Php 如何在开始日和结束日之间每天输出?
我想在开始日和结束日之间每天输出,例如:Php 如何在开始日和结束日之间每天输出?,php,date,Php,Date,我想在开始日和结束日之间每天输出,例如: $start_day = '20150530'; $end_day = '20150602'; //the output should be array('20150530', '20150531', '20150601', '20150602'); print_r(output_days($start_day, $end_day)); function output_days($start_day, $end_day) { // any i
$start_day = '20150530';
$end_day = '20150602';
//the output should be array('20150530', '20150531', '20150601', '20150602');
print_r(output_days($start_day, $end_day));
function output_days($start_day, $end_day) {
// any idea?
}
谢谢。
<?php
$date_from = strtotime("10 September 2000");
$date_to = strtotime("15 September 2000");
$day_passed = ($date_to - $date_from); //seconds
$day_passed = ($day_passed/86400); //days
$counter = 1;
$day_to_display = $date_from;
while($counter < $day_passed){
$day_to_display += 86400;
echo $day_to_display;
$counter++;
}
?>
$start_day='20150530';
$end_day='20150602'代码>
//输出应该是数组('20150530','20150531','20150601','20150602')代码>
print_r(输出日($start_day,$end_day))代码>
函数输出天数($start\u day,$end\u day){
$strotime\u start=strotime($start\u day);
$strotime\u end=strotime($end\u day);
$totaldays=$strotime\u end-$strotime\u start;
$x=0
$dates[0]=日期('Ymd',$STROTTIME\U star)
而($x这也许是你想要的
function alldays($start_date,$end_date,$format_date="Y-m-d") {
$seconds = strtotime($end_date) - strtotime($start_date);
for ($i=0; $i < $seconds ; $i+=(60*60*24)) {
$dates[] = date($format_date, strtotime($start_date)+$i);
}
return $dates;
}
//Tiny test
print_r(alldays('1999-01-01','1999-02-01'));
function alldays($start\u date,$end\u date,$format\u date=“Y-m-d”){
$seconds=strotime($end\u date)-strotime($start\u date);
对于($i=0;$i<$s;$i+=(60*60*24)){
$dates[]=日期($format_date,strottime($start_date)+$i);
}
返回$日期;
}
//微小试验
打印(所有日期('1999-01-01','1999-02-01');
$start_day='20150530';
$end_day='20150602';
功能输出日($start日,$end日){
而(strotime($start\u day)为什么人们写得这么复杂
$start = mktime(0,0,0,9,10,2000);
$end = mktime(0,0,0,9,15,2000);
while($start<=$end)
{
$output[]=date("Ymd",$start);
$start+=86400;
}
$start=mktime(0,0,0,9,102000);
$end=mktime(0,0,0,9,152000);
虽然这是个好问题,但这应该是最好的答案
$start = mktime(0,0,0,9,10,2000);
$end = mktime(0,0,0,9,15,2000);
while($start<=$end)
{
$output[]=date("Ymd",$start);
$start+=86400;
}