=$array;
//print_r($array2);

$updated=0;

foreach($array as $key=>$value)
{
$countsame=0;
foreach($array2 as $k=>$v)
{
    if($v>=$value) 
    $countsame++; 
    else
    break;
}
$countless = $totalcount - $countsame;

reset($array2);

$percentile=round($countless/$totalcount*100,2);

$updatep1q=mysql_query("UPDATE kmarks set percentile1=$percentile where   username='.$key.' and testcode='.$_REQUEST[testcode].'");

if(!$updatep1q)
  echo "<br />Could not update p1 for username: ".$key;
else
    $updated++;


}

echo "<br />Updated ".$updated." records in kmarks db, out of ".$totalcount." records for  testcode ".$_REQUEST[testcode];




}
}

Answer 1

这段代码存在多个严重问题 - 甚至没有触及功能......

1 PHP syntax

$_REQUEST[testcode]

不好，总是用括号！

$_REQUEST['testcode']

2 Injection proneness

你对SQL Injection和HTML/Javascript注射也很开放

echo "<br />Got testcode: ".$_REQUEST[testcode]; //HTML injection...
//SQL injection
$getsortedq=mysql_query("SELECT username, section1right as m from kmarks where testcode='.$_REQUEST[testcode].' order by section1right DESC"); 

始终使用适当的清理（mysql(i)_real_escape_string($_REQUEST['testcode'])取决于使用的mysql_或mysqli）。甚至更好：在SQL案例中准备好的语句......

3 Deprecation

从PHP 5.5开始，不推荐使用必需的mysql_ *警告：mysql_函数。不要使用它们：使用PDO或至少mysqli_函数...

功能

这是罪魁祸首：

$updatep1q=mysql_query("UPDATE kmarks set percentile1=$percentile where   username='.$key.' and testcode='.$_REQUEST[testcode].'");

生成的查询将显示为：

UPDATE kmarks set percentile1=<somevalue>  --this is OK
where username='.<somevalue>.' and testcode='.$_REQUEST[testcode].'
                ^           ^                ^^^^^^^^^^^^^^^^^^^^^

突出问题......有不需要的点，以及整个不好的部分。我想你想要这样的东西

UPDATE kmarks set percentile1=<somevalue>  
where username='<somevalue>' and testcode='<somevalue>'

改为使用它（当然是消毒!!!）：

//WARNING! STILL HAS SQL INJECTION --apply sanitization from #2 to make it safer...
$updatep1q=mysql_query("UPDATE kmarks set percentile1=$percentile where username='".$key."' and testcode='".$_REQUEST[testcode]."'");

数组不能在字符串文字中使用，并且在普通变量的情况下不需要.连接运算符...

Answer 2

好像很多代码。你可以这样做：

$results = $db->query("SELECT * FROM your_table ORDER BY sort_field");

$data = array();
while($row = $results->fetch_assoc()){
    $data[] = $row;
}

$chunks = array_chunk($data,ceil((count($data)/100)));
foreach($chunks as $key => $dataset){ 
$percentile = 99 - $key;

    foreach($dataset as $row){
        $db->query("UPDATE your_table SET percentile={$percentile} WHERE id={$row['id']}");
    }
}