Php 按天选择多行数的平均值

我正在为我运行的服务编写一个统计脚本。我想计算出过去3天，每周、每月和每年每天的平均行数。我的基本想法是这样的：抓住平均数：

<?php
function getDayNumber($day)
{
    $start = strtotime("last monday");
    $q = mysql_query(sprintf("SELECT COUNT(*) as count FROM mytable WHERE dayname(datetime) = "%s" AND unix_timestamp(datetime) > %s",$day,$start));
    $row = mysql_fetch_assoc($q);
    return $row['count'];
}

$monday = getDayNumber("Monday");
$tuesday = getDayNumber("Tuesday");
$wednesday = getDayNumber("Wednesday");
$thursday = getDayNumber("Thursday");
$friday = getDayNumber("Friday");

$average = ($monday+$tuesday+$wednesday+$thursday+$friday)/5;

?>

以上只是我头脑中编写的psuedo代码。但是我想不出更好的方法了

这可以通过嵌套/联接查询实现吗？我需要能够做到每天，每周，每月。。。etc基础。我想如果我能找到一种方法来提取MySQL中的所有数据，我可以将所有count（*）加在一起，然后通过响应的数量来划分它们

这是否可行

---

感谢您的帮助，谢谢大家

我建议找到每个数据集的日期范围，然后运行查询。希望这有帮助

<?php
/**
 * Get the average number of rows that were created
 * during the given date range.
 * 
 * @param  string    $range    The date range to look for
 * @return integer             The average number of rows created in the given range
 */
function getAverageRows($range)
{
    $date_begin = null;
    $date_end   = null;

    switch($range)
    {
        case 'year':
            // Calculate the beginning and end datetime of the past year
            $date_begin = FIRST_DAY_OF_YEAR . ' 00:00:00';
            $date_end   = LAST_DAY_OF_YEAR  . ' 23:59:59';
        break;

        case 'month':
            // Calculate the beginning and end datetime of the past month
            $date_begin = FIRST_DAY_OF_MONTH . ' 00:00:00';
            $date_end   = LAST_DAY_OF_MONTH  . ' 23:59:59';
        break;

        case 'week':
            // Calculate the beginning and end datetime of the past week
            $date_begin = FIRST_DAY_OF_WEEK . ' 00:00:00';
            $date_end   = LAST_DAY_OF_WEEK  . ' 23:59:59';
        break;

        default:
            // Calculate the beginning and end datetime of the past three days
            $date_begin = SEVENTY_SIX_HOURS_AGO;
            $date_end   = now();
        break;
    }

    $query = "SELECT COUNT(*) FROM mytable WHERE datetime >= '{$date_begin}' AND datetime <= '{$date_end}'";

    // Get and return query results
}

echo "Average Rows:<br/>";
echo "Past three days: " . getAverageRows()        . "<br/>";
echo "Past week: "       . getAverageRows('week')  . "<br/>";
echo "Past Month: "      . getAverageRows('month') . "<br/>";
echo "Past Year: "       . getAverageRows('year')  . "<br/>";

---

添加具有预期结果的示例数据可能会有帮助。对不起，您想要什么类型的数据？表中的数据是不相关的，对吗？我只想获取计数，并根据当天计算平均值。这里有一个名为“datetime”的字段，它是一个标准的日期/时间。对不起，我不认为我完全理解它是如何工作的。你使用的常数，我猜它们是在别处定义的？还有，这是如何获取平均值的？从我看到的情况来看，这将获取这些时段之间的总行数。我需要计算出给定时间段内的平均行数。