如何在php javascript中在数组中存储日期和时间序列？

我有一个数据库，其中存储在一个字段中的数据为“startdata”，第二个字段中的时间为“starttime”，第二个字段中的频率为：“每2天”或“每2小时”，最后是第三个字段中的“Update”（+4天）和一个字段中的“UptimeTime”（+48小时）

我想要一个php页面，当今天是范围startdate-->Update中的一个日期时显示警报，如果不是，则告诉我下一个警报还有多少天/小时”

我第一次想到的只是考虑数据字段。 我将从startdate开始到“Update”字段的所有数据/时间以“while”周期存储在一个数组中（即startdate 07/13和+4天-->范围为：07/13-->07/15-->07/17）

我计算enddata：

$endg = strtotime ( "+". $dataupto . " days", strtotime ( $row['startdate'] ) ) ; 

从startdata开始，以enddata结束，在每个步骤中，我都执行以下操作：

while (startdata  up to enddata){
    tempdata = startdata + frequence
    arraymonitor[] = tempdata
}

我的代码是：

$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/>";                     


        }

和订购：

rsort($arraytemp); 

print(" </br> ordered array");
print_r($arraytemp);

rsort（$arraytemp）；
打印（“
有序数组”）；
打印（$arraytemp）；

在这里，我在arraymonitor中搜索并比较今天的日期与其他日期，并计算距离第二天还有多少天：

if ( in_array($today, $arraymonitor) ){
    echo "Today u have to monitor";

}else{
    //calculate how many days remaining to the next day


    $arraytemp = $arraymonitor;             
    $arraytemp[]= strtotime($today);   //I add today date in the array      
    rsort($arraytemp); //I ordered the array    
    print(" </br> orderedarray ");
    print_r($arraytemp);
    $keytoday = array_search($oggi, $arraystore);   // I find the data-today key    
    $proxmonitor = $arraystore[$chiaveoggi+1];   // I add "1" that it is the next day, **right**?

if（在数组中（$today$arraymonitor））{
回声“今天你必须监视”；
}否则{
//计算下一天还有多少天
$arraytemp=$arraymonitor；
$arraytemp[]=strotime（$today）；//我在数组中添加了今天的日期
rsort（$arraytemp）；//我对数组进行了排序
打印（“
orderedarray”）；
打印（$arraytemp）；
$keytoday=array_search（$oggi，$arraystore）；//我找到了data today键
$proxmonitor=$arraystore[$chiaveoggi+1]；//我加上“1”表示这是第二天，**对吧**？

我在这里打印结果：

    echo " </br> prox day monitor ";
    print(date("Y/m/d",$proxmonitor));  

    $timedayforthenextdata = $proxmonitor - $today;   
    $timedayforthenextdata = (int) $timedayforthenextdata / 60/ 60/ 24;  
    echo "</br> Miss " . $timedayforthenextdata . " days/hours to the next monitoring";

echo“
prox日间监视器”；
打印（日期（“Y/m/d”，“proxmonitor”）；
$timedayforthenextdata=$proxmonitor-$today；
$timedayforthenextdata=（int）$timedayforthenextdata/60/60/24；
echo“
未命中“$timedayforthenextdata.”到下一次监视的天数/小时数”；

我在“while”循环中出错，因为它没有存储所有数据！ 为什么？我哪里错了？还有别的解决办法吗

我的所有代码：

<?php   
$oggi = date("Y/m/d");

foreach($db->query("SELECT startdate, starthour dataupto, hourupto dataevery, hourevery FROM event ") as $row) { 
    #echo $row['annotazione'].' '.$row['Prescrizione']; //etc...



    $monitevery = $row['dataevery'];
    $monitupto = $row['dataupto'];


    $start = $row['startdata'];

    $today = date('Y/m/d');


    $today2 = strtotime($row['startdata']);
    $today2 = strtotime($today);

    echo "start monitor ";
    echo $row['startdate'];
    echo "</br> oggi ";
    echo $today;                    
    $ris = $today2 - $start2;
    $rishour =  (int) ($ris/ 60)/ 60;   // result in hours
    $risdays =  (int) $ris/ 60/ 60/ 24;   // result in hours
    echo "</br> result ";
    echo $risdays;
    //echo gettype($a);
    //echo gettype($b);                 


    $elapseddays = (int) ( strtotime($today) - strtotime($row['startdate'])    )/60/ 60/ 24 ;  //  //today less startdate i find how is passed
    $daysremained = ($row['startdate'] + $monitupto  ) - $today;   // ( startmonit+every = end day )- today =  how may days remains

    echo " <br/>";
    echo "today " .$oggi . "<br/>";
    echo "monitor start " . $row['stardate']. "<br/>";
    echo " days elapsed ";
    echo $elapseddays;   
    if ($elapseddays > 0){ 

        $firstcycleelapsed = strtotime($row['startdate']) + strtotime($row['dataevery']]);  // 20 + 2 = 22


        $dataupto = $row['dataupto'];  
        echo "</br>  dataupto ";
        echo $dataupto;

        $tempgSTR = strtotime($row['startdate']) ; 

        //end day
        $endg = strtotime ( "+". $dataupto . " days", strtotime ( $row['startdate'] ) ) ;     



        $endg = date ( 'Y/m/d' , $endg ); 
        $finegSTR = strtotime($fineg);
        echo " </br> tempg1 " . $row['startdate'] . "<br/>";
        echo " </br> tempg1STR " . $tempgSTR . "<br/>";
        echo " endg1 " . $endg . "<br/>";
        echo " endg1STR " . $endgSTR . "<br/>";


    }

        $arraymonitor = array();

        $tempg = strtotime($row['startdate']);  // i start with the startdate
        echo "</br> after IF dayselapsed>0, before  WHILE:  </br> tempg ";
        echo $tempg . " </br> endg " ;  //tempg e instrotime
        echo $finegSTR;

        $tempgdate = $row['startdate'] ;
        while( $tempg < $finegSTR){   

            $arraymonitor[] = strtotime(  date($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/>";                     


        }
        echo "</br>END WHILE:  array date: ";
        print_r($arraymonitor);


        if ( in_array($today, $arraymonitor) ){
            echo "otoday u have to monitor";

        }else{
            //calculate how many days remaining to the next day


            $arraytemp = $arraymonitor;

            $arraytemp[]= strtotime($today);   //i add today date in the array                      
            rsort($arraytemp); //I order the array

            print(" </br> orderedarray ");
            print_r($arraytemp);
            $keytoday = array_search($otoday, $arraytemp);   // i look for the key of today date

            $proxmonitor = $arraytemp[$keytoday+1];   // i add "1" to know what day is the next for showing the alert


            echo " </br> next day of monitor ";
            print(date("Y/m/d",$proxmonitor)); 

            $howmanydaysremain = $proxmonitor - $today;   
            $howmanydaysremain = (int) $howmanydaysremain  / 60/ 60/ 24;  
            echo "</br> reamining " . $howmanydaysremain  . " days  at next monitor";



        }  // end else

    }


}// end for each

?>

---

在第一个代码中，更改

while(){
    ...
    $tempg = $tempg + (strtotime ( " +1 days", $tempg  ) );
    ...
}

到

---

然后它将正确计算“+1天”，并将该日期存储到数组中

$arraymonitor

这其中的JavaScript部分是什么？我用php做了这件事，但如果我也能用JavaScript解决它，那就好了！

while(){
    ...
    $tempg = $tempg + (strtotime ( " +1 days", $tempg  ) );
    ...
}

while(){
    ...
    $tempg = strtotime ( " +1 days", $tempg  );
    ...
}