如何在php中管理日期顺序以显示距离下一个日期还有多少天？

我有一个开始日期为07/18/2014，结束日期为07/24/2014，周期日期为每2天。

所以，从07/18开始，每2天我向用户发出警告，提醒他们要做某件事情，如果不是正确的日期，我会提醒用户距离下一个警告还有多少天。

我该怎么做？如果还要加上时间和每2小时的周期呢？

我考虑先将所有的警告日期存储在一个数组中[07-18/2014, 07-20-2014, 07-22-2014, 07/24/2014]

我的代码如下，但它不起作用。可能我使用strtotime的方式不正确：

$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 "
 tempg in while:";
        echo $tempg .  " ";
        echo "
 tempg in while 2:";
        echo  date ( 'Y/m/d' , $tempg ) .  "";                     
    }

我也接受其他建议！

更新解决方案：

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"; // 警告后的一天
testWarning( $today3, $end, $delay );
$diffmin = 1000;                                        
for ($i = 0 ; $i<= count($ardata)-1; $i++){
    //print_r($ardata);
    echo "
 
 ardata ";
    echo $ardata[$i];
    $dataseq = date ( 'm/d/Y' , strtotime($ardata[$i]) );       
    $diffdata = dayDiff( strtotime($today3), strtotime($dataseq) );
    echo "

 DIFFDATA: ";
    echo $diffdata;
    echo " DIFFMIN ";
    echo $diffmin;
    if ($diffdata > 0){  // 避免今天之前的日期
        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){    // 今天的日期，所以警告
        echo "warning    WARNING 
 
";
    }else if ($diffmin > 0 ){
        echo " oggi ". $today3."  prossimo:". $nextdata. " 

";
        testWarning( $today3, $nextdata, $delay );   
    }else{
        // 错误情况
        echo " 
 
 ERRORE nel calcolo non può essere negativo";
    }                       
}