Question

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']);

以后的地方：

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

正如您现在所知，时间保存在unix时间戳中。

此代码无法正常运行。

让我们说数据年份是2011年和第39周。

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

如何让它选择并显示几周？

Answer 1

$max个月正在使用$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

为xx周选择全部，不能正常运行php

1 个答案: