这样就能转换String nowTime="2008-01-18";
Date date = java.sql.Date.valueOf(nowTime);
System.out.println(date);
但是,带上时间就不行String nowTime="2008-01-18 09:10:10";
Date date = java.sql.Date.valueOf(nowTime);
System.out.println(date);
怎么把String nowTime="2008-01-18 09:10:10";带时间也转换成date类型??
转换为这样格式的SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date date = java.sql.Date.valueOf(nowTime);
System.out.println(date);
但是,带上时间就不行String nowTime="2008-01-18 09:10:10";
Date date = java.sql.Date.valueOf(nowTime);
System.out.println(date);
怎么把String nowTime="2008-01-18 09:10:10";带时间也转换成date类型??
转换为这样格式的SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date day = df.parse(nowTime);
Fri Jan 18 09:10:10 CST 2008
可以是2008-01-18 09:10:10吗
看java.sql.Date.valueOf(String)的原码
public static Date valueOf(String s) {
int year;
int month;
int day;
int firstDash;
int secondDash; if (s == null)
throw new java.lang.IllegalArgumentException(); firstDash = s.indexOf('-');
secondDash = s.indexOf('-', firstDash + 1);
if ((firstDash > 0) & (secondDash > 0) & (secondDash < s.length() - 1)) {
year = Integer.parseInt(s.substring(0, firstDash)) - 1900;
month = Integer.parseInt(s.substring(firstDash + 1, secondDash)) - 1;
day = Integer.parseInt(s.substring(secondDash + 1));
} else {
throw new java.lang.IllegalArgumentException();
} return new Date(year, month, day);
}
使用SimpleDateFormat转没问题啊
SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
try {
Date data = df.parse("2008-01-18 09:10:10");
System.out.println(data);
} catch (ParseException e) {
e.printStackTrace();
}