I have a start date like 07/18/2014, a end date like 07/24/2014, and a period date: every 2 days. So every 2 days starting by 07/18 i warn the user that something have to do, if is not the right day i warn the user about how left days remaining to the next warning. How can i do? And if add also time and a period of every 2 hours?
我的开始日期为2014年7月18日,结束日期为2014年7月24日,期限为每2天一次。因此,从07/18开始每隔2天我会警告用户必须要做的事情,如果不是正确的一天,我会警告用户剩下的剩余天数到下一个警告。我能怎么做?如果还加上时间和每2个小时的一段时间?
I thought about store first all the dates warning in an array [07-18/2014, 07-20-2014, 07-22-2014, 07/24/2014]
我想到了存储在阵列中的所有日期警告[07-18 / 2014,07-20-2014,07-22-2014,07 / 24/2014]
My code is the following but it does not work. Maybe it is not correct like I use strtotime
我的代码如下,但它不起作用。也许这不像我使用strtotime那样正确
$endg = strtotime ( "+". $dataupto . " days", strtotime ( $row['startdate'] ) ) ;
$endg = date ( 'Y/m/d' , $endg );
$endgSTR = strtotime($endg);
$tempg = strtotime($row['startdate']);
while( $tempg < $endgSTR){
$arraymonitor[] = strtotime( date($tempgdate ) );
$tempg = $tempg + (strtotime ( " +1 days", $tempg ) );
$arraymonitor[] = $tempg;
echo "</br> tempg in while:";
echo $tempg . " ";
echo "</br> tempg in while 2:";
echo date ( 'Y/m/d' , $tempg ) . "<br/>";
}
I accept also other suggestions!
我也接受其他建议!
UPDATE SOLUTION
header( "content-type: text-plain" );
function dayDiff($start, $end){
$timeleft = $end - $start;
$daysleft = round((($timeleft/24)/60)/60);
return $daysleft;
}
function testWarning($today, $end, $delay){
$endDate = strtotime($end);
$warningDate = $endDate;
$todayDate = strtotime($today);
if( $todayDate == $warningDate ){
echo "Oggi c'è un controllo da fare";
}elseif( $todayDate < $warningDate ){
echo "Miss " . dayDiff($todayDate, $warningDate) . " days";
}else{
echo "warning was " . abs(dayDiff($todayDate, $warningDate)) . " giorni fa";
}
echo"\n";
}
$ardata = [07/18/2014, 07/20/2014, 07/22/2014, 07/24/2014];
$today3 = "07/21/2014"; // the day after warning
testWarning( $today3, $end, $delay );
$diffmin = 1000;
for ($i = 0 ; $i<= count($ardata)-1; $i++){
//print_r($ardata);
echo "</br> </br> ardata ";
echo $ardata[$i];
$dataseq = date ( 'm/d/Y' , strtotime($ardata[$i]) );
$diffdata = dayDiff( strtotime($today3), strtotime($dataseq) );
echo "</br></br> DIFFDATA: ";
echo $diffdata;
echo " DIFFMIN ";
echo $diffmin;
if ($diffdata > 0){ // avoiding days before today
if ($diffdata < $diffmin){
$diffmin = $diffdata;
$nextdata = $ardata[$i];
}else{
if ($diffdata == -1){
echo "error array empty";
$nextdata = $today3;
}
}
}else{
echo "monitorterminato";
$nextdata = -1;
}
}
if ($nextdata != -1){
if ($diffmin == 0){ // giorno di oggi quindi avviso
echo "warning WARNING </br> </br>";
}else if ($diffmin > 0 ){
echo " oggi ". $today3." prossimo:". $nextdata. " </br></br>";
testWarning( $today3, $nextdata, $delay );
}else{
// caso errato
echo " </br> </br> ERRORE nel calcolo non può essere negativo";
}
}
2 个解决方案
#1
0
Here is an example on how to calculate your warning dates and days left before warning.
以下是有关如何计算警告日期和警告前几天的示例。
<?php
header( "content-type: text-plain" );
$end = "09/18/2014";
$delay = "- 2 days"; // warning will be on 16th
$warningDay = date('m-d-Y', strtotime($delay, strtotime($end)));
echo "Warning is on " . $warningDay . "\n";
$today1 = "09/13/2014"; // 3 days before warning
testWarning( $today1, $end, $delay );
$today2 = "09/16/2014"; // warning day
testWarning( $today2, $end, $delay );
$today3 = "09/17/2014"; // the day after warning
testWarning( $today3, $end, $delay );
function testWarning($today, $end, $delay){
$endDate = strtotime($end);
$warningDate = strtotime($delay, $endDate);
$todayDate = strtotime($today);
echo "today : " . date('m-d-Y', $todayDate) . "->";
if( $todayDate == $warningDate ){
echo "Warning is today";
}elseif( $todayDate < $warningDate ){
echo "Warning in " . dayDiff($todayDate, $warningDate) . " days";
}else{
echo "Warning was " . abs(dayDiff($todayDate, $warningDate)) . " days ago";
}
echo"\n";
}
// $start and $end are timestamps
function dayDiff($start, $end){
$timeleft = $end - $start;
$daysleft = round((($timeleft/24)/60)/60);
return $daysleft;
}
#2
0
As seen here
如此处所见
$future = strtotime('21 July 2012'); //Future date.
$timefromdb = //source time
$timeleft = $future-$timefromdb;
$daysleft = round((($timeleft/24)/60)/60);
echo $daysleft;
#1
0
Here is an example on how to calculate your warning dates and days left before warning.
以下是有关如何计算警告日期和警告前几天的示例。
<?php
header( "content-type: text-plain" );
$end = "09/18/2014";
$delay = "- 2 days"; // warning will be on 16th
$warningDay = date('m-d-Y', strtotime($delay, strtotime($end)));
echo "Warning is on " . $warningDay . "\n";
$today1 = "09/13/2014"; // 3 days before warning
testWarning( $today1, $end, $delay );
$today2 = "09/16/2014"; // warning day
testWarning( $today2, $end, $delay );
$today3 = "09/17/2014"; // the day after warning
testWarning( $today3, $end, $delay );
function testWarning($today, $end, $delay){
$endDate = strtotime($end);
$warningDate = strtotime($delay, $endDate);
$todayDate = strtotime($today);
echo "today : " . date('m-d-Y', $todayDate) . "->";
if( $todayDate == $warningDate ){
echo "Warning is today";
}elseif( $todayDate < $warningDate ){
echo "Warning in " . dayDiff($todayDate, $warningDate) . " days";
}else{
echo "Warning was " . abs(dayDiff($todayDate, $warningDate)) . " days ago";
}
echo"\n";
}
// $start and $end are timestamps
function dayDiff($start, $end){
$timeleft = $end - $start;
$daysleft = round((($timeleft/24)/60)/60);
return $daysleft;
}
#2
0
As seen here
如此处所见
$future = strtotime('21 July 2012'); //Future date.
$timefromdb = //source time
$timeleft = $future-$timefromdb;
$daysleft = round((($timeleft/24)/60)/60);
echo $daysleft;