How can I convert a String to an int in Java?
My String contains only numbers, and I want to return the number it represents.
For example, given the string "1234" the result should be the number 1234.
转载于:https://stackoverflow.com/questions/5585779/how-do-i-convert-a-string-to-an-int-in-java
For example, here are two ways:
Integer x = Integer.valueOf(str);
// or
int y = Integer.parseInt(str);
There is a slight difference between these methods:
valueOf returns a new or cached instance of java.lang.IntegerparseInt returns primitive int.The same is for all cases: Short.valueOf/parseShort, Long.valueOf/parseLong, etc.
Well, a very important point to consider is that the Integer parser throws NumberFormatException as stated in Javadoc.
int foo;
String StringThatCouldBeANumberOrNot = "26263Hello"; //will throw exception
String StringThatCouldBeANumberOrNot2 = "26263"; //will not throw exception
try {
foo = Integer.parseInt(StringThatCouldBeANumberOrNot);
} catch (NumberFormatException e) {
//Will Throw exception!
//do something! anything to handle the exception.
}
try {
foo = Integer.parseInt(StringThatCouldBeANumberOrNot2);
} catch (NumberFormatException e) {
//No problem this time, but still it is good practice to care about exceptions.
//Never trust user input :)
//Do something! Anything to handle the exception.
}
It is important to handle this exception when trying to get integer values from split arguments or dynamically parsing something.
Do it manually:
public static int strToInt( String str ){
int i = 0;
int num = 0;
boolean isNeg = false;
//Check for negative sign; if it's there, set the isNeg flag
if (str.charAt(0) == '-') {
isNeg = true;
i = 1;
}
//Process each character of the string;
while( i < str.length()) {
num *= 10;
num += str.charAt(i++) - '0'; //Minus the ASCII code of '0' to get the value of the charAt(i++).
}
if (isNeg)
num = -num;
return num;
}
You can also begin by removing all non-numerical characters and then parsing the int:
string mystr = mystr.replaceAll( "[^\\d]", "" );
int number= Integer.parseInt(mystr);
But be warned that this only works for non-negative numbers.
Converting a string to an int is more complicated than just convertig a number. You have think about the following issues:
Currently I'm doing an assignment for college, where I can't use certain expressions, such as the ones above, and by looking at the ASCII table, I managed to do it. It's a far more complex code, but it could help others that are restricted like I was.
The first thing to do is to receive the input, in this case, a string of digits; I'll call it String number, and in this case, I'll exemplify it using the number 12, therefore String number = "12";
Another limitation was the fact that I couldn't use repetitive cycles, therefore, a for cycle (which would have been perfect) can't be used either. This limits us a bit, but then again, that's the goal. Since I only needed two digits (taking the last two digits), a simple charAtsolved it:
// Obtaining the integer values of the char 1 and 2 in ASCII
int semilastdigitASCII = number.charAt(number.length()-2);
int lastdigitASCII = number.charAt(number.length()-1);
Having the codes, we just need to look up at the table, and make the necessary adjustments:
double semilastdigit = semilastdigitASCII - 48; //A quick look, and -48 is the key
double lastdigit = lastdigitASCII - 48;
Now, why double? Well, because of a really "weird" step. Currently we have two doubles, 1 and 2, but we need to turn it into 12, there isn't any mathematic operation that we can do.
We're dividing the latter (lastdigit) by 10 in the fashion 2/10 = 0.2 (hence why double) like this:
lastdigit = lastdigit/10;
This is merely playing with numbers. We were turning the last digit into a decimal. But now, look at what happens:
double jointdigits = semilastdigit + lastdigit; // 1.0 + 0.2 = 1.2
Without getting too into the math, we're simply isolating units the digits of a number. You see, since we only consider 0-9, dividing by a multiple of 10 is like creating a "box" where you store it (think back at when your first grade teacher explained you what a unit and a hundred were). So:
int finalnumber = (int) (jointdigits*10); // Be sure to use parentheses "()"
And there you go. You turned a String of digits (in this case, two digits), into an integer composed of those two digits, considering the following limitations:
I'm have a solution, but I do not know how effective it is. But it works well, and I think you could improve it. On the other hand, I did a couple of tests with JUnit which step correctly. I attached the function and testing:
static public Integer str2Int(String str) {
Integer result = null;
if (null == str || 0 == str.length()) {
return null;
}
try {
result = Integer.parseInt(str);
}
catch (NumberFormatException e) {
String negativeMode = "";
if(str.indexOf('-') != -1)
negativeMode = "-";
str = str.replaceAll("-", "" );
if (str.indexOf('.') != -1) {
str = str.substring(0, str.indexOf('.'));
if (str.length() == 0) {
return (Integer)0;
}
}
String strNum = str.replaceAll("[^\\d]", "" );
if (0 == strNum.length()) {
return null;
}
result = Integer.parseInt(negativeMode + strNum);
}
return result;
}
Testing with JUnit:
@Test
public void testStr2Int() {
assertEquals("is numeric", (Integer)(-5), Helper.str2Int("-5"));
assertEquals("is numeric", (Integer)50, Helper.str2Int("50.00"));
assertEquals("is numeric", (Integer)20, Helper.str2Int("$ 20.90"));
assertEquals("is numeric", (Integer)5, Helper.str2Int(" 5.321"));
assertEquals("is numeric", (Integer)1000, Helper.str2Int("1,000.50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int("0.50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int(".50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int("-.10"));
assertEquals("is numeric", (Integer)Integer.MAX_VALUE, Helper.str2Int(""+Integer.MAX_VALUE));
assertEquals("is numeric", (Integer)Integer.MIN_VALUE, Helper.str2Int(""+Integer.MIN_VALUE));
assertEquals("Not
is numeric", null, Helper.str2Int("czv.,xcvsa"));
/**
* Dynamic test
*/
for(Integer num = 0; num < 1000; num++) {
for(int spaces = 1; spaces < 6; spaces++) {
String numStr = String.format("%0"+spaces+"d", num);
Integer numNeg = num * -1;
assertEquals(numStr + ": is numeric", num, Helper.str2Int(numStr));
assertEquals(numNeg + ": is numeric", numNeg, Helper.str2Int("- " + numStr));
}
}
}
We can use the parseInt(String str) method of the Integer wrapper class for converting a String value to an integer value.
For example:
String strValue = "12345";
Integer intValue = Integer.parseInt(strVal);
The Integer class also provides the valueOf(String str) method:
String strValue = "12345";
Integer intValue = Integer.valueOf(strValue);
We can also use toInt(String strValue) of NumberUtils Utility Class for the conversion:
String strValue = "12345";
Integer intValue = NumberUtils.toInt(strValue);
You can use this code also, with some precautions.
Option #1: Handle the exception explicitly, for example, showing a message dialog and then stop the execution of the current workflow. For example:
try
{
String stringValue = "1234";
// From String to Integer
int integerValue = Integer.valueOf(stringValue);
// Or
int integerValue = Integer.ParseInt(stringValue);
// Now from integer to back into string
stringValue = String.valueOf(integerValue);
}
catch (NumberFormatException ex) {
//JOptionPane.showMessageDialog(frame, "Invalid input string!");
System.out.println("Invalid input string!");
return;
}
Option #2: Reset the affected variable if the execution flow can continue in case of an exception. For example, with some modifications in the catch block
catch (NumberFormatException ex) {
integerValue = 0;
}
Using a string constant for comparison or any sort of computing is always a good idea, because a constant never returns a null value.
Integer.decodeYou can also use public static Integer decode(String nm) throws NumberFormatException.
It also works for base 8 and 16:
// base 10
Integer.parseInt("12"); // 12 - int
Integer.valueOf("12"); // 12 - Integer
Integer.decode("12"); // 12 - Integer
// base 8
// 10 (0,1,...,7,10,11,12)
Integer.parseInt("12", 8); // 10 - int
Integer.valueOf("12", 8); // 10 - Integer
Integer.decode("012"); // 10 - Integer
// base 16
// 18 (0,1,...,F,10,11,12)
Integer.parseInt("12",16); // 18 - int
Integer.valueOf("12",16); // 18 - Integer
Integer.decode("#12"); // 18 - Integer
Integer.decode("0x12"); // 18 - Integer
Integer.decode("0X12"); // 18 - Integer
// base 2
Integer.parseInt("11",2); // 3 - int
Integer.valueOf("11",2); // 3 - Integer
If you want to get int instead of Integer you can use:
Unboxing:
int val = Integer.decode("12");
intValue():
Integer.decode("12").intValue();
int foo=Integer.parseInt("1234");
Make sure there is no non-numeric data in the string.
As mentioned Apache Commons NumberUtils can do it. Which return 0 if it cannot convert string to int.
You can also define your own default value.
NumberUtils.toInt(String str, int defaultValue)
example:
NumberUtils.toInt("3244", 1) = 3244
NumberUtils.toInt("", 1) = 1
NumberUtils.toInt(null, 5) = 5
NumberUtils.toInt("Hi", 6) = 6
NumberUtils.toInt(" 32 ", 1) = 1 //space in numbers are not allowed
NumberUtils.toInt(StringUtils.trimToEmpty( " 32 ",1)) = 32;
Apart from these above answers, I would like to add several functions:
public static int parseIntOrDefault(String value, int defaultValue) {
int result = defaultValue;
try {
result = Integer.parseInt(value);
} catch (Exception e) {
}
return result;
}
public static int parseIntOrDefault(String value, int beginIndex, int defaultValue) {
int result = defaultValue;
try {
String stringValue = value.substring(beginIndex);
result = Integer.parseInt(stringValue);
} catch (Exception e) {
}
return result;
}
public static int parseIntOrDefault(String value, int beginIndex, int endIndex, int defaultValue) {
int result = defaultValue;
try {
String stringValue = value.substring(beginIndex, endIndex);
result = Integer.parseInt(stringValue);
} catch (Exception e) {
}
return result;
}
And here are results while you running them:
public static void main(String[] args) {
System.out.println(parseIntOrDefault("123", 0)); // 123
System.out.println(parseIntOrDefault("aaa", 0)); // 0
System.out.println(parseIntOrDefault("aaa456", 3, 0)); // 456
System.out.println(parseIntOrDefault("aaa789bbb", 3, 6, 0)); // 789
}
Guava has tryParse(String), which returns null if the string couldn't be parsed, for example:
Integer fooInt = Ints.tryParse(fooString);
if (fooInt != null) {
...
}
Just for fun: You can use Java 8's Optional for converting a String into an Integer:
String str = "123";
Integer value = Optional.of(str).map(Integer::valueOf).get();
// Will return the integer value of the specified string, or it
// will throw an NPE when str is null.
value = Optional.ofNullable(str).map(Integer::valueOf).orElse(-1);
// Will do the same as the code above, except it will return -1
// when srt is null, instead of throwing an NPE.
Here we just combine Integer.valueOf and Optinal. Probably there might be situations when this is useful - for example when you want to avoid null checks. Pre Java 8 code will look like this:
Integer value = (str == null) ? -1 : Integer.parseInt(str);
Here we go
String str="1234";
int number = Integer.parseInt(str);
print number;//1234
You can use new Scanner("1244").nextInt(). Or ask if even an int exists: new Scanner("1244").hasNextInt()
In programming competitions, where you're assured that number will always be a valid integer, then you can write your own method to parse input. This will skip all validation related code (since you don't need any of that) and will be a bit more efficient.
For valid positive integer:
private static int parseInt(String str) {
int i, n = 0;
for (i = 0; i < str.length(); i++) {
n *= 10;
n += str.charAt(i) - 48;
}
return n;
}
For both positive and negative integers:
private static int parseInt(String str) {
int i=0, n=0, sign=1;
if(str.charAt(0) == '-') {
i=1;
sign=-1;
}
for(; i<str.length(); i++) {
n*=10;
n+=str.charAt(i)-48;
}
return sign*n;
}
If you are expecting a whitespace before or after these numbers, then make sure to do a str = str.trim() before processing further.
This is Complete program with all conditions positive, negative without using library
import java.util.Scanner;
public class StringToInt {
public static void main(String args[]) {
String inputString;
Scanner s = new Scanner(System.in);
inputString = s.nextLine();
if (!inputString.matches("([+-]?([0-9]*[.])?[0-9]+)")) {
System.out.println("Not a Number");
} else {
Double result2 = getNumber(inputString);
System.out.println("result = " + result2);
}
}
public static Double getNumber(String number) {
Double result = 0.0;
Double beforeDecimal = 0.0;
Double afterDecimal = 0.0;
Double afterDecimalCount = 0.0;
int signBit = 1;
boolean flag = false;
int count = number.length();
if (number.charAt(0) == '-') {
signBit = -1;
flag = true;
} else if (number.charAt(0) == '+') {
flag = true;
}
for (int i = 0; i < count; i++) {
if (flag && i == 0) {
continue;
}
if (afterDecimalCount == 0.0) {
if (number.charAt(i) - '.' == 0) {
afterDecimalCount++;
} else {
beforeDecimal = beforeDecimal * 10 + (number.charAt(i) - '0');
}
} else {
afterDecimal = afterDecimal * 10 + number.charAt(i) - ('0');
afterDecimalCount = afterDecimalCount * 10;
}
}
if (afterDecimalCount != 0.0) {
afterDecimal = afterDecimal / afterDecimalCount;
result = beforeDecimal + afterDecimal;
} else {
result = beforeDecimal;
}
return result * signBit;
}
}
Simply you can try this:
Integer.parseInt(your_string); to convert a String to intDouble.parseDouble(your_string); to convert a String to doubleString str = "8955";
int q = Integer.parseInt(str);
System.out.println("Output>>> " + q); // Output: 8955
String str = "89.55";
double q = Double.parseDouble(str);
System.out.println("Output>>> " + q); // Output: 89.55
Methods to do that:
1. Integer.parseInt(s)
2. Integer.parseInt(s, radix)
3. Integer.parseInt(s, beginIndex, endIndex, radix)
4. Integer.parseUnsignedInt(s)
5. Integer.parseUnsignedInt(s, radix)
6. Integer.parseUnsignedInt(s, beginIndex, endIndex, radix)
7. Integer.valueOf(s)
8. Integer.valueOf(s, radix)
9. Integer.decode(s)
10. NumberUtils.toInt(s)
11. NumberUtils.toInt(s, defaultValue)
Integer.valueOf produces Integer object, all other methods - primitive int.
Last 2 methods from commons-lang3 and big article about converting here.
Use Integer.parseInt(yourString)
Remember following things:
Integer.parseInt("1"); // ok
Integer.parseInt("-1"); // ok
Integer.parseInt("+1"); // ok
Integer.parseInt(" 1"); // Exception (blank space)
Integer.parseInt("2147483648"); // Exception (Integer is limited to a maximum value of 2,147,483,647)
Integer.parseInt("1.1"); // Exception (. or , or whatever is not allowed)
Integer.parseInt(""); // Exception (not 0 or something)
There is only one type of exception: NumberFormatException
Use Integer.parseInt() and put it inside a try...catch block to handle any errors just in case a non-numeric character is entered, for example,
private void ConvertToInt(){
String string = txtString.getText();
try{
int integerValue=Integer.parseInt(string);
System.out.println(integerValue);
}
catch(Exception e){
JOptionPane.showMessageDialog(
"Error converting string to integer\n" + e.toString,
"Error",
JOptionPane.ERROR_MESSAGE);
}
}