计算数据库中没有新行的百分比 - Calculating Percentage of no new rows in database

我做了一个脚本，看看我们管理的PC上是否有任何活动(我在c#中创建了一个程序，每5分钟检查一次活动)。数据库如下所示:

服务器:

readip, time(datetime), user, computeruse (int, 0=no 1=yes), ID

我已经运行了一个cronjob，用于创建每天之后计算机使用的百分比。但如果客户的电脑关闭，这将不算数。

是否有任何方法来检查是否有没有新的行添加到数据库的一段时间?

我的代码是:

'$result2 = mysql_query("SELECT * FROM server WHERE serverip >='$ipfrom' AND    serverip<='$ipto' AND DATE(time) = CURDATE() GROUP BY serverip ORDER BY time");
while($row2 = mysql_fetch_array($result2)) {
    $serverip = $row2['serverip'];
    $queryfornumrowsno = mysql_query("SELECT * From server WHERE serverip = '$serverip'") or die(mysql_error());
    $numrowstotal = mysql_num_rows($queryfornumrowsno) . ' ';
    $queryfornumrowsyes = mysql_query("SELECT * FROM server WHERE serverip = '$serverip' AND computeruse = 'Yes'");
    $numrowsyes = mysql_num_rows($queryfornumrowsyes);
    $percentage = $numrowsyes * 100 / $numrowstotal;
    $percentage = round($percentage);
    $percentagetotal[] = $percentage;

    $datenow = $row2['time'];

}
$totaalPercentage = 0;
foreach( $percentagetotal as $totaal )
{
    $totaalPercentage += $totaal;
}
$percentagecount = count($percentagetotal);

if($percentagecount == "0"){
    $totaal = "0";
}
else{
    $totaal =  $totaalPercentage / $percentagecount;
}
echo $datenow . ' @ ' . $totaal . '% van ' . $lokaal . '<br />'; 
if($datenow){

}

对不起，我的英语不好，这不是我的母语。

解决方案:

include 'connect.php';
$result = mysql_query("SELECT * FROM iprange");
while($row = mysql_fetch_array($result)) {
$ipfrom = $row['iprangefrom'];
$lokaal = $row['lokaal'];
$ipto = $row['iprangeto'];
$ipfrom = ip2long($ipfrom);
$ipto = ip2long($ipto);
$listpc = 0;
$result = mysql_query("SELECT * FROM server WHERE serverip >='$ipfrom' AND            serverip<='$ipto' AND DATE(time) = CURDATE() AND computeruse = 'Yes'");
$countpc = mysql_query("SELECT * FROM server WHERE serverip >='$ipfrom' AND serverip<='$ipto' AND DATE(time) = CURDATE() GROUP BY serverip ORDER BY time") or die(mysql_error());
$countpcrows = mysql_num_rows($countpc);

while($row = mysql_fetch_array($result)) {
    $listpc = $listpc + 1;
    }
echo $countpcrows . ' ' . $listpc . '<br />';
$percentage = $listpc / 84 * 100 / $countpcrows;
    //84 = 5 minutes.
echo round($percentage) . '%';
}

获取最近10分钟内没有活动的服务器列表:-

SELECT a.serverip
FROM
(
    SELECT DISTINCT serverip 
    FROM server 
    WHERE serverip >='$ipfrom' 
    AND   serverip<='$ipto' 
) a
LEFT OUTER JOIN
(
    SELECT DISTINCT serverip 
    FROM server 
    WHERE serverip >='$ipfrom' 
    AND   serverip<='$ipto' 
    AND time > DATE_ADD(CURDATE(), INTERVAL -10 MINUTE) 
) b
ON a.serverip = b.serverip
WHERE b.serverip IS NULL

第一个子查询获取您感兴趣的范围内的服务器列表。第二个获取最近10分钟内所有活动的服务器。LEFT JOIN将获得两者的组合，包括那些在过去10分钟内没有活动的服务器。WHERE子句然后将排除所有这些，除了那些在过去10分钟内有任何活动的。

基于此，您应该能够计算百分比。您希望百分比作为所有服务器的数量，还是只是在您感兴趣的范围内?

假设在您感兴趣的范围内所有服务器的百分比:-

SELECT (COUNT(DISTINCT a.serverip) / c.all_server) * 100 AS percentage_not_used
FROM server a
CROSS JOIN
(
    SELECT COUNT(*) AS all_server
    FROM server
    WHERE serverip >='$ipfrom' 
    AND   serverip <='$ipto' 
) c
LEFT OUTER JOIN
(
    SELECT DISTINCT serverip 
    FROM server 
    WHERE serverip >='$ipfrom' 
    AND   serverip<='$ipto' 
    AND time > DATE_ADD(CURDATE(), INTERVAL -10 MINUTE) 
) b
ON a.serverip = b.serverip
WHERE a.serverip >='$ipfrom' 
AND   a.serverip <='$ipto' 
AND   b.serverip IS NULL

EDIT -计算每个服务器在过去24小时内被使用的时间百分比(假设24小时内可以使用的最大次数为84次):-

SELECT a.serverip, (COUNT(b.serverip) / 84) * 100 AS percentage_not_used
FROM server a
LEFT OUTER JOIN server b
ON a.serverip = b.serverip
AND b.serverip >='$ipfrom' 
AND b.serverip<='$ipto' 
AND b.time > DATE_ADD(CURDATE(), INTERVAL -24 HOUR) 
GROUP BY a.serverip

将服务器表与自身连接起来。表的第一次出现用于获取serverip(无论是否在过去24小时内使用)，而第二次出现用于获取它在过去24小时内被使用的次数(计数将忽略null)。