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number format exception when trying to parse a string

I want to convert Integer.MAX_VALUE to binary and want the represenation to be of type int. I passed it to Integer.tobinarystring() and wraped that with Integer.parseint but i get numberformatexception.
Here is the code
System.out.println(
Integer.parseInt(
Integer.toBinaryString(Integer.MAX_VALUE)
)
);
Here is the exception
Exception in thread "main" java.lang.NumberFormatException: For input string: "1111111111111111111111111111111"

Integer.MAX_VALUE is 2,147,483,647
In binary this is:
1111111111111111111111111111111
If we treat that like an integer again, which is 1,111,111,111,111,111,111,111,111,111,111 you can probably see that it is much larger than the max value.
You probably want to look into BigInteger if you really need to deal with that as a int.

If you want to get the integer value of the binary string
1111111111111111111111111111111
you must use another signature of parseInt() that takes as 2nd parameter the radix, in this case of a binary string the radix is 2
String str = Integer.toBinaryString(Integer.MAX_VALUE);
int number = Integer.parseInt(str, 2);
System.out.println(number);
it will print:
2147483647

Convert hexadecimal string to hexadecimal integer in java

I have hexadecimal String eg. "0x103E" , I want to convert it into integer.
Means String no = "0x103E";
to int hexNo = 0x103E;
I tried Integer.parseInt("0x103E",16); but it gives number format exception.
How do I achieve this ?

You just need to leave out the "0x" part (since it's not actually part of the number).
You also need to make sure that the number you are parsing actually fits into an integer which is 4 bytes long in Java, so it needs to be shorter than 8 digits as a hex number.

No need to remove the "0x" prefix; just use Integer.decode instead of Integer.parseInt:
int x = Integer.decode("0x103E");
System.out.printf("%X%n", x);

Problems parsing String to long - what am I doing wrong?

Here is the code I am working with in a main method:
String numbers = "12345678900";
long upc = Integer.parseInt(numbers);
System.out.println(upc);
gives me:
Exception in thread "main" java.lang.NumberFormatException: For input string: "12345678900"
at java.lang.NumberFormatException.forInputString...
at java.lang.Integer.parseInt....
at java.lang.Integer.parseInt...
at testRun.main...
I cannot use a double, they need to be stored as values without a decimal. I am trying to get the string of numbers from a string into a variable that holds numbers (no decimals)

To parse a long, use Long.parseLong, not Integer.parseInt. That way, you get access to the full range of long values (whereas with parseInt, you only get the rather more restricted range of int values).

Use Long.parseLong()
String numbers = "12345678900";
long upc = Long.parseLong(numbers);
System.out.println(upc);

use
String numbers = "12345678900";
long upc = Long.parseLong(numbers);
System.out.println(upc);

The number you are passing is outside the range of integer which is from -2,147,483,648 to 2,147,483,647.
Try using the static function parseLong

Your number is enough long to not fit in Integer. So, you can't cast in integer. You can cast it in Long by calling Long.parseLong(number).

Use
long upc = Long.parseLong(numbers);

Java Double to String conversion without formatting

I have the number 654987. Its an ID in a database. I want to convert it to a string.
The regular Double.ToString(value) makes it into scientific form, 6.54987E5. Something I dont want.
Other formatting functions Ive found checks the current locale and adds appropriate thousand separators and such. Since its an ID, I cant accept any formatting at all.
How to do it?
[Edit] To clarify: Im working on a special database that treats all numeric columns as doubles. Double is the only (numeric) type I can retrieve from the database.

Use a fixed NumberFormat (specifically a DecimalFormat):
double value = getValue();
String str = new DecimalFormat("#").format(value);
alternatively simply cast to int (or long if the range of values it too big):
String str = String.valueOf((long) value);
But then again: why do you have an integer value (i.e. a "whole" number) in a double variable in the first place?

Use Long:
long id = 654987;
String str = Long.toString(id);

If it's an integer id in the database, use an Integer instead. Then it will format as an integer.

How about String.valueOf((long)value);

What about:
Long.toString(value)
or
new String(value)

Also you can use
double value = getValue();
NumberFormat f = NumberFormat.getInstance();
f.setGroupingUsed(false);
String strVal = f.format(value);

If what you are storing is an ID (i.e. something used only to identify another entity, whose actual numeric value has no significance) then you shouldn't be using Double to store it. Precision will almost certainly screw you.
If your database doesn't allow integer values then you should stored IDs as strings. If necessary make the string the string representation of the integer you want to use. With appropriate use of leading zeros you can make the alphabetic order of the string the same as the numeric order of the ints.
That should get you round the issue.

What about Long.toString((long)value) ?

double d = 56789;
String s = d+"";

Is J2ME's Integer.parseInt() broken?

While writing a game for J2ME we ran into an issue using java.lang.Integer.parseInt()
We have several constant values defined as hex values, for example:
CHARACTER_RED = 0xFFAAA005;
During the game the value is serialized and is received through a network connection, coming in as a string representation of the hex value. In order to parse it back to an int we unsuccesfully tried the following:
// Response contains the value "ffAAA005" for "characterId"
string hexValue = response.get("characterId");
// The following throws a NumberFormatException
int value = Integer.parseInt(hexValue, 16);
Then I ran some tests and tried this:
string hexValue = Integer.toHexString(0xFFAAA005);
// The following throws a NumberFormatException
int value = Integer.parseInt(hexValue, 16);
This is the exception from the actual code:
java.lang.NumberFormatException: ffaaa005
at java.lang.Integer.parseInt(Integer.java:462)
at net.triadgames.acertijo.GameMIDlet.startMIDlet(GameMIDlet.java:109)
This I must admit, baffled me. Looking at the parseInt code the NumberFormatException seems to be thrown when the number being parsed "crosses" the "negative/positive boundary" (perhaps someone can edit in the right jargon for this?).
Is this the expected behavior for the Integer.parseInt function? In the end I had to write my own hex string parsing function, and I was quite displeased with the provided implementation.
In other words, was my expectation of having Integer.parseInt() work on the hex string representation of an integer misguided?
EDIT: In my initial posting I wrote 0xFFFAAA005 instead of 0xFFAAA005. I've since corrected that mistake.

The String you are parsing is too large to fit in an int. In Java, an int is a signed, 32-bit data type. Your string requires at least 36 bits.
Your (positive) value is still too large to fit in a signed 32-bit int.

Do realize that your input (4289372165) overflows the maximum size of an int (2147483647)?
Try parsing the value as a long and trim the leading "0x" off the string before you parse it:
public class Program {
public static void main(String[] args) {
String input = "0xFFFAAA005";
long value = Long.parseLong(input.substring(2), 16);
System.out.print(value);
}
}

I'm not a java dev, but I'd guess parseInt only works with integers. 0xFFFAAA005 has 9 hex digits, so it's a long, not an int. My guess is it's complaining because you asked it to parse a number that's bigger than it's result data type.

Your number seems to be too large to fit in an int, try using Long.parseLong() instead.
Also, the string doesn't seem to get parsed if you have 0x in your string, so try to cut that off.