PHP当中如何判断一个周的起始日期与结束日期

[吉林小江南]

数据库表:
id, usid, weekNum, monday-----sunday.

PHP当中如何判断一个周的起始日期与结束日期

[foolbirdflyfirst]

你要根据什么来判断?
weekNum存的是什么数据?
本年内的第几个周?

[吉林小江南]

weekNum存的是本年内的第X个周

[foolbirdflyfirst]

涉及一点算法,有点意思,参考以下.你再测试一下,看有什么问题没有.
===========================================================
//function:
/**
* get week's first and last day
*
* @param string $year //年份
* @param string $weekAtYear // 周数
* @return array $days
*/
function getDays($year,$weekAtYear)
{
//找出年份第一天是星期几(1-7)
$firstDay = date('N',strtotime("$year-01-01"));
//找出第一个星期剩余天数,用于计数.比如2008-01-01号是星期二,则剩余6天到第一个星期天.
$leftDayInWeek = 8 - $firstDay;
//星期一的日期(Y-m-d)
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+($weekAtYear-1)*$leftDayInWeek*24*60*60);
if($weekAtYear == 1 && $firstDay != 1)
{//第一周,且第一天不是星期一
$days['sunday'] = date("Y-m-0$leftDayInWeek",strtotime($days['monday']));
}
else $days['sunday'] = date("Y-m-d",strtotime($days['monday'])+6*24*60*60);
return $days;
}
//use:
$date = getDays('2007',2);
//output:
print_r($date);

[多菜鸟]

<?php
/*
* 功能：取得给定日期所在周的开始日期和结束日期
* 参数：$gdate 日期，默认为当天，格式：YYYY-MM-DD
* $first 一周以星期一还是星期天开始，0为星期天，1为星期一
* 返回：数组array("开始日期", "结束日期");
* 作者：多菜鸟
* Email: kingerq AT msn.com
* 来源：http://blog.csdn.net/kingerq
*/
function aweek($gdate = "", $first = 0){
if(!$gdate) $gdate = date("Y-m-d");
$w = date("w", strtotime($gdate));//取得一周的第几天,星期天开始0-6
$dn = $w ? $w - $first : 6;//要减去的天数
$st = date("Y-m-d", strtotime("$gdate -".$dn." days"));
$en = date("Y-m-d", strtotime("$st +6 days"));
return array($st, $en);//返回开始和结束日期
}
echo implode("|", aweek("", 1));
?>


Trackback: http://tb.blog.csdn.net/TrackBack.aspx?PostId=326856

[Aylazhang]

一个比较简单的：
function get_spdate($s) {
$g = strftime("%u",strtotime($s));
$sdate = strftime("%Y-%m-%d",strtotime($s) - ($g-1)*86400);
$edate = strftime("%Y-%m-%d",strtotime($s) + (7-$g)*86400);
echo "Start Date:$sdate End Date:$edate \n";
}

$s = "2007-02-08";
get_spdate($s);

[foolbirdflyfirst]

T_T...
我的有错.....
参考以上两位

[吉林小江南]

看到代码了：）好感动．找了好久．．．
总之非常谢谢大家．．．．．

[foolbirdflyfirst]

你是根据周数来判断的哦?
你确定答案就是你想要的.
我的改了一下,计数错料.:-)

//function:
/**
* get week's first and last day
*
* @param string $year //year
* @param string $weekAtYear // the week num in given year
* @return array $days
*/
function getDays($year,$weekAtYear)
{
//找出年份第一天是星期几(1-7)
$firstDay = date('N',strtotime("$year-01-01"));
//找出第一个星期剩余天数,用于计数.比如2008-01-01号是星期二,则剩余6天到第一个星期天.
$leftDayInFirstWeek = 8 - $firstDay;
//星期一的日期(Y-m-d)
if($weekAtYear <= 2)
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+($weekAtYear-1)*$leftDayInFirstWeek*24*60*60);
}
else
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+$leftDayInFirstWeek*24*60*60+($weekAtYear-2)*7*24*60*60);
}
if($weekAtYear == 1 && $firstDay != 1)
{//第一周,且第一天不是星期一
$days['sunday'] = date("Y-m-0$leftDayInFirstWeek",strtotime($days['monday']));
}
else $days['sunday'] = date("Y-m-d",strtotime($days['monday'])+6*24*60*60);

return $days;
}
//use:
$date = getDays('2006',date("W"));
//output:
print_r($date);

[吉林小江南]

/**
# @param $year 年份
# @param $week 周次
# @return 例如：ＸＸＸＸ年第Ｘ周的起始日期和结束日期
*/
function getWeekStartTimeAndEndTime($year, $week)
{

}

[吉林小江南]

朋友们顶．．．．送出所有分也没问题．．

参数１　只有年份（２００７　｜　２００８　｜　２００９）
参数２　周次(第一周．．．．．)

[吉林小江南]

因为数据库表里是这样的：

id, usID ,year(年份), weekNum(周次), 等

在前台显示的时候我需要了解ＸＸＸＸ年ＸＸ周的起始结束日期．

[foolbirdflyfirst]

参考我最后给你的答案,测试一下.:-)

[吉林小江南]

表格
第一周：

　　星期一，星期二...等
　　ＸＸ号，ＸＸ号．．．
　　数据１，数据２

[吉林小江南]

foolbirdflyfirst(湖水清澈)：兄弟

好像是差了一天．

[吉林小江南]

$date = getDays('2007',3);
//output:
print_r($date);

[monday] => 2007-01-16　＝＝＞＜这个是星期二＞
[sunday] => 2007-01-22

[foolbirdflyfirst]

我说了你看看我最后发给你的,我改过啦!
======================================
//function:
/**
* get week's first and last day
*
* @param string $year //year
* @param string $weekAtYear // the week num in given year
* @return array $days
*/
function getDays($year,$weekAtYear)
{
//找出年份第一天是星期几(1-7)
$firstDay = date('N',strtotime("$year-01-01"));
//找出第一个星期剩余天数,用于计数.比如2008-01-01号是星期二,则剩余6天到第一个星期天.
$leftDayInFirstWeek = 8 - $firstDay;
//星期一的日期(Y-m-d)
if($weekAtYear <= 2)
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+($weekAtYear-1)*$leftDayInFirstWeek*24*60*60);
}
else
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+$leftDayInFirstWeek*24*60*60+($weekAtYear-2)*7*24*60*60);
}
if($weekAtYear == 1 && $firstDay != 1)
{//第一周,且第一天不是星期一
$days['sunday'] = date("Y-m-0$leftDayInFirstWeek",strtotime($days['monday']));
}
else $days['sunday'] = date("Y-m-d",strtotime($days['monday'])+6*24*60*60);

return $days;
}
//use:
$date = getDays('2007',3);
//output:
print_r($date);

=========
Array ( [monday] => 2007-01-15 [sunday] => 2007-01-21 )

[吉林小江南]

<?php
//function:
/**
* get week's first and last day
*
* @param string $year //year
* @param string $weekAtYear // the week num in given year
* @return array $days
*/
function getDays($year,$weekAtYear)
{
//找出年份第一天是星期几(1-7)
$firstDay = date('N',strtotime("$year-01-01"));
//找出第一个星期剩余天数,用于计数.比如2008-01-01号是星期二,则剩余6天到第一个星期天.
$leftDayInFirstWeek = 8 - $firstDay;
//星期一的日期(Y-m-d)
if($weekAtYear <= 2)
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+($weekAtYear-1)*$leftDayInFirstWeek*24*60*60);
}
else
{
$days['monday'] = date("Y-m-d",strtotime("$year-01-01")+$leftDayInFirstWeek*24*60*60+($weekAtYear-2)*7*24*60*60);
}
if($weekAtYear == 1 && $firstDay != 1)
{//第一周,且第一天不是星期一
$days['sunday'] = date("Y-m-0$leftDayInFirstWeek",strtotime($days['monday']));
}
else $days['sunday'] = date("Y-m-d",strtotime($days['monday'])+6*24*60*60);

return $days;
}
//use:
$date = getDays('2007',1);
//output:
print_r($date);
?>
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

X-Powered-By: PHP/4.4.4
Content-type: text/html

Array
(
[monday] => 2007-01-01
[sunday] => 2007-01-08===>应该为７Ｑ
)
还是错误．代码我已经换成你修改的了

[foolbirdflyfirst]

.....

N
ISO-8601
格式数字表示的星期中的第几天（PHP 5.1.0 新加） 1（表示星期一）到 7（表示星期天）

php5.1.0新加,我用php5,没问题.

//找出年份第一天是星期几(1-7)
$firstDay = date('N',strtotime("$year-01-01"));
============
改成

[foolbirdflyfirst]

$firstDay = date('w',strtotime("$year-01-01"));

[foolbirdflyfirst]

没php4,你再测试一下

$firstDay = date('N',strtotime("$year-01-01"));
============================
改成 $firstDay = date('w',strtotime("$year-01-01"));

php5下测试了一下,是通过的.