Java time difference

Question

从mysql获取时间戳，它返回类似于"2014-03-19 12:00:43"的东西，在java中，我想从时间戳和当前时间中获得差异。

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

date_def

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

date_default_timezone_set('America/New_York');
$server = timesince($row['timestamp']);
function timesince($time){
    $seconds = strtotime($time) - time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到java新手和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

date_default_timezone_set('America/New_York');
$server = timesince($row['timestamp']);

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

date_default_timezone_set('America/New_York');
$server = timesince($row['timestamp']);
function timesince($time){
    $seconds = strtotime($time) - time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;
function timesince($time){
    $seconds = strtotime($time) - time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;
    $seconds = strtotime($time) - time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;ault_timezone_set('America/New_York');

新的java和非常不熟悉，这是我在php虽然....任何类似的东西都可以达到相同的效果

date_default_timezone_set('America/New_York');
$server = timesince($row['timestamp']);
function timesince($time){
    $seconds = strtotime($time) - time();
    $days = floor($seconds/ -86400);
    $seconds %= 86400;
    $minutes = floor($seconds / -60);
    $seconds %= 60;
    $seconds *= -1;
    if ($days > 360){
        return "Not voted here yet.";
    }
    if($days > 1){
        $days = $days." days, ";
    } elseif ($days == 1){
        $days = $days." day, ";
    } else {
        $days  = "";
    }
    if($hours > 1){
        $hours = $hours." hours, ";
    }elseif ($hours == 1){
        $hours = $hours." hour, ";
    } else {
        $hours  = "";
    }
    if($minutes > 1){
        $minutes = $minutes." minutes and ";
    }elseif ($minutes == 1){
        $minutes = $minutes." minute and ";
    }else {
        $minutes  = "";
    }
    if($seconds > 1){
        $seconds = $seconds." seconds";
    }elseif ($seconds == 1){
        $seconds = $seconds." second";
    }
    return $days.$hours.$minutes.$seconds." ago";
}

我在java中没有看到任何类似于strtotime的东西，也不想使用任何额外的库。

同样在php脚本中我在顶部有

date_default_timezone_set('America/New_York');

所以它得到正确的时差与时区的考虑。

从哪里开始在java中做类似的事情呢?

Answer 1

Javascript的原生Date对象能够非常容易地复制php的strtodate函数。要调用它，只需使用:

var d = new Date(dateString);

(参见http://www.w3schools.com/jsref/jsref_obj_date.asp了解更多信息)因此，您可以使用如下函数:

function timesince( dateString ) {
    var originalTime = new Date();          // The time you provided in milliseconds converted to javascript object
    var currentTime = new Date();                       // The current time as a javascript object
    return millisecondDifference = parseFloat(currentTime.getTime()) - parseFloat(originalTime.getTime());
    // returns the difference in milliseconds between the current time and the original time
    // parseFloat is used to ensure that the values returned are read by javascript as floats (numbers) and thus can have math operations performed on them
}

以毫秒为单位返回差值。然后，如果您想将其转换为"人类可读"格式，您可以使用一个新的Date对象，如下所示:

var diffRaw = timesince( '2014-03-19 12:00:43' );
var diff = new Date(diffRaw);
var seconds = diff.getSeconds() // Seconds from 0 - 59
var minutes = diff.getMinutes() // Minutes from 0 - 59
var hours = diff.getHours()     // Hours from 0 - 23
var day = diff.getDay()         // Days from 0 - 6
var date = diff.getDate()       // Date from 1 - 31
var month = diff.getMonth()     // Months from 0 - 11
var year = diff.getFullYear()   // XXXX year starting from 1970

要完全复制您所编写的函数，您可以使用以下函数:

function timesince( dateString ) {
    var originalTime = new Date(dateString);
    var currentTime = new Date();
    var diffRaw = millisecondDifference = parseFloat(currentTime.getTime()) - parseFloat(originalTime.getTime());
    var diff = new Date(diffRaw);
    var seconds = diff.getSeconds();
    var minutes = diff.getMinutes();
    var hours = diff.getHours();
    var day = diff.getDay();
    var date = diff.getDate() - 1;
    var month = diff.getMonth();
    var year = diff.getFullYear();
    if( year !== 1970 ) {
        return "Not voted here yet.";
    }
    var returnString = "";
    if(month > 1 || month == 0) {
        returnString += month + " months, ";
    } else {
        returnString += month + " month, ";
    }
    if(date > 1 || date == 0) {
        returnString += date + " days, ";
    } else {
        returnString += date + " day, ";
    }
    if(hours > 1 || hours == 0) {
        returnString += hours + " hours, ";
    } else {
        returnString += hours + " hour, ";
    }
    if(minutes > 1 || minutes == 0) {
        returnString += minutes + " minutes, ";
    } else {
        returnString += minutes + " minute, ";
    }
    if(seconds > 1 || seconds == 0) {
        returnString += seconds + " seconds, ";
    } else {
        returnString += seconds + " second, ";
    }
    return returnString;
}

你可以在这里找到一个工作示例:http://jsfiddle.net/MQPPw/

Answer 2

在Java(或PHP)中不要这样做。这是在MySQL中完成的，使用它的日期/时间函数。

SELECT UNIX_TIMESTAMP() - UNIX_TIMESTAMP(my_timestamp_col) AS seconds_ago ...

第一个方法返回当前时间作为unix时间戳(即从epoch开始的秒数)。第二个方法返回my_timestamp_col时间作为unix时间戳。这两个相减，你就得到了my_timestamp_col在多少秒前。

EDIT:正如Alexander指出的那样，当您考虑时区因素时，情况会变得更加复杂。一种选择是使用CONVERT_TZ函数将my_timestamp_col转换为UTC，然后调用UNIX_TIMESTAMP:

UNIX_TIMESTAMP(CONVERT_TZ(timestamp, '-5:00', '+0:00')) 

另一个选择是在Java端这样做，毕竟:获取字段作为Date，它应该为您做时区转换;然后调用它的getTime()方法来获得自epoch以来的毫秒数，并从System.currentTimeMillis()中减去它。它给出了事件发生的毫秒数