php日期差異

我試圖在兩個日期之間獲得天數差異。第一天是當天。另一天是他們的生日。我想計算今天和他們下一個生日之間的差異。目前，如果用戶今天輸入他們的bday，它會正確輸出，但是其他任何東西都會被混淆。我想按天計算一下。php日期差異

例如，如果我輸入我的bday爲01/11/1988爲例，我的程序將輸出0 Years, 1 Months, 21 Days。如果我能夠正確輸出年，月，日，那麼我可以計算總天數，但截至目前，它只是不能正常工作。謝謝參觀。這裏是我的代碼：

if (isset($_POST["birthday"])){ 
    $birthday=$_POST["birthday"]; 
    if (!empty($birthday)){ 
     $birthdayArray = explode("/", $birthday); 
     if (count($birthdayArray) != 3 || strlen($birthday) != 10){ 
      echo "<b>Please follow the format (ex. 09/15/1988).</b><br/>"; 
      }else{ 
       $month = $birthdayArray[0]; 
       $day = $birthdayArray[1]; 
       $year = $birthdayArray[2]; 
       echo "Your bday is: ".$month."/".$day."/".$year; 
       echo "<br/>"; 
       if (strcmp($currentMonth,$month) == 0 && strcmp($currentDay,$day) == 0){ 
        echo "Your birthday is today.<br/>"; 
       }else{ 
        $date1 = new DateTime("$currentYear-$currentMonth-$currentDay"); 
        if (strcmp($currentMonth, $month) <= 0 && strcmp($currentDay, $day) < 0){ 
         $currentYear = $currentYear + 1; 
         $date2 = new DateTime("$currentYear-$month-$day"); 
        }else{ 
         $date2 = new DateTime("$currentYear-$month-$day"); 
        } 
        $interval = $date1->diff($date2); 
        echo $interval->y." Years, ".$interval->m." Months, ".$interval->d." Days."; 

       } 
      } 
    }else{ 
     echo "<b>Please enter your birthday.</b><br/>"; 
    } 
}

來源

2012-03-01 stytown

這需要進行基本的調試。什麼時候出問題了？測試輸出是什麼'month'，''day'和'year'顯示？ – 2012-03-01 08:34:11

它們顯示用戶輸入的日期，例如，如果我輸入09/15/1988，那麼$ month將是09，$ day將是15，$ year將是1988. – stytown 2012-03-01 08:36:21

好吧，然後進一步調試。每次操作後都要測試每個值的輸出。它在哪一點脫軌？ – 2012-03-01 08:49:06

if (strcmp($currentMonth, $month) <= 0 && strcmp($currentDay, $day) < 0){ 
      $currentYear = $currentYear + 1;

這是什麼情況實際上在做什麼？

如果當前月份小於生日月份，則表示當前年份增加。但你應該以相反的方式做。

可能是你想實現這一點：

//assume BIRTHDAY : 03/24/1990  
if(...) 
    echo "today birthday "; 
    else 
    { 
    if(currentMonth > $month) // may be today : 04/21/2012 
    { 
    $currentYear++; 
    } 
    else if(currentMonth == $month) // may be today : 03/29/2012 
    { 
    if($currentDay > $day) // may be today : 03/29/2012 
    { $currentYear++; } 
    } 
    } 
    $date2 = new DateTime("$currentYear-$month-$day");

來源

2012-03-01 09:18:48 Makesh

請看看這段代碼，它給了你的願望。

if (isset($_POST["birthday"])){ 
    $birthday=$_POST["birthday"]; 
    if (!empty($birthday)){ 
     $birthdayArray = explode("/", $birthday); 
     if (count($birthdayArray) != 3 || strlen($birthday) != 10){ 
      echo "<b>Please follow the format (ex. 09/15/1988).</b><br/>"; 
      }else{ 
       $month = $birthdayArray[0]; 
       $day = $birthdayArray[1]; 
       $year = $birthdayArray[2]; 
       $currentMonth = date('m'); 
       $currentDay = date('d'); 
       $currentYear = date('Y'); 
       echo "Your bday is: ".$month."/".$day."/".$year; 
       echo "<br/>"; 
       if (strcmp($currentMonth,$month) == 0 && strcmp($currentDay,$day) == 0){ 
        echo "Your birthday is today.<br/>"; 
       }else{ 
        $date1 = new DateTime("$currentYear-$currentMonth-$currentDay"); 
        if (strcmp($currentMonth, $month) <= 0 && strcmp($currentDay, $day) < 0){ 
         $currentYear = $currentYear + 1; 
         $date2 = new DateTime("$currentYear-$month-$day"); 
        }else{ 
         $date2 = new DateTime("$currentYear-$month-$day"); 
        } 
        $interval = $date1->diff($date2); 
        echo $interval->y." Years, ".$interval->m." Months, ".$interval->d." Days."; 

       } 
      } 
    }else{ 
     echo "<b>Please enter your birthday.</b><br/>"; 
    } 
}

來源

2012-03-01 10:02:15 Kamal

回答

相關問題