如何修改此功能以考慮“月”和“年”？

Question

function prettyDate(time){
    var date = time,
        diff = (((new Date()).getTime() - date.getTime()) / 1000),
        day_diff = Math.floor(diff / 86400);
    if ( isNaN(day_diff) || day_diff < 0 || day_diff >= 31 ){
        return;
    }
    return day_diff == 0 && (
            diff < 60 && Math.floor(diff) + " seconds" ||
            diff < 120 && "1 minute" ||
            diff < 3600 && Math.floor( diff / 60 ) + " min" ||
            diff < 7200 && "1 hour" ||
            diff < 86400 && Math.floor( diff / 3600 ) + " hours") ||
        day_diff == 1 && "1 day" ||
        day_diff < 7 && day_diff + " days" ||
        day_diff < 31 && Math.ceil( day_diff / 7 ) + " weeks";
}

我的朋友幫我編寫了將日期轉換為“漂亮日期”的功能。 問題是，現在它無法處理幾個月。 如果您查看該代碼，則當天數差超過31天時，它將不返回任何內容。

我要怎么做才能使這項工作幾個月又幾年？

最后一行是否會解決這個問題？

Math.ceil( day_diff / 31 ) + " months";

Answer 1

您需要在return表達式的末尾添加一些子句，並同時刪除day_diff >= 31保護day_diff >= 31 ：

function prettyDate(time){
    var date = time,
        diff = (((new Date()).getTime() - date.getTime()) / 1000),
        day_diff = Math.floor(diff / 86400);
    if ( isNaN(day_diff) || day_diff < 0){
        return;
    }
    return day_diff == 0 && (
            diff < 60 && Math.floor(diff) + " seconds" ||
            diff < 120 && "1 minute" ||
            diff < 3600 && Math.floor( diff / 60 ) + " min" ||
            diff < 7200 && "1 hour" ||
            diff < 86400 && Math.floor( diff / 3600 ) + " hours") ||
        day_diff == 1 && "1 day" ||
        day_diff < 7 && day_diff + " days" ||
        day_diff < 31 && Math.ceil( day_diff / 7 ) + " weeks" ||
        day_diff < 365 && Math.ceil( day_diff / 31 ) + " months" ||
        Math.ceil( day_diff / 365 ) + " years";
}

看到它在行動 。