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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);

converts parts of a string to int in java

I was wondering how can I take some numbers in a string and convert them to an integer type? for example if a user entered 12:15pm how can I get 1 and 2 and make an int with value 12?

Given the example above, you could try something like this:
final int value = Integer.parseInt(input.substring(0, input.indexOf(':'))); //value = 12
Where input = 12:15pm in this case.
Generally speaking, just use a combination of String#indexOf(String), String#substring(int, int) and Integer.parseInt(String).

Read the String and Integer API's
You can use the String.split() to get the two numeric strings
You can use Integer.parseInt(...) to convert the String to an int.
Edit: Using the split() you can do something like:
String time = "12:34pm";
int hour = Integer.parseInt( time.split(":")[0] );

Exception while Parsing Hexadecimal to Short in Java

In file data.hex, the data is given in hexadecimal form where the first digit of hexadecimal number is always less than 8. Eg.
01FC 04BF 04C0 04C1 04C2 24C3 04C4 34C5 ...
To parse this file and store the values in shrt[] array i have written this code
void read_hex_short(String filename, short[] shrt, int x, int y) throws Exception
{
String str;
Scanner s=new Scanner(new BufferedReader(new FileReader(filename)));
for(int i=0;i<height*width;i++)
{
str= s.next(); // i have tried str="0x"+s.next() but it didn't work
image[i]=(short)Integer.parseInt(str);
}
s.close();
}
But i am getting NumberFormatException which arises while passing the first string i.e. 01FC only.
How can i parse these hexadecimal values and store them in the shrt[] array?

You should be using Integer.parseInt(str, 16) to tell it to use hex.
You should also be aware that any values greater than 0x7FFF will end up being negative in your array: Java doesn't have any unsigned numeric types (unless you count char).

You can use Short.parseShort(str, 16) to parse HEX. This will avoid the need to cast it to short from an int.

Give parseInt method a radix parameter. Like this:
Integer.parseInt(str, 16)

How can I convert an int number from decimal to binary?

How can I convert an int number from decimal to binary. For example:
int x=10; // radix 10
How can I make another integer has the binary representation of x, such as:
int y=1010; // radix 2
by using c only?

An integer is always stored in binary format internally -- saying that you want to convert int x = 10 base 10 to int y = 1010 base 2 doesn't make sense. Perhaps you want to convert it to a string representing the binary format of the integer, in which case you can use Integer.toBinaryString.

First thing you should understand is that a value is an abstract notion, that is not bounded to any representation. For example, if you have 20 apples, the number of apples will be the same regardless of the representation. So, dec("10") == bin("1010").
The value of an int reffers to this abstract notion of value, and it does not have any form until you with to print it. This means that the notion of base is important only for conversions from string to int and back.

String s = Integer.toBinaryString(10);
http://download.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html

Whether it's binary or decimal doesn't really have anything to do with the integer itself. Binary or decimal is a property of a physical representation of the integer, i.e. a String. Thus, the methods you should look at are Integer.toString() and Integer.valueOf() (the versions that take a radix parameter).
BTW, internally, all Java integers are binary, but literals in the source code are decimal (or octal).

Your question is a bit unclear but I'll do my best to try to make sense of it.
How can I make another integer has the binary representation of x such as: int y=1010 radix 2?
From this it looks like you wish to write a binary literal in your source code. Java doesn't support binary integer literals. It only supports decimal, hexadecimal and octal.
You can write your number as a string instead and use Integer.parseInt with the desired radix:
int y = Integer.parseInt("1010", 2);
But you should note that the final result is identical to writing int y = 10;. The integer 10 that was written as a decimal literal in the source code is identical in every way to one which was parsed from the binary string "1010". There is no difference in their internal representation if they are both stored as int.
If you want to convert an existing integer to its binary representation as a string then you can use Integer.toBinaryString as others have already pointed out.

Both integers will have the same interior representation, you can however display as binary via Integer.toBinaryString(i)

Use Integer.toBinaryString()
String y = Integer.toBinaryString(10);

Converting an integer to another base (string representation):
int num = 15;
String fifteen = Integer.toString(num, 2);
// fifteen = "1111"
Converting the string back into an integer
String fifteen = "1111";
int num = Integer.valueOf(fifteen, 2);
// num = 15
This covers the general case for any base. There's no way to explicitly assign an integer as binary (only decimal, octal, and hexadecimal)
int x = 255; // decimal
int y = 0377; // octal (leading zero)
int z = 0xFF; // hex (prepend 0x)

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.