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.Integer
• parseInt 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;
    }

An alternate solution is to use Apache Commons' NumberUtils:

int num = NumberUtils.toInt("1234");

The Apache utility is nice because if the string is an invalid number format then 0 is always returned. Hence saving you the try catch block.

Apache NumberUtils API Version 3.4

Integer.decode

You can also use public static Integer decode(String nm) throws NumberFormatException.

It also works for bases 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:

1. Unboxing:

    int val = Integer.decode("12"); 
   

2. intValue():

    Integer.decode("12").intValue();
   

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:

• No repetitive cycles
• No "Magic" Expressions such as parseInt

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 an Integer object and all other methods a primitive int.

The last two methods are from commons-lang3 and a big article about converting here.

Whenever there is the slightest possibility that the given String does not contain an Integer, you have to handle this special case. Sadly, the standard Java methods Integer::parseInt and Integer::valueOf throw a NumberFormatException to signal this special case. Thus, you have to use exceptions for flow control, which is generally considered bad coding style.

In my opinion, this special case should be handled by returning an empty Optional<Integer>. Since Java does not offer such a method, I use the following wrapper:

private Optional<Integer> tryParseInteger(String string) {
    try {
        return Optional.of(Integer.valueOf(string));
    } catch (NumberFormatException e) {
        return Optional.empty();
    }
}

Example usage:

// prints "12"
System.out.println(tryParseInteger("12").map(i -> i.toString()).orElse("invalid"));
// prints "-1"
System.out.println(tryParseInteger("-1").map(i -> i.toString()).orElse("invalid"));
// prints "invalid"
System.out.println(tryParseInteger("ab").map(i -> i.toString()).orElse("invalid"));

While this is still using exceptions for flow control internally, the usage code becomes very clean. Also, you can clearly distinguish the case where -1 is parsed as a valid value and the case where an invalid String could not be parsed.

Use Integer.parseInt(yourString).

Remember the 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

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);

I 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:

    public static 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));
        }
    }
}

You can also begin by removing all non-numerical characters and then parsing the integer:

String mystr = mystr.replaceAll("[^\\d]", "");
int number = Integer.parseInt(mystr);

But be warned that this only works for non-negative numbers.

Google Guava has Ints.tryParse(String), which returns null if the string couldn't be parsed, for example:

Integer fooInt = Ints.tryParse(fooString);
if (fooInt != null) {
  ...
}

Apart from the previous answers, I would like to add several functions. These are results while you use 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
}

Implementation:

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;
}

As mentioned, Apache Commons' NumberUtils can do it. It returns 0 if it cannot convert a string to an 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;

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 the number will always be a valid integer, then you can write your own method to parse the input. This will skip all validation-related code (since you don't need any of that) and will be a bit more efficient.

1. 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;
        }

2. 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;
        }

3. If you are expecting whitespace before or after these numbers, then make sure to do a str = str.trim() before processing further.

For a normal String, you can use:

int number = Integer.parseInt("1234");

For a StringBuilder and StringBuffer, you can use:

Integer.parseInt(myBuilderOrBuffer.toString());

I am a little bit surprised that nobody mentioned the Integer constructor that takes String as a parameter.

So, here it is:

String myString = "1234";
int i1 = new Integer(myString);

Java 8 - Integer(String).

Of course, the constructor will return type Integer, and an unboxing operation converts the value to int.

Note 1: It's important to mention: This constructor calls the parseInt method.

public Integer(String var1) throws NumberFormatException {
    this.value = parseInt(var1, 10);
}

Note 2: It's deprecated: @Deprecated(since="9") - JavaDoc.

The two main ways to do this are using the method valueOf() and method parseInt() of the Integer class.

Suppose you are given a String like this

String numberInString = "999";

Then you can convert it into integer by using

int numberInInteger = Integer.parseInt(numberInString);

And alternatively, you can use

int numberInInteger = Integer.valueOf(numberInString);

But the thing here is, the method Integer.valueOf() has the following implementation in Integer class:

public static Integer valueOf(String var0, int var1) throws NumberFormatException {
    return parseInt(var0, var1);
}

As you can see, the Integer.valueOf() internally calls Integer.parseInt() itself. Also, parseInt() returns an int, and valueOf() returns an Integer

There are different ways of converting a string int value into an Integer data type value. You need to handle NumberFormatException for the string value issue.

1. Integer.parseInt

    foo = Integer.parseInt(myString);
   

2. Integer.valueOf

    foo = Integer.valueOf(myString);
   

3. Using Java 8 Optional API

    foo = Optional.ofNullable(myString).map(Integer::parseInt).get();
   

For Android developers ending up here these are various solutions for Kotlin:

// Throws exception if number has bad form
val result1 = "1234".toInt()

// Will be null if number has bad form
val result2 = "1234".toIntOrNull()

// Will be the given default if number has bad form
val result3 = "1234".toIntOrNull() ?: -1

// Will be return of the run block if number has bad form
val result4 = "1234"
    .toIntOrNull()
    ?: run {
        // some code
        // return an Int
    }

// Ignores any none-digit character in string
val result5 = "12abc34"
    .filter(Char::isDigit)
    .toIntOrNull()

Note that String.toInt etc. accept non-English (non-ASCII) digits too (same behavior as Kotlin myChar.isDigit() and Java Character.isDigit(myChar)):

val result = "12۳۴۵".toInt() // 12345

As I write on GitHub:

public class StringToInteger {
    public static void main(String[] args) {
        assert parseInt("123") == Integer.parseInt("123");
        assert parseInt("-123") == Integer.parseInt("-123");
        assert parseInt("0123") == Integer.parseInt("0123");
        assert parseInt("+123") == Integer.parseInt("+123");
    }

    /**
     * Parse a string to integer
     *
     * @param s the string
     * @return the integer value represented by the argument in decimal.
     * @throws NumberFormatException if the {@code string} does not contain a parsable integer.
     */
    public static int parseInt(String s) {
        if (s == null) {
            throw new NumberFormatException("null");
        }
        boolean isNegative = s.charAt(0) == '-';
        boolean isPositive = s.charAt(0) == '+';
        int number = 0;
        for (int i = isNegative ? 1 : isPositive ? 1 : 0, length = s.length(); i < length; ++i) {
            if (!Character.isDigit(s.charAt(i))) {
                throw new NumberFormatException("s=" + s);
            }
            number = number * 10 + s.charAt(i) - '0';
        }
        return isNegative ? -number : number;
    }
}

You can have your own implementations for this, like:

public class NumericStringToInt {

    public static void main(String[] args) {
        String str = "123459";

        int num = stringToNumber(str);
        System.out.println("Number of " + str + " is: " + num);
    }

    private static int stringToNumber(String str) {

        int num = 0;
        int i = 0;
        while (i < str.length()) {
            char ch = str.charAt(i);
            if (ch < 48 || ch > 57)
                throw new NumberFormatException("" + ch);
            num = num * 10 + Character.getNumericValue(ch);
            i++;
        }
        return num;
    }
}

This function accepts any param types as input

• then try to convert it toString()
• then extract integer via regex
• and safe convert string to int

    public int toInt(Object o) {

        // input param is an integer :|
        if (o instanceof Integer)
            return (int) o;

        // input param is (null) so return zero
        if (o == null)
            return 0;

        // input param is boolean, so false = 0 \ true = 1
        if (o instanceof Boolean)
            return Boolean.TRUE.equals(o) ? 1 : 0;

        // convert object to string
        String str = "0";
        if (o instanceof String) {
            str = (String) o;
        } else { 
            try {
                str = o.toString();
            } catch (Exception e) {}
        }

        // return zero if the string is empty
        if (str == "")
            return 0;

        // detect and export numbers from the string
        try {
            Pattern p = Pattern.compile("\\d+");
            Matcher m = p.matcher(str);
            if ( m.find() ) {
                str = m.group(0);
            } else { // string not contains any numbers
                str = "0";
            }
        } catch (Exception e) {
            str = "0";
        }
        
        // java stores integers in 32-bit, so can not store more than 10 digits
        if (str.length() > 19) {
            str = str.substring(0, 19);
        }

        // convert string to integer
        int result = 0;
        try {
            result = Integer.parseInt(str);
        } catch (Exception e) {}

        return result;
    }

You can change

catch (Exception e) {}

to

catch (Exception e) { e.printStackTrace(); }

to show more detailed data about errors in logcat

Can handle inputs such as :

• false
• ""
• "00004"
• " 51"
• "74.6ab.cd"
• "foo 654 bar"

Warning

This function will change (string)"ab2cd3ef4" to (int)234

import java.util.*;

public class strToint {

    public static void main(String[] args) {

        String str = "123";
        byte barr[] = str.getBytes();

        System.out.println(Arrays.toString(barr));
        int result = 0;

        for (int i = 0; i < barr.length; i++) {
            // System.out.print(barr[i] + " ");
            int ii = barr[i];
            char a = (char) ii;
            int no = Character.getNumericValue(a);
            result = result * 10 + no;
            System.out.println(result);
        }

        System.out.println("result:" + result);
    }
}

NumberFormat

An instance of NumberFormat should be used to parse (and eventually format) numbers in a localized manner:

var locale = Locale.of("de", "DE");       // Locale.FRANCE, Locale.UK, Locale.ROOT, ...
var format = NumberFormat.getIntegerInstance(locale);  // getIntegerInstance() for default

int i = format.parse("1234").intValue();  // parse() returns a Number instance (Long)

Localized

That means that Locale specific decimal point, thousands-separator, and even digits will be parsed and used for formatting.
For integer parsing, as asked:

• thousands-separators are ignored;
• decimal point and decimal values are discarded for integer formatters, but used by the other formatters (e.g. getNumberInstance(), getPercentInstance(), ...)
• Unicode digits are accepted

Example:

int i;
NumberFormat format;

format = NumberFormat.getIntegerInstance(Locale.ROOT);
i = format.parse("123,456.789").intValue();  // i = 123456
i = format.parse("١٢٣٤").intValue();         // i = 1234

format = NumberFormat.getIntegerInstance(Locale.GERMANY);
i = format.parse("123.456,789").intValue();  // i = 123456 - note: changed separators
i = format.parse("١٢٣٤").intValue();         // i = 1234

Formatting with NumberFormat will use the digits as specified by the given Locale (if supported):

var locale = Locale.forLanguageTag("jv-ID-u-nu-java");
var formatter = NumberFormat.getIntegerInstance(locale);
var str = formatter.format(123456);  // "꧑꧒꧓,꧔꧕꧖"

locale = Locale.forLanguageTag("th-TH-u-nu-thai");
//     = new Locale.Builder().setUnicodeLocaleKeyword("nu", "thai").build(); // alternative
formatter = NumberFormat.getIntegerInstance(locale);
str = formatter.format(123456);  // "๑๒๓,๔๕๖"

Leniency

NumberFormat, by default, is set to parse leniently. This will hide eventual errors in the input string. Use setStrict(true) to change this behaviour if needed.

var format = NumberFormat.getIntegerInstance(Locale.ROOT);
format.parse("123xyz456");  // just 123

format.setStrict(true);
format.parse("123xyz456");  // throws ParseException

Please refer to the documentation of NumberFormat, Locale and related classes for more details.

With Java 11, there are several ways to convert an int to a String type:

1) Integer.parseInt()

String str = "1234";
int result = Integer.parseInt(str);

2) Integer.valueOf()

String str = "1234";
int result = Integer.valueOf(str).intValue();

3) Integer constructor

  String str = "1234";
  Integer result = new Integer(str);

4) Integer.decode

String str = "1234";
int result = Integer.decode(str);

As already answered above by Rob Hruska.

You can convert a String to an int in Java using Integer.parseInt(). Here's a quick example:

String value = "1234";
int number = Integer.parseInt(value);

Make sure the string contains a valid integer, otherwise a NumberFormatException will be thrown. If you're unsure about the input, wrap it with a try-catch block to handle errors safely.