I've a String
representing a date.
String date_s = "2011-01-18 00:00:00.0";
I'd like to convert it to a Date
and output it in YYYY-MM-DD
format.
2011-01-18
How can I achieve this?
Okay, based on the answers I retrieved below, here's something I've tried:
String date_s = " 2011-01-18 00:00:00.0";
SimpleDateFormat dt = new SimpleDateFormat("yyyyy-mm-dd hh:mm:ss");
Date date = dt.parse(date_s);
SimpleDateFormat dt1 = new SimpleDateFormat("yyyyy-mm-dd");
System.out.println(dt1.format(date));
But it outputs 02011-00-1
instead of the desired 2011-01-18
. What am I doing wrong?
Use LocalDateTime#parse()
(or ZonedDateTime#parse()
if the string happens to contain a time zone part) to parse a String
in a certain pattern into a LocalDateTime
.
String oldstring = "2011-01-18 00:00:00.0";
LocalDateTime datetime = LocalDateTime.parse(oldstring, DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss.S"));
Use LocalDateTime#format()
(or ZonedDateTime#format()
) to format a LocalDateTime
into a String
in a certain pattern.
String newstring = datetime.format(DateTimeFormatter.ofPattern("yyyy-MM-dd"));
System.out.println(newstring); // 2011-01-18
Or, when you're not on Java 8 yet, use SimpleDateFormat#parse()
to parse a String
in a certain pattern into a Date
.
String oldstring = "2011-01-18 00:00:00.0";
Date date = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.S").parse(oldstring);
Use SimpleDateFormat#format()
to format a Date
into a String
in a certain pattern.
String newstring = new SimpleDateFormat("yyyy-MM-dd").format(date);
System.out.println(newstring); // 2011-01-18
Update: as per your failed attempt: the patterns are case sensitive. Read the java.text.SimpleDateFormat
javadoc what the individual parts stands for. So stands for example M
for months and m
for minutes. Also, years exist of four digits yyyy
, not five yyyyy
. Look closer at the code snippets I posted here above.