Convert string to int exception java

Если строка, обозначенная переменной myString , является допустимым целым числом, например «1», «200», Java спокойно преобразует её в примитивный тип данных int . Если по какой-либо причине это не удается, подобное действие может вызвать исключение NumberFormatException , поэтому чтобы программа работала корректно для любой строки, нам нужно немного больше кода. Программа, которая демонстрирует метод преобразования Java String в int , управление для возможного NumberFormatException :

 public class JavaStringToIntExample < public static void main (String[] args) < // String s = "fred"; // используйте это, если вам нужно протестировать //исключение ниже String s = "100"; try < // именно здесь String преобразуется в int int i = Integer.parseInt(s.trim()); // выведем на экран значение после конвертации System.out.println("int i = " + i); >catch (NumberFormatException nfe) < System.out.println("NumberFormatException: " + nfe.getMessage()); >> 

Обсуждение

Когда вы изучите пример выше, вы увидите, что метод Integer.parseInt (s.trim ()) используется для превращения строки s в целое число i , и происходит это в следующей строке кода:

int i = Integer.parseInt (s.trim ())

• Integer.toString (int i) используется для преобразования int в строки Java.
• Если вы хотите преобразовать объект String в объект Integer (а не примитивный класс int ), используйте метод valueOf () для класса Integer вместо метода parseInt () .
• Если вам нужно преобразовать строки в дополнительные примитивные поля Java, используйте такие методы, как Long.parseLong () и ему подобные.

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Java String to Int – How to Convert a String to an Integer

Thanoshan MV

String objects are represented as a string of characters.

If you have worked in Java Swing, it has components such as JTextField and JTextArea which we use to get our input from the GUI. It takes our input as a string.

If we want to make a simple calculator using Swing, we need to figure out how to convert a string to an integer. This leads us to the question – how can we convert a string to an integer?

In Java, we can use Integer.valueOf() and Integer.parseInt() to convert a string to an integer.

1. Use Integer.parseInt() to Convert a String to an Integer

This method returns the string as a primitive type int. If the string does not contain a valid integer then it will throw a NumberFormatException.

So, every time we convert a string to an int, we need to take care of this exception by placing the code inside the try-catch block.

Let's consider an example of converting a string to an int using Integer.parseInt() :

 String str = "25"; try < int number = Integer.parseInt(str); System.out.println(number); // output = 25 >catch (NumberFormatException ex)

Let's try to break this code by inputting an invalid integer:

 String str = "25T"; try < int number = Integer.parseInt(str); System.out.println(number); >catch (NumberFormatException ex)

As you can see in the above code, we have tried to convert 25T to an integer. This is not a valid input. Therefore, it must throw a NumberFormatException.

Here's the output of the above code:

java.lang.NumberFormatException: For input string: "25T" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at java.lang.Integer.parseInt(Integer.java:615) at OOP.StringTest.main(StringTest.java:51)

Next, we will consider how to convert a string to an integer using the Integer.valueOf() method.

2. Use Integer.valueOf() to Convert a String to an Integer

This method returns the string as an integer object. If you look at the Java documentation, Integer.valueOf() returns an integer object which is equivalent to a new Integer(Integer.parseInt(s)) .

We will place our code inside the try-catch block when using this method. Let us consider an example using the Integer.valueOf() method:

 String str = "25"; try < Integer number = Integer.valueOf(str); System.out.println(number); // output = 25 >catch (NumberFormatException ex)

Now, let's try to break the above code by inputting an invalid integer number:

 String str = "25TA"; try < Integer number = Integer.valueOf(str); System.out.println(number); >catch (NumberFormatException ex)

Similar to the previous example, the above code will throw an exception.

Here's the output of the above code:

java.lang.NumberFormatException: For input string: "25TA" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at java.lang.Integer.valueOf(Integer.java:766) at OOP.StringTest.main(StringTest.java:42)

We can also create a method to check if the passed-in string is numeric or not before using the above mentioned methods.

I have created a simple method for checking whether the passed-in string is numeric or not.

public class StringTest < public static void main(String[] args) < String str = "25"; String str1 = "25.06"; System.out.println(isNumeric(str)); System.out.println(isNumeric(str1)); >private static boolean isNumeric(String str) < return str != null && str.matches("[0-9.]+"); >>

The isNumeric() method takes a string as an argument. First it checks if it is null or not. After that we use the matches() method to check if it contains digits 0 to 9 and a period character.

This is a simple way to check numeric values. You can write or search Google for more advanced regular expressions to capture numerics depending on your use case.

It is a best practice to check if the passed-in string is numeric or not before trying to convert it to integer.

You can connect with me on Medium.

Happy Coding!

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How to convert String to int in Java

In this article, we are going to see how we can convert from a String data type into integer data type in Java.

Conversion Modes

There are two ways in which String data can be converted into integer. They are:

• Using the static method parseInt(String) of the java.lang.Integer wrapper class
• Using the static method valueOf(String) of the java.lang.Integer wrapper class
• Using the constructor of Integer wrapper class that takes String as its parameter

Let us see both the modes and how to do them.

parseInt(String) method

The class java.lang.Integer has a static method called parseInt(String) that allows the programmer to easily convert a String containing integer type data(i.e. numbers with no decimal points or post decimal values) into the corresponding Integer value. The function returns a value of primitive ‘int’ type.

Let us see a code snippet to illustrate the same:

The above code on execution provides the following output:

What if String is not convertible to int

If String is not convertible to int, you will get below exception.

Exception in thread “main” java.lang.NumberFormatException: For input string: “45.1”
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.valueOf(Integer.java:983)
at org.arpit.java2blog.java8.ConvertStringToInteger.main(ConvertStringToInteger.java:8)

Using valueOf(String) method

You can also use valueOf(String) method to convert String to int in java. You can call Integer’s intValue() to convert it to primitive type.

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