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好的，这就是我所拥有的：

function getFirstDayOfWeek($iYear, $iWeekNumber)
{
    if ( is_null($iYear) ) $iYear = date('Y');
    if ( $iWeekNumber < 10 ) $iWeekNumber = '0'.$iWeekNumber;

    $iTime = strtotime($iYear.'W'.$iWeekNumber);

    return $iTime;
}

$firstdayofweek = getFirstDayOfWeek($data['year'], $data['week']);
$mDate = date('m', $firstdayofweek);
$dDate = date('d', $firstdayofweek);
$min = mktime(0, 0, 0, $mDate, $dDate, $data['year']);
$max = mktime(23, 59, 59, $dDate, $dDate+6, $data['year']);

它后来在哪里：

SELECT id
FROM training_activities
WHERE date time >= {$min} 
  AND time <= {$max}

到目前为止，时间都保存在 unix 时间戳中。

此代码无法正常工作。

假设 $data 年份是 2011 年和第 39 周。

这显示我的日期是 26/09-04/10，8 天，而第 39 周是 29/09 - 02/10

我怎样才能让它选择并显示周数？

4

1 回答 1

1

该$max月使用$dDate而不是$mDate; 以下作品：

<?php

function getFirstDayOfWeek($iYear, $iWeekNumber)
{
    if ( is_null($iYear) ) $iYear = date('Y');
    if ( $iWeekNumber < 10 ) $iWeekNumber = '0'.$iWeekNumber;

    $iTime = strtotime($iYear.'W'.$iWeekNumber);

    return $iTime;
}

for ($i = 1; $i <= 52; $i++) {
    $firstdayofweek = getFirstDayOfWeek(2011, $i);
    $mDate = date('m', $firstdayofweek);
    $dDate = date('d', $firstdayofweek);
    $min = mktime(0, 0, 0, $mDate, $dDate, 2011);
    $max = mktime(23, 59, 59, $mDate, ($dDate+6), 2011);
    echo date('m/d/Y',$min)." - ".date('m/d/Y',$max)." ($dDate - ".($dDate+6).")\n";
}

?>

http://codepad.org/7hMJQFhq

于 2011-10-02T18:36:16.570 回答