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【转】php 得到上周,本周,上月,本月,本季度,上季度

2013-01-10 14:33 232 查看
转自:http://www.the8m.com/blog/article/phpa/phpwek.html

注意强大的strtotime()函数和mktime()函数。

<?php
echo date("Ymd",strtotime("now")), "\n";
echo date("Ymd",strtotime("-1 week Monday")), "\n";
echo date("Ymd",strtotime("-1 week Sunday")), "\n";
echo date("Ymd",strtotime("+0 week Monday")), "\n";
echo date("Ymd",strtotime("+0 week Sunday")), "\n";

echo "*********第几个月:";
echo date('n');
echo "*********本周周几:";
echo date("w");
echo "*********本月天数:";
echo date("t");
echo "*********";

echo '<br>上周:<br>';
echo date("Y-m-d H:i:s",mktime(0, 0 , 0,date("m"),date("d")-date("w")+1-7,date("Y"))),"\n";
echo date("Y-m-d H:i:s",mktime(23,59,59,date("m"),date("d")-date("w")+7-7,date("Y"))),"\n";
echo '<br>本周:<br>';
echo date("Y-m-d H:i:s",mktime(0, 0 , 0,date("m"),date("d")-date("w")+1,date("Y"))),"\n";
echo date("Y-m-d H:i:s",mktime(23,59,59,date("m"),date("d")-date("w")+7,date("Y"))),"\n";

echo '<br>上月:<br>';
echo date("Y-m-d H:i:s",mktime(0, 0 , 0,date("m")-1,1,date("Y"))),"\n";
echo date("Y-m-d H:i:s",mktime(23,59,59,date("m") ,0,date("Y"))),"\n";
echo '<br>本月:<br>';
echo date("Y-m-d H:i:s",mktime(0, 0 , 0,date("m"),1,date("Y"))),"\n";
echo date("Y-m-d H:i:s",mktime(23,59,59,date("m"),date("t"),date("Y"))),"\n";

$season = ceil((date('n'))/3);//当月是第几季度
echo '<br>本季度:<br>';
echo date('Y-m-d H:i:s', mktime(0, 0, 0,$season*3-3+1,1,date('Y'))),"\n";
echo date('Y-m-d H:i:s', mktime(23,59,59,$season*3,date('t',mktime(0, 0 , 0,$season*3,1,date("Y"))),date('Y'))),"\n";

$season = ceil((date('n'))/3)-1;//上季度是第几季度
echo '<br>上季度:<br>';
echo date('Y-m-d H:i:s', mktime(0, 0, 0,$season*3-3+1,1,date('Y'))),"\n";
echo date('Y-m-d H:i:s', mktime(23,59,59,$season*3,date('t',mktime(0, 0 , 0,$season*3,1,date("Y"))),date('Y'))),"\n";
?>

php两个日期相减得天数
<?php
function count_days($a,$b){
$a_dt=getdate($a);
$b_dt=getdate($b);
$a_new=mktime(12,0,0,$a_dt['mon'],$a_dt['mday'],$a_dt['year']);
$b_new=mktime(12,0,0,$b_dt['mon'],$b_dt['mday'],$b_dt['year']);
return round(abs($a_new-$b_new)/3600/24);
}
function count_days($formdate,$todate){
return round(abs(strtotime($formdate)-strtotime($todate))/3600/24);
}
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