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Parsing from SimpleDateFormat to Date not working?

SimpleDateFormat df = new SimpleDateFormat();
Date lastLogin = null;
try {
String troubleChild = lineScanner.next();
lastLogin = df.parse(troubleChild);
} catch (ParseException e) {
System.out.println("ohnoes");
}
Hi I'm quite new to using the date functions and I've come up with a problem. I have a file that is being parsed into various variables and they all work except this one i can never get it so that it passes the try/catch clause i've looked up similar problems but none of them work on my code.(The date i am inputting is in the format: Mon, Oct 30 22:20:11 GMT 2017) please can I get some help and thanks for it!

Solution: java.time
Please don’t take the trouble with the long outmoded classes Date and SimpleDateFormat. Instead use java.time, the modern Java date and time API also known as JSR-310:
DateTimeFormatter dtf
= DateTimeFormatter.ofPattern("E, MMM d H:mm:ss z uuuu", Locale.UK);
String inputDate = "Mon, Oct 30 22:20:11 GMT 2017";
ZonedDateTime lastLogin = ZonedDateTime.parse(inputDate, dtf);
System.out.println(lastLogin);
This prints
2017-10-30T22:20:11Z[GMT]
Since dates and times may come in so many different textual formats, I am using a format pattern string to specify your particular format. For which letters you may use, and what difference it makes whether you use 1, 3 or 4 of the same letter, see the documentation. Beware that format pattern strings are case sensitive.
Problem: SimpleDateFormat
You used the no-arg SimpleDateFormat constructor. The way I read the documentation, this gives you the default date format for your locale. If your JVM is running UK locale, I believe the format goes like 28/11/17 10:57 — not much like the input format you were trying to parse. You can use System.out.println(df.format(new Date())); to find out. The usual SimpleDateFormat constructor to use would be SimpleDateFormat(String, Locale) so that you may again supply a format pattern string and a locale.

Is there an xml timestamp format like the java timestamp?

My java timestamp has the following format:
YYYY-MM-DD hh:mm:ss.ms
2016-01-08 15:16:44.554
I got it using the following method:
private String getCurrentTimeStamp() {
Date date= new java.util.Date();
return((new Timestamp(date.getTime())).toString());
}
Is there a standardized xml date and Time format for timestamp? The xs: dateTime has the following format: "YYYY-MM-DDThh: mm: SS" And it is not taking into consideration milliseconds.

XML itself does not define any timestamp formats.
XML Schema Part 2: Datatypes Second Edition incorporates ISO 8601 formats by reference. The dateTime format allows but does not require a decimal point followed by arbitrary fractions of a second. For example, 2016-01-08T15:16:44.554

In XML Schema (XSD), all formats of dates and times are well defined.
The java.time framework built into Java 8 and later supports these formats. See Tutorial.
Here are some examples:
// Dates in XML: YYYY-MM-DD
DateTimeFormatter.ISO_LOCAL_DATE.parse("2002-09-24");
// Dates with TimeZone in XML: YYYY-MM-DDZ
DateTimeFormatter.ISO_DATE.parse("2002-09-24Z");
// Dates with TimeZone in XML: YYYY-MM-DD-06:00 or YYYY-MM-DD+06:00
DateTimeFormatter.ISO_DATE.parse("2002-09-24-06:00");
// Times in XML: hh:mm:ss
DateTimeFormatter.ISO_TIME.parse("09:00:00");
DateTimeFormatter.ISO_TIME.parse("09:00:00.5");
// DateTimes in XML: YYYY-MM-DDThh:mm:ss (with an optional TimeZone)
DateTimeFormatter.ISO_DATE_TIME.parse("2002-05-30T09:00:00");
DateTimeFormatter.ISO_DATE_TIME.parse("2002-05-30T09:30:10.5");
DateTimeFormatter.ISO_DATE_TIME.parse("2002-05-30T09:00:00Z");
DateTimeFormatter.ISO_DATE_TIME.parse("2002-05-30T09:30:10.5-06:00");
Durations and Periods however are not perfectly compatible, because they are split in Durations and Periods in Java. Here are however some examples:
Period.parse("P5Y");
Period.parse("P5Y2M10D");
Duration.parse("PT15H");
Duration.parse("-P10D");

If you can't change the schema, you have to change your function
private String getCurrentTimeStamp() {
Date date = new java.util.Date();
return new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS").format(date);
}

Maybe you mean the standart ISO for Date.
On this thread
Convert Java Date...
See also:
What is Jaxb and why would i use it

There seems to be a format as part of XSD 1.1:
The type xsd:dateTimeStamp represents a specific date
and time in the format CCYY-MM-DDThh:mm:ss.sss.
http://www.datypic.com/sc/xsd11/t-xsd_dateTimeStamp.html

Convert ical time to milliseconds

I am trying to parse an iCal-file and write the info to a database for my Android application.
I am using the ical4j-library to parse the data, and as output I get, for example, a date formatted like this: 20140116T121500Z.
I want to convert that date into milliseconds. I tried using Joda, but couldn't get it to work:
Caused by: java.lang.IllegalArgumentException: Invalid format: "20140110T091500Z" is malformed at "1500Z"

Using Joda-Time API for DateTimeFormat:
DateTimeFormatter formatter = DateTimeFormat.forPattern("yyyyMMdd'T'HHmmssZ");
System.out.println(formatter.parseDateTime("20140116T121500Z").getMillis());
Output:
1389874500000
I am able to parse your date through java.text.SimpleDateFormat by escaping 'Z' representing the timezone (or omit Z altogether):
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyyMMdd'T'HHmmss'Z'");
System.out.println(dateFormat.parse("20140116T121500Z").getTime());
Output:
1389854700000
Note: The difference in the 2 times is introduced by diluting the timezone in java.text.SimpleDateFormat, which can be overcome by setting an explicit GMT timezone below:
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyyMMdd'T'HHmmss'Z'");
dateFormat.setTimeZone(TimeZone.getTimeZone("GMT"));
System.out.println(dateFormat.parse("20140116T121500Z").getTime());
Output:
1389874500000

Trying to convert yyyy-MM-ddTHH:mm:ss-07:00 to correct time in Android

Okay, so here's my issue in Android right now. On our Database there's a timestamp in this format 8/15/2013 2:00:48 PM and through a .NET WebService I get that same time like this in Android: 2013-08-15T14:00:48-07:00. Now I want to convert this format into a Date Time format that I can use for comparison (for example this webservice provides every instance where a device failed at logging in so we want to check the amount of time between occurances to see if there's any issues). Below I have this code where I'm trying to use JODA Time but it's still not returning the correct format:
public static Date convertStringToDate(String input) {
String pattern = "yyyy-MM-dd'T'HH:mm:ssZ";
DateTimeFormatter formatter = DateTimeFormat.forPattern(pattern);
DateTime dateTime = formatter.parseDateTime(input);
return dateTime.toDate();
//printout shows: Thu Aug 15 17:00:48 EDT 2013
}
I know that the server is returning some crappy time format that is hard to work with (it took a while to get this to work in the iOS App we have, and even there it's still rather clunky) so I don't mind changing the webservice or the query if that would make things easier.

I have a very similar format, and I parse it using SimpleDateFormat, try this:
SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZZZ", Locale.US);
Date dateTime = format .parse(value);

What i understand is that you have your correct instance of date already and what you need is to parse it to String.
I suggest you use:
DateFormat formatter = new SimpleDateFormat("d/MM/yyyy hh:mm:ss a");
//this will give you the format '8/15/2013 2:00:48 PM'
String d = formatter.format(date);
Hope this helps.
EDIT:
Also seams you want to have your date instance in -07:00 timezone
So you can change your line
DateTime dateTime = formatter.parseDateTime(input);
for
DateTime dateTime = formatter.withZone(DateTimeZone.forID("-07:00")).parseDateTime(input);

JSON parse date and time?

Im having a little issue with parsing json date.
Here is what I would like to parse:
{"driver": "247","firstName": "XXXXX","lastName": "XXXXX","lastLatitudeUpdate": "5/21/2012 4:49:17 PM","suspended": "false","checkedin": "0"}
I am having trouble parsing "lastLatitudeUpdate" is it because there are spaces in between? Thanks in advance for the help.

Assuming you are on Android and therefore working with java (yes you don't mention that, only the tag in your question suggests it...)
Like mentioned here (and in various other places) you can parse a date in java using the SimpleDateFormat class:
SimpleDateFormat parserSDF=new SimpleDateFormat("M/d/yyyy h:m:s a");
Date d = parserSDF.parse(dateField,0);
Of course you have to first parse you json input with some library (e.g. standard library from json.org or Google gson) and then parse the string you'll get there for the field into a date.

Short answer: No, there is no way for the JSON engine to recognize a string as a Date object.
Long answer:
There is no 'date' type in JSON. However, this JSON is fine, the catch is that lastLatitudeUpdate will be parsed as a string. In order to convert this to a date you should try something like
var my_object= JSON.parse({"driver": "247","firstName": "XXXXX","lastName": "XXXXX","lastLatitudeUpdate": "5/21/2012 4:49:17 PM","suspended": "false","checkedin": "0"});
my_object.lastLatitudeUpdate= Date.parse(my_object.lastLatitudeUpdate)
This function will give a timestamp. However, you have to make sure the string is correctly recognized, you may have to do some extra work.

Some links for hints
http://docs.oracle.com/javase/1.4.2/docs/api/java/text/SimpleDateFormat.html
http://www.java-samples.com/showtutorial.php?tutorialid=406

How are you parsing the date? In Chrome this seems to work fine:
new Date("5/21/2012 4:49:17 PM");
Mon May 21 2012 16:49:17 GMT-0400 (US Eastern Daylight Time)