js,java时间处理

1.JS获取时间格式为“yyyy-MM-dd HH:mm:ss”的字符串

function getTimeStr(){var myDate = new Date();var year = myDate.getFullYear(); //获取完整的年份(4位,1970-????)var month = myDate.getMonth(); //获取当前月份(0-11,0代表1月)month = month > 9 ? month : "0"+month;var day =  myDate.getDate(); //获取当前日(1-31)day = day > 9 ? day : "0"+day;var hours =  myDate.getHours(); //获取当前小时数(0-23)hours = hours > 9 ? hours : "0"+hours;var minutes =  myDate.getMinutes(); //获取当前分钟数(0-59)minutes = minutes > 9 ? minutes : "0"+minutes;var seconds =  myDate.getSeconds(); //获取当前秒数(0-59)seconds = seconds > 9 ? seconds : "0"+seconds;return year+"-"+month+"-"+day+" "+hours+":"+minutes+":"+seconds;
}

2.java字符串转时间“字符串可以是多种格式”

 public static void main(String[] args) {String strDate="2005年04月22日";//注意：SimpleDateFormat构造函数的样式与strDate的样式必须相符SimpleDateFormat simpleDateFormat=new SimpleDateFormat("yyyy年MM月dd日 ");SimpleDateFormat sDateFormat=new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); //加上时间//必须捕获异常try {Date date=simpleDateFormat.parse(strDate);System.out.println(date);} catch(ParseException px) {px.printStackTrace();}}

3.java 计算两个时间之间的差多久

public static String getDatePoor(Date endDate, Date nowDate) {long nd = 1000 * 24 * 60 * 60;long nh = 1000 * 60 * 60;long nm = 1000 * 60;// long ns = 1000;// 获得两个时间的毫秒时间差异long diff = endDate.getTime() - nowDate.getTime();// 计算差多少天long day = diff / nd;// 计算差多少小时long hour = diff % nd / nh;// 计算差多少分钟long min = diff % nd % nh / nm;// 计算差多少秒//输出结果// long sec = diff % nd % nh % nm / ns;return day + "天" + hour + "小时" + min + "分钟";
}