所以这是我的目标,我想知道多久以前某个日期是一个非常易读的字符串。类似于:'4天3小时前'。我正在使用datejs并且我的日期目前都是时代。
我知道我可以从另一个中减去一个纪元,取差异然后手动计算要输出的字符串。但这感觉就像datejs会处理的东西。但是我没有看到任何文件(任何js lib的文件)表示将日期对象或时期转换为一定时间的方法。
关于如何处理这个问题的任何想法?
更新
让我更清楚一点。当我说人类可读意味着
我最终做了什么
prettyAgo = (ago) ->
ago = ago.getTime() if typeof ago is 'object'
diff = ((new Date()).getTime() - ago) / 1000
dayDiff = Math.floor(diff / 86400)
return false if isNaN(dayDiff) or dayDiff < 0
writer = ''
if dayDiff is 0
if diff < 1 then writer = 'just now'
else if diff < 60 then writer = Math.floor(diff)+' seconds ago'
else if diff < 120 then writer = '1 minute ago'
else if diff < 3600 then writer = Math.floor(diff / 60)+' minutes ago'
else if diff < 7200 then writer = '1 hour ago'
else if diff < 86400 then writer = Math.floor(diff / 3600)+' hours ago'
else if dayDiff is 1 then writer = 'yesterday'
else if dayDiff < 7 then writer = dayDiff+' days ago'
else if dayDiff is 7 then writer = '1 week ago'
else if dayDiff < 31 then writer = Math.ceil( dayDiff / 7 )+' weeks ago'
else writer = new Date(ago).toString 'MMM yyyy'
return writer
答案 0 :(得分:2)
DateJS通过包含一个名为TimeSpan的类(在time.js中定义,而不是在基本JS文件中)为您进行计算来处理此问题。它没有格式化,所以你需要这样的东西:
function showDiff(date1, date2) {
var diff, rv, yrs;
diff = new TimeSpan(date2 - date1);
if (diff.days > 0) {
// A day or more difference
if (diff.days === 1) {
rv = "yesterday";
}
else if (diff.days >= 365) {
yrs = diff.days / 365;
if (yrs === 1) {
rv = "1 year ago";
}
else {
rv = yrs + " years ago";
}
}
else {
rv = diff.days + " ago";
}
}
else {
// Less than 1 day difference
if (diff.hours > 0) {
if (diff.hours === 1) {
rv = "1 hour ago";
}
else {
rv = diff.hours + " hours ago";
}
}
else if (diff.minutes > 0) {
if (diff.minutes === 1) {
rv = "1 minute ago";
}
else {
rv = diff.minutes + " minutes ago";
}
}
else {
rv = "just now";
}
return rv;
}
显然这并不完美,并没有处理闰年等等,但它应该给你这个想法。
答案 1 :(得分:1)
实际上你可以用纯JavaScript做到这一点。
function format(num) {
var date = new Date(num),
time = [];
with(time) {
push(date.getUTCDate() - 1);
push(date.getUTCHours());
push(date.getUTCMinutes());
push(date.getUTCSeconds());
}
if (time[0] > 29) {
time[0] = Math.floor(date / 86400000);
}
return time[0] + " day " + time[1] + " hours " + time[2] + " mintues " + time[3] + " seconds ago";
}