php - 将日期转换为日期时间时出错

Question

我使用 jQuery 日期选择器在我的输入日期中选择一个日期。

我在日期选择器脚本中使用的格式是:dateFormat: 'DD, d MM, yy',

所以我在输入中得到了这个日期:“Quinta-feira, 1 Maio, 2014”(葡萄牙语日期)。

但现在我需要转换这个日期以像 mysql 中的日期时间一样保存。

如果日期是英文，我只需要使用下面的代码:

$date = DateTime::createFromFormat('l, j F, Y', $_POST['date']);
echo $date->format('Y-m-d');    

但我的日期不是英语，所以我需要进行转换，我正在尝试使用下面的“convertDate”函数来完成此操作。

但是当我调用该函数时，我会像这样传递输入日期值:convertDate($_POST['date']);

我得到一个错误“在非对象上调用成员函数 format() 在这一行:$day= $date->format ("l");

你看到这里有什么不对吗？因为这个功能对我来说似乎不错!

function convertDate($myDate){

        $date = DateTime::createFromFormat('Ymd', $myDate);
        $day    = $date->format("l");
        $daynum = $date->format("j");
        $month  = $date->format("F");
        $year   = $date->format("Y");

        switch($day)
        {
            case "Segunda-Feira":    $day = "Monday";  break;
            case "Terça-Feira":      $day = "Tuesday"; break;
            case "Quarta-Feira":     $day = "Wednesday";  break;
            case "Quinta-Feira":     $day = "Thursday"; break;
            case "Sexta-Feira":      $day = "Friday";  break;
            case "Sábado":           $day = "Saturday";  break;
            case "Domingo":          $day = "Sunday";  break;
            default:                 $day = "Unknown"; break;
        }

        switch($month)
        {
            case "Janeiro":    $month = "January";    break;
            case "Fevereiro":  $month = "February";   break;
            case "Março":      $month = "March";     break;
            case "Abril":      $month = "April";     break;
            case "Maio":       $month = "May";       break;
            case "Junho":      $month = "June";      break;
            case "Julho":      $month = "July";      break;
            case "Agosto":     $month = "August";    break;
            case "Setembro":   $month = "September"; break;
            case "Outubro":    $month = "October";   break;
            case "Novembro":   $month = "November";  break;
            case "Dezembro":   $month = "December";  break;
            default:           $month = "Unknown";   break;
        }

    echo $daynum . ", " . $month . ", " . $year;
}

尝试使用 str_ireplace:

$english = array("Segunda-Feira","Terça-Feira","Quarta-Feira","Quinta-Feira","Sexta-Feira","Sábado","Domingo");
$portuguese = array("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"); 
$result= str_ireplace ($english , $portuguese, $_POST['date']);
$date = DateTime::createFromFormat('l, j F, Y', $result);
echo $date->format('Y-m-d');

我在 echo $date->format('Y-m-d') 中遇到了同样的错误

Answer 1

您可以让 JQuery 将某种 UTC 放入隐藏的表单字段中，而不是尝试用可能的各种语言翻译日期，然后使用它:

这将返回一个 JS 日期对象:

   var currentDate = $( ".selector" ).datepicker( "getDate" );

这将从它生成一个 PHP UTC(在 JS 中是微秒...)

   var phputc = Math.ceil((currentDate.getTime()/1000));

这将以隐藏形式放入:

   $(".selector" ).change(
        function()
        {
             var currentDate = $( ".selector" ).datepicker( "getDate" );
             var phputc = Math.ceil((currentDate.getTime()/1000));
             document.forms[0].nameofhiddenformfield.value = phputc;
        }
   );

确保你的表单中有这样的字段

   <input type="hidden" name="nameofhiddenformfield" />

然后在 PHP 端，您可以执行以下操作:

   $dateUTC = $_POST['nameofhiddenformfield'];
   echo date('Y-m-d',$dateUTC);