I have an incoming string which should be a date in format "yyyyMMdd".(e.g today's Date - 20200507)
But sometimes input string can be an invalid format and Date parser should give an exception(parse exception), but it is not except it is returning a date object.
Sample Code in case when string is wrong or alphanumeric like below:
class Demo {
public static void main(String args[]) throws Exception {
String inputString = "9450524Q";
SimpleDateFormat formatter = new SimpleDateFormat("yyyyMMdd", Locale.ENGLISH);
System.out.println(formatter.parse(inputString));
}}
Output:
Tue Apr 04 00:00:00 IST 9454
From the JavaDoc of DateFormat from which SimpleDateFormat directly inherits:
By default, parsing is lenient: If the input is not in the form used by this object's format method but can still be parsed as a date, then the parse succeeds. Clients may insist on strict adherence to the format by calling setLenient(false).
java.time
I recommend that you use java.time, the modern Java date and time API, for your date work. Advantages in your particular case include that the formatter you need is already built in and it does throw the exception that you ask for.
For demonstration I am using this auxiliary method:
public static void tryToParse(String dateString) {
try {
LocalDate date
= LocalDate.parse(dateString, DateTimeFormatter.BASIC_ISO_DATE);
System.out.println(dateString + " -> " + date);
} catch (DateTimeParseException dtpe) {
System.out.println(dateString + ": " + dtpe.getMessage());
}
}
Trying it out:
// Valid date
tryToParse("20200507");
// Invalid date value
tryToParse("20210229");
// Text after date
tryToParse("20200623Q");
// Example from question
tryToParse("9450524Q");
Output is:
20200507 -> 2020-05-07
20210229: Text '20210229' could not be parsed: Invalid date 'February 29' as '2021' is not a leap year
20200623Q: Text '20200623Q' could not be parsed, unparsed text found at index 8
9450524Q: Text '9450524Q' could not be parsed at index 6
Please also enjoy the precise and helpful exception messages. What happened in the last case was: year 9450 and month 52 were parsed, but since 4Q is not a valid two-digit date, the exception was thrown (before validating whether 52 is a valid month number).
What happened in your code
The SimpleDateFormat class is a notorious troublemaker. You have discovered a central design problem with it, but certainly not the only one. What it did was: It parsed 4-digit year, 9450, and two-digit month, 52. There are 12 months in a year, but a SimpleDateFormat with standard settings doesn’t care. It converts 48 of the months to 4 years and ends up in the 4th month 4 years later. Finally 4 is parsed as day of month. The remainder of the text, the Q, is ignored.
As I see it your example exhibits three flaws of SimpleDateFormat:
With standard settings it is lenient, it accepts an invalid month number.
When asked to parse two digits and finding only one, it settles with that one digit without reporting the error.
In the case of unparseable text after the parsed text it does not report any error either.
Links
Oracle tutorial: Date Time explaining how to use java.time.
Related questions:
SimpleDateFormat giving wrong date instead of error
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I'm trying to parse two different dates and calculate the difference between them, but the next error appears:
java.time.format.DateTimeParseException: Text '103545' could not be parsed at index 2
Here's the code:
String thisDate= mySession.getVariableField(myVariable).toString().trim();
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("ddMMyyyy");
LocalDate theDate= LocalDate.parse(thisDate, formatter);
That’s as expected (approximately).
Your format pattern string, ddMMyyyy specifies two digits day of month, two digits month and (at least) four digits year for a total of (at least) eight (8) digits. So when you give it a string consisting of only 6 digits, parsing will necessarily fail.
If your user or another system is required to give you a date in ddMMyyyy format and they give you 103545, they are making an error. Your validation caught the error, which is a good thing. You will probably want to give them a chance to try again and give you a string like for example 10112021 (for 10 November 2021).
In case (just guessing) 103545 was meant to denote a time of day, 10:35:45 then you need to use the LocalTime class for it, and you also need to change the format pattern string to specify hours, minutes and seconds instead of year, month and date.
String thisDate = "103545";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("HHmmss");
LocalTime theTime = LocalTime.parse(thisDate, formatter);
System.out.println(theTime);
Output from this snippet is:
10:35:45
The problem here is the date parser has to recieve a date in the format specified (in this case "ddMMyyyy")
For example, this is what you would need to input for the parser to return a valid date:
String thisDate = '25Sep2000';
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("ddMMyyyy");
LocalDate theDate = LocalDate.parse(thisDate, formatter);
I think what you want is to convert a date in milliseconds to a date with a specific format. This is what you can do:
//Has to be made long because has to fit higher numbers
long thisDate = 103545; //Has to be a valid date in milliseconds
DateFormat formatter = new SimpleDateFormat("ddMMyyyy"); //You can find more formatting documentation online
Date theDate = new Date(thisDate);
String finalDate = formatter.format(theDate);
While validating the given string text is of Hmm format getting this error
java.text.ParseException: Unparseable date: "1637"
but this is working for text "747".
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
public class Validation{
public void validateDateFormat(String format,String text){
SimpleDateFormat sdfrmt = new SimpleDateFormat(format);
sdfrmt.setLenient(false);
Date testDate = null;
try {
testDate = sdfrmt.parse(text);
} catch (ParseException e) {
System.out.println("dateFormat Exception :");
e.printStackTrace();
}
}
public static void main(String []args){
Validation val = new Validation();
val.validateDateFormat("Hmm","747"); //working
val.validateDateFormat("Hmm","1637");//not working
}
}
This is for validating the given columns from the uploaded file.So wrote this to be dynamic based on the format written in the config for each column.
Well, it should work.
But you should use the newer Java Date and Time API (JSR 310) available in the java.time package.
If you then replace Date with LocalTime, SimpleDateFormat with DateTimeFormatter (using the ofPattern factory method) and ParseException with DateTimeParseException, it'll work.
This is happening because the SimpleDateFormat is unable to parse the given date using the format you have given.
Let's understand - your format is Hmm and you have given the date as 747 then, of course during parsing, the first letter 7 of the date is mapped to the first letter "H" of the format i.e. hours and 47 is mapped to mm i.e. minutes and hence it is able to convert correctly but for the next date 1637 it is failing because it does not know which letter to assign to H.
Here are some options you can try to make it more generic, choose format such as HHmm and always give the date of length 4, for e.g. for 747 enter the input as 0747 and it should work.
Or choose a format that is more clear for the parser to map for e.g. you can choose the format as H:mm and give the inputs as 7:47 or 16:37 since there is a delimiter : between hours and minutes, the parser will be able to parse all types of time irrespective of the length of the given input 3 or 4.
This question already has answers here:
SimpleDateFormat parsing date with 'Z' literal [duplicate]
(12 answers)
Closed 4 years ago.
I am using the below code to format millisecond resolution date strings. It works for 2018-09-14T13:05:21.329Z but not 2018-09-14T13:05:21.3Z. Can anybody suggest the reason and how to correct it?
DateFormat formatter = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSX");
SimpleDateFormat sdfDestination = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
try {
Date parsedDate = formatter.parse(date);
String destDate = sdfDestination.format(parsedDate);
return destDate;
} catch (java.text.ParseException parseException) {
logger.error("Parse Exception occured while converting publication time to date "
+ "format 'yyyy-MM-dd HH:mm:ss'", parseException);
}
I get below exception:
java.text.ParseException: Unparseable date: "2018-09-14T13:05:21.3Z"
at java.text.DateFormat.parse(Unknown Source) ~[na:1.8.0_181]
at com.noordpool.api.implementation.utility.Utility.parseDate(Utility.java:136) [classes/:na]
at com.noordpool.api.implementation.utility.Utility.parseMessage(Utility.java:77) [classes/:na]
Your only problem is that you are using a wrong pattern for SimpleDateFormat, you need to change:
DateFormat formatter = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSX");
To:
DateFormat formatter = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'");
Because the Z used in the date string means "zero hour offset" so you just need to pass it as 'Z' in your pattern.
This is a working demo with the right pattern.
Edit:
And to make things work with different Locales and Timezones, you need to use the appropriate Locale when you are creating the SimpleDateFormat instance, this is how should be the code:
DateFormat formatter = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'", Locale.US);
The only possible issue I can see is that you're passing in milliseconds incorrectly and the program doesn't know what to do about it.
So the last part of the formatter indicates with milliseconds and a timezone as .SSSX
But how does it evaluate 3Z for the input into this? I mean, do you say it's 300 timezone Z, or say it's 003 timezone Z, or worse, try and parse it as 3Z, which hopefully you see that you cannot turn '3Z' into a number.
To remedy this, I'd validate your input 'date' and ensure the milliseconds part is always 3 digits long, this removes the ambiguity and the program always knows that you mean '300 milliseconds, timezone Z'.
There is a problem in java 8 where the number of characters that you specified with the formatter should be an exact match (which is not specified in the documentation).
You can use three different Formatters and use nested exception as follows:
DateFormat format1 = new SimpleDateFormat("y-M-d'T'H:m:s.SX");
DateFormat format2 = new SimpleDateFormat("y-M-d'T'H:m:s.SSX");
DateFormat format3 = new SimpleDateFormat("y-M-d'T'H:m:s.SSSX");
Date parsedDate;
try {
// Parsing for the case - 2018-09-14T13:05:21.3Z
parsedDate = format1.parse(date);
} catch (ParseException e1) {
try {
// Parsing for the case - 2018-09-14T13:05:21.32Z
parsedDate = format2.parse(date);
} catch (ParseException e2) {
try {
// Parsing for the case - 2018-09-14T13:05:21.329Z
parsedDate = format3.parse(date);
} catch (ParseException e2) {
//The input date format is wrong
logger.error("Wrong format for date - " + date);
}
}
}
java.time
DateTimeFormatter dtfDestination
= DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss");
String date = "2018-09-14T13:05:21.3Z";
String destDate = Instant.parse(date)
.atZone(ZoneId.of("Indian/Comoro"))
.format(dtfDestination);
System.out.println(destDate);
Output from this snippet is:
2018-09-14 16:05:21
Please substitute your correct time zone if it didn’t happen to be Indian/Comoro, since correct output depends on using the correct time zone. If you want to use your JVM’s default time zone, specify ZoneId.systemDefault(), but be aware that the default can be changed at any time from other parts of your program or other programs running in the same JVM.
I am exploiting the fact that your string, "2018-09-14T13:05:21.3Z", is in ISO 8601 format, the format that the classes of java.time parse as their default, that is, without any explicit formatter. Instant.parse accepts anything from 0 through 9 decimals on the seconds, so there is no problem giving it a string with just 1 decimal, as you did. In comparison there is no way that an old-fashioned SimpleDateFormat can parse 1 decimal on the seconds with full precision since it takes pattern letter (uppercase) S to mean milliseconds, so .3 will be parsed as 3 milliseconds, not 3 tenths of a second, as it means.
Jahnavi Paliwal has already correctly diagnosed and explained the reason for the exception you got.
The date-time classes that you used, DateFormat, SimpleDateFormat and Date, are all long outdated and SimpleDateFormat in particular is notoriously troublesome. Since you seem to be using Java 8 (and even if you didn’t), I suggest you avoid those classes completely and use java.time instead.
Links
Oracle tutorial: Date Time explaining how to use java.time.
Wikipedia article: ISO 8601
I am getting issues with Date Format. I am using SimpleDateFormat to parse date and using MM/dd/yyyy format to parse date, its working properly with correct input like 11/11/2011 and its returning correct result (Fri Nov 11 00:00:00 IST 2011)
But if we enter 11/11/11 as input then its not working properly (Wed Nov 11 00:00:00 IST 11), nor giving parse error.
public static void main(String[] args) {
String format = "MM/dd/yyyy";
String dt = "11/11/11";
Date date = null;
try {
date = TestDate.parDate(dt, format);
} catch (ParseException e) {
e.printStackTrace();
}
System.out.println(date);
}
public static Date parDate(String value, String dateFormat) throws ParseException {
Date date = null;
date = new SimpleDateFormat(dateFormat).parse(value);
return date;
}
If you can use Java 8 or above, use DateTimeFormatter and LocalDate classes to parse date values. It will throw an error if the input is not in the expected format.
DateTimeFormatter formatter1 = DateTimeFormatter.ofPattern("MM-dd-yyyy");
LocalDate date1 = LocalDate.parse("11-11-2011", formatter1);
System.out.println(date1);
Above will work as expected but if you try to parse "11-11-11" with the same formatter object you will get an exception like
Exception in thread "main" java.time.format.DateTimeParseException: Text '11-11-11' could not be parsed at index 6
at java.time.format.DateTimeFormatter.parseResolved0(Unknown Source)
at java.time.format.DateTimeFormatter.parse(Unknown Source)
at java.time.LocalDate.parse(Unknown Source)
at com.is.TestDate.main(TestDate.java:14)
Use MM/dd/yy for both 11/11/11 and 11/11/2011.
If you want to put a strict check, that would not be possible with SimpleDateFormat. You can instead put a pattern check on the date string:
if (!dt.matches("\\d{2}/\\d{2}/\\d{4}")) {
//throw exception
}
It's not an issue. It's correct accroding to Java API of SimpleDateFormat. It says regarding the pattern of Year:
For parsing, if the number of pattern letters is more than 2, the year
is interpreted literally, regardless of the number of digits. So using
the pattern "MM/dd/yyyy", "01/11/12" parses to Jan 11, 12 A.D.
Thus there is no exception thrown as you provided a valid input.
If you have to deal with two different date formats for a year yyyy and yy convert one version to the other or use two formatter - if you have to use SimpleDateFormat at all.
That's the way since Java 8: How to parse/format dates with LocalDateTime? (Java 8)
please consider the following snippet:
SimpleDateFormat parser = new SimpleDateFormat("MMdd");
parser.setLenient(false);
try {
Date date = parser.parse("0229");
System.out.println(date);
} catch (ParseException e) {
e.printStackTrace();
}
I have a text field which should contain a date in the format MMdd (no year, as it should always default to the current year).
This year, 2012, a leap year, I found myself in the strange situation of not being able to parse the valid date February 29th.
The calendar used by the code above always defaults to 1970, which, though luck, was not a leap year. Hence trying to parse "0229" always throws a parse exception.
Any idea on how to parse this?
Append the current year and use a format of "MMddyyyy"?