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Add leading Zeros to converted Integer Java

I get gpnString as a String and I have to parse it to an int.
I was using Integer.parseInt(gpnString) but gpnString contains leading zeros and they get removed with Integer.parseInt().
The gpnString has allways the length of 8 so I tried to add so many 0 as needed but the int detects it as NULL so it does not add the number 0. This is my Code:
int parseToInt(String gpnString) {
int gpn = Integer.parseInt(gpnString);
for (int addZero = 8 - String.valueOf(gpn).length(); addZero != 0; addZero--) {
gpn = 0 + gpn;
}
return gpn;
}
/*
Input : gpnString = "00012345"
Output: 12345
*/
I have looked this up on Stackoverflow but I only found an answer like this:
String gpn = String.format("%08d" , gpnSring);
I have tried it but you get a String so I would have to parse it again and this would lead to the exact same Problem as before.
Edit:
I heard an Integer or int can not have leading zeros...
I use the gpn for personal identification in a Database. Would the best Idea be to edit the grafical output so it just looks like it has the 0? Or is there a better way to solve this?

An integer can't have leading zeros.
As leading zeros add no mathematical significance for an integer, they will not be stored as such. Leading zeros most likely only add to the readability for the human viewing / processing the value. For that usage a string can be used in the way you already found yourself.

error while parsing an int from a char in String

import java.util.*;
public class test
{
public static void main(String[] args)
{
int i = 123;
String s = Integer.toString(i);
int Test = Integer.parseInt(s, s.charAt(0)); // ERROR!
}
}
I want to parse the input string based on char position to get the positional integer.
Error message:
Exception in thread "main" java.lang.NumberFormatException: radix 49 greater than Character.MAX_RADIX
at java.lang.Integer.parseInt(Unknown Source)
at test.main(test.java:11)

That method you are calling parseInt(String, int) expects a radix; something that denotes the "number system" you want to work in, like
parseInt("10", 10)
(10 for decimal)! Instead, use
Integer.parseInt(i)
or
Integer.parseInt(i, 10)
assuming you want to work in the decimal system. And to explain your error message - lets have a look at what your code is actually doing. In essence, it calls:
Integer.parseInt("123", '1')
and that boils down to a call
Integer.parseInt("123", 49) // '1' --> int --> 49!
And there we go - as it nicely lines up with your error message; as 49 isn't a valid radix for parsing numbers.
But the real answer here: don't just blindly use some library method. Study its documentation, so you understand what it is doing; and what the parameters you are passing to it actually mean.
Thus, turn here and read what parseInt(String, int) is about!

Integer.parseInt(parameter) expects the parameter to be a String.
You could try Integer.parseInt(s.charAt(0) + ""). The +"" is to append the character to an empty String thereby casting the char to String and this is exactly what the method expects.
Another method to parse Characters to Integers (and in my opinion much better!) is to use Character.getNumericValue(s.charAt(0));
Check this post for further details on converting char to int

Need to convert String.valueOf(s.charAt(0)) to String.valueOf(s.charAt(0)) i.e. Char to String.
import java.util.*;
public class test
{
public static void main(String[] args)
{
int i = 123;
String s = Integer.toString(i);
int Test = Integer.parseInt(String.valueOf(s.charAt(0)));
}
}

Let use what we have here.
To parse one digit from a String into an integer. Use getNumericValue(char)
In your case, to get the first character into a number :
int n = Character.getNumericValue(s.charAt(0));
Be aware that you should take the absolute value if you integer can be negative.

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);

Remove decimals after a numeric value in java

Hi I have a excel file reading application which reads every cell in the file.
whenever a cell contains a numeric value the app is treating it a numeric cell.
For example the cell contains (40002547) the application will treat this as numeric cell. I cab get the value by using this code:
SONum = String.valueOf(cellSONum.getNumericCellValue());
Well that works fine. My Problem is it appends decimal at the end of the string. it will be (40002547.0). I need it to be as is. Thanks in advance

It's because cellSONum.getNumericCellValue() is returning a floating point type. If you force it to an integer before calling valueOf(), you should get the string representation in an integral form, if indeed that's what you want for all possibilities:
SONum = String.valueOf((int)cellSONum.getNumericCellValue());
You can see this in the following code:
class Test {
public static void main(String[]args) {
double d = 1234;
System.out.println(String.valueOf(d));
System.out.println(String.valueOf((int)d));
}
}
which outputs:
1234.0
1234
However, if you want to just get rid of .0 at the end of any string but allow non-integral values to survive, you can just remove the trailing text yourself:
class Test {
public static void main(String[]args) {
double d1 = 1234;
double d2 = 1234.567;
System.out.println(String.valueOf(d1).replaceFirst("\\.0+$", ""));
System.out.println(String.valueOf(d2).replaceFirst("\\.0+$", ""));
}
}
That snippet outputs:
1234
1234.567

Try with split().
SONum = String.valueOf(cellSONum.getNumericCellValue());
SONum = SONum.split("\\.")[0];
When you split 40002547.0 with . ,the split function returns two parts and the first one you need.

If you want to be sure you are not cutting of any valid decimals, you can use regexp also:
String pattern = "\.0+"; // dot followed by any number of zeros
System.out.println(String.valueOf(cellSONum.getNumericCellValue()).replaceAll(pattern, ""));
More on java regexp for example: http://www.vogella.com/articles/JavaRegularExpressions/article.html

As PaxDiablo also mentions, cellSONum.getNumericCellValue() returns a floating point.
You can cast this to Long or int to get rid of all behind the '.'
String SONum = String.valueOf(cellSONum.getNumericCellValue().longValue());
used as example:
String SONum = String.valueOf((new Double(0.5)).longValue());

SONum = ""+cellSONum.getNumericCellValue().split(".")[0];

try
double value = 23.0;
DecimalFormat df = new DecimalFormat("0.##");
System.out.println("bd value::"+ df.format(value))

Consider using BigDecimal.
You could simply say
BigDecimal scaledDecimal = new BigDecimal(value).setScale(0, RoundingMode.HALF_EVEN);

This will help in case your input is String and you need result also in String
1). Convert the string to Double using Double.parseDouble,
2). Type cast to int, then convert to string using String.valueOf()
private String formatText(String text) {
try {
return String.valueOf((int) Double.parseDouble(text));
} catch (NumberFormatException e) {
return text;
}
}

You can do Explicit type casting to remove the decimals,
double desvalue = 3.586;
int value = (int)desvalue;

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.