问题

I have the following instruction:

TimeZone zone = TimeZone.getTimeZone("Asia/Toyo");

Obviously, it should return null, but it will return the default timeZone, which is not the desired behaviour for my case. From Java Doc:

Returns the specified TimeZone, or the GMT zone if the given ID cannot be understood.

Is there a way to get corresponding TimeZone and null value if the String does not indicate a valid TimeZone?

I don't find a good solution to get all TimeZones and iterate over them.

回答1:

Instead of iterating you could also simply use:

boolean exists = Arrays.asList(TimeZone.getAvailableIDs()).contains("Asia/Toyo");

or:

boolean exists = Arrays.stream(TimeZone.getAvailableIDs()).anyMatch("Asia/Toyo"::equals);

If you need to do that often, putting the available IDs in a HashSet would be more efficient.

回答2:

This should answer your need TimeZone validation in Java

String[] validIDs = TimeZone.getAvailableIDs();
for (String str : validIDs) {
      if (str != null && str.equals("yourString")) {
        System.out.println("Valid ID");
      }
}

回答3:

`java.time.ZoneId`

The old date-time classes are now legacy, supplanted by the java.time classes.

Instead of java.util.TimeZone, you should be using ZoneId. For parsing a proper time zone name, call ZoneId.of. If the name you pass in unrecognized, the method throws a ZoneRulesException.

So, to catch a misspelling of Asia/Tokyo as Asia/Toyo, trap for the ZoneRulesException.

ZoneId z;
try {
    z = ZoneId.of ( "Asia/Toyo" );  // Misspell "Asia/Tokyo" as "Asia/Toyo".
} catch ( ZoneRulesException e ) {
    System.out.println ( "Oops! Failed to recognize your time zone name. ZoneRulesException message: " + e.getMessage () );
    return;
}
ZonedDateTime zdt = ZonedDateTime.now ( z );

See this code run live at IdeOne.com.

About java.time

The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.

The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.

To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.

Where to obtain the java.time classes?

Java SE 8 and SE 9 and later
- Built-in.
- Part of the standard Java API with a bundled implementation.
- Java 9 adds some minor features and fixes.
Java SE 6 and SE 7
- Much of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
Android
- The ThreeTenABP project adapts ThreeTen-Backport (mentioned above) for Android specifically.
- See How to use ThreeTenABP….

The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.

回答4:

Just check the returned TimeZone:

public static TimeZone getTimeZone(String timeZoneID) {
    TimeZone timeZone = TimeZone.getTimeZone(timeZoneID);
    if (timeZone.getID().equals(timeZoneID))
        return timeZone;
    throw new IllegalArgumentException("Unknown time zone: " + timeZoneID);
}

In Java 8, use the new ZoneId.of(String zoneId), which will fail for unknown time zones, and TimeZone.getTimeZone(ZoneId zoneId) if you need the old TimeZone:

ZoneId zoneId = ZoneId.of("Asia/Toyo");
TimeZone timeZone = TimeZone.getTimeZone(zoneId);

回答5:

If you don't want to iterate over TimeZone.getAvailableIDs(). You can use.

boolean validateTimeZone(String timeZoneStr) {
    return TimeZone.getTimeZone(timeZoneStr).getID().matches(timeZoneStr);
}

来源：https://stackoverflow.com/questions/33373442/java-gettimezone-without-returning-a-default-value

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java

timezone

Java: getTimeZone without returning a default value

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