How to convert HH:mm:ss.SSS to milliseconds?
Solution 1:
You can use SimpleDateFormat
to do it. You just have to know 2 things.
- All dates are internally represented in UTC
-
.getTime()
returns the number of milliseconds since 1970-01-01 00:00:00 UTC.
package se.wederbrand.milliseconds;
import java.text.SimpleDateFormat;
import java.util.Date;
import java.util.TimeZone;
public class Main {
public static void main(String[] args) throws Exception {
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.SSS");
sdf.setTimeZone(TimeZone.getTimeZone("UTC"));
String inputString = "00:01:30.500";
Date date = sdf.parse("1970-01-01 " + inputString);
System.out.println("in milliseconds: " + date.getTime());
}
}
Solution 2:
If you want to parse the format yourself you could do it easily with a regex such as
private static Pattern pattern = Pattern.compile("(\\d{2}):(\\d{2}):(\\d{2}).(\\d{3})");
public static long dateParseRegExp(String period) {
Matcher matcher = pattern.matcher(period);
if (matcher.matches()) {
return Long.parseLong(matcher.group(1)) * 3600000L
+ Long.parseLong(matcher.group(2)) * 60000
+ Long.parseLong(matcher.group(3)) * 1000
+ Long.parseLong(matcher.group(4));
} else {
throw new IllegalArgumentException("Invalid format " + period);
}
}
However, this parsing is quite lenient and would accept 99:99:99.999 and just let the values overflow. This could be a drawback or a feature.
Solution 3:
Using JODA:
PeriodFormatter periodFormat = new PeriodFormatterBuilder()
.minimumParsedDigits(2)
.appendHour() // 2 digits minimum
.appendSeparator(":")
.minimumParsedDigits(2)
.appendMinute() // 2 digits minimum
.appendSeparator(":")
.minimumParsedDigits(2)
.appendSecond()
.appendSeparator(".")
.appendMillis3Digit()
.toFormatter();
Period result = Period.parse(string, periodFormat);
return result.toStandardDuration().getMillis();