Java ошибка integer parseint

Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
    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 set07102.Cards.main(Cards.java:68)
C:\Users\qasim\AppData\Local\NetBeans\Cache\8.1\executor-snippets\run.xml:53: Java returned: 1

means:

There was an error. We try to give you as much information as possible
It was an Exception in main thread. It's called NumberFormatException and has occurred for input "Ace of Clubs".
at line 65th of NumberFormatException.java which is a constructor,
which was invoked from Integer.parseInt() which is in file Integer.java in line 580,
which was invoked from Integer.parseInt() which is in file Integer.java in line 615,
which was invoked from method main in file Cards.java in line 68.

It has resulted in exit code 1

In other words, you tried to parse "Ace of Clubs" to an int what Java can’t do with method Integer.parseInt. Java has provided beautiful stacktrace which tells you exactly what the problem is. The tool you’re looking for is debugger and using breakpoints will allow you to inspect the state of you application at the chosen moment.

The solution might be the following logic in case you want to use parsing:

if (cards[index].startsWith("Ace")) 
    value = 1;
else if (cards[index].startsWith("King"))
    value = 12;
else if (cards[index].startsWith("Queen"))
    value = 11;
...
else {
    try {
        Integer.parseInt(string.substring(0, cards[index].indexOf(" "))); 
    } catch (NumberFormatException e){
        //something went wrong
    }
}

What is an Exception in Java?

An exception is an event, which occurs during the execution of a
program, that disrupts the normal flow of the program’s instructions.

-Documentation

Constructors and usage in Integer#parseInt

static NumberFormatException forInputString(String s) {
    return new NumberFormatException("For input string: \"" + s + "\"");
}

public NumberFormatException (String s) {
    super (s);
}

They are important for understanding how to read the stacktrace. Look how the NumberFormatException is thrown from Integer#parseInt:

if (s == null) {
    throw new NumberFormatException("null");
}

or later if the format of the input String s is not parsable:

throw NumberFormatException.forInputString(s); 

What is a NumberFormatException?

Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.

-Documentation

NumberFormatException extends IllegalArgumentException. It tells us that it’s more specialized IllegalArgumentException. Indeed, it’s used for highlighting that although, the argument type was correct (String) the content of the String wasn’t numeric (a,b,c,d,e,f are considered digits in HEX and are legal when needed).

How do I fix it?
Well, don’t fix the fact that it’s thrown. It’s good that it’s thrown. There are some things you need to consider:

1. Can I read the stacktrace?
2. Is the String which causes an Exception a null?
3. Does it look like a number?
4. Is it ‘my string’ or user’s input?
5. to be continued

Ad. 1.

The first line of a message is an information that the Exception occurred and the input String which caused the problem. The String always follows : and is quoted ("some text"). Then you become interested in reading the stacktrace from the end, as the first few lines are usually NumberFormatException‘s constructor, parsing method etc. Then at the end, there is your method in which you made a bug. It will be pointed out in which file it was called and in which method. Even a line will be attached. You’ll see. The example of how to read the stacktrace is above.

Ad. 2.

When you see, that instead of "For input string:" and the input, there is a null (not "null") it means, that you tried to pass the null reference to a number. If you actually want to treat is as 0 or any other number, you might be interested in my another post on StackOverflow. It’s available here.

The description of solving unexpected nulls is well described on StackOverflow thread What is a NullPointerException and how can I fix it?.

Ad. 3.

If the String that follows the : and is quoted looks like a number in your opinion, there might be a character which your system don’t decode or an unseen white space. Obviously " 6" can’t be parsed as well as "123 " can’t. It’s because of the spaces. But it can occure, that the String will look like "6" but actually it’s length will be larger than the number of digits you can see.

In this case I suggest using the debugger or at least System.out.println and print the length of the String you’re trying to parse. If it shows more than the number of digits, try passing stringToParse.trim() to the parsing method. If it won’t work, copy the whole string after the : and decode it using online decoder. It’ll give you codes of all characters.

There is also one case which I have found recently on StackOverflow, that you might see, that the input looks like a number e.g. "1.86" and it only contains those 4 characters but the error still exists. Remember, one can only parse integers with #Integer#parseInt#. For parsing decimal numbers, one should use Double#parseDouble.

Another situation is, when the number has many digits. It might be, that it’s too large or too small to fit int or long. You might want to try new BigDecimal(<str>).

Ad. 4.

Finally we come to the place in which we agree, that we can’t avoid situations when it’s user typing «abc» as a numeric string. Why? Because he can. In a lucky case, it’s because he’s a tester or simply a geek. In a bad case it’s the attacker.

What can I do now? Well, Java gives us try-catch you can do the following:

try {
    i = Integer.parseInt(myString);
} catch (NumberFormatException e) {
    e.printStackTrace();
    //somehow workout the issue with an improper input. It's up to your business logic.
}

Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
    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 set07102.Cards.main(Cards.java:68)
C:\Users\qasim\AppData\Local\NetBeans\Cache\8.1\executor-snippets\run.xml:53: Java returned: 1

means:

There was an error. We try to give you as much information as possible
It was an Exception in main thread. It's called NumberFormatException and has occurred for input "Ace of Clubs".
at line 65th of NumberFormatException.java which is a constructor,
which was invoked from Integer.parseInt() which is in file Integer.java in line 580,
which was invoked from Integer.parseInt() which is in file Integer.java in line 615,
which was invoked from method main in file Cards.java in line 68.

It has resulted in exit code 1

In other words, you tried to parse "Ace of Clubs" to an int what Java can’t do with method Integer.parseInt. Java has provided beautiful stacktrace which tells you exactly what the problem is. The tool you’re looking for is debugger and using breakpoints will allow you to inspect the state of you application at the chosen moment.

The solution might be the following logic in case you want to use parsing:

if (cards[index].startsWith("Ace")) 
    value = 1;
else if (cards[index].startsWith("King"))
    value = 12;
else if (cards[index].startsWith("Queen"))
    value = 11;
...
else {
    try {
        Integer.parseInt(string.substring(0, cards[index].indexOf(" "))); 
    } catch (NumberFormatException e){
        //something went wrong
    }
}

What is an Exception in Java?

An exception is an event, which occurs during the execution of a
program, that disrupts the normal flow of the program’s instructions.

-Documentation

Constructors and usage in Integer#parseInt

static NumberFormatException forInputString(String s) {
    return new NumberFormatException("For input string: \"" + s + "\"");
}

public NumberFormatException (String s) {
    super (s);
}

They are important for understanding how to read the stacktrace. Look how the NumberFormatException is thrown from Integer#parseInt:

if (s == null) {
    throw new NumberFormatException("null");
}

or later if the format of the input String s is not parsable:

throw NumberFormatException.forInputString(s); 

What is a NumberFormatException?

Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.

-Documentation

NumberFormatException extends IllegalArgumentException. It tells us that it’s more specialized IllegalArgumentException. Indeed, it’s used for highlighting that although, the argument type was correct (String) the content of the String wasn’t numeric (a,b,c,d,e,f are considered digits in HEX and are legal when needed).

How do I fix it?
Well, don’t fix the fact that it’s thrown. It’s good that it’s thrown. There are some things you need to consider:

1. Can I read the stacktrace?
2. Is the String which causes an Exception a null?
3. Does it look like a number?
4. Is it ‘my string’ or user’s input?
5. to be continued

Ad. 1.

The first line of a message is an information that the Exception occurred and the input String which caused the problem. The String always follows : and is quoted ("some text"). Then you become interested in reading the stacktrace from the end, as the first few lines are usually NumberFormatException‘s constructor, parsing method etc. Then at the end, there is your method in which you made a bug. It will be pointed out in which file it was called and in which method. Even a line will be attached. You’ll see. The example of how to read the stacktrace is above.

Ad. 2.

When you see, that instead of "For input string:" and the input, there is a null (not "null") it means, that you tried to pass the null reference to a number. If you actually want to treat is as 0 or any other number, you might be interested in my another post on StackOverflow. It’s available here.

The description of solving unexpected nulls is well described on StackOverflow thread What is a NullPointerException and how can I fix it?.

Ad. 3.

If the String that follows the : and is quoted looks like a number in your opinion, there might be a character which your system don’t decode or an unseen white space. Obviously " 6" can’t be parsed as well as "123 " can’t. It’s because of the spaces. But it can occure, that the String will look like "6" but actually it’s length will be larger than the number of digits you can see.

In this case I suggest using the debugger or at least System.out.println and print the length of the String you’re trying to parse. If it shows more than the number of digits, try passing stringToParse.trim() to the parsing method. If it won’t work, copy the whole string after the : and decode it using online decoder. It’ll give you codes of all characters.

There is also one case which I have found recently on StackOverflow, that you might see, that the input looks like a number e.g. "1.86" and it only contains those 4 characters but the error still exists. Remember, one can only parse integers with #Integer#parseInt#. For parsing decimal numbers, one should use Double#parseDouble.

Another situation is, when the number has many digits. It might be, that it’s too large or too small to fit int or long. You might want to try new BigDecimal(<str>).

Ad. 4.

Finally we come to the place in which we agree, that we can’t avoid situations when it’s user typing «abc» as a numeric string. Why? Because he can. In a lucky case, it’s because he’s a tester or simply a geek. In a bad case it’s the attacker.

What can I do now? Well, Java gives us try-catch you can do the following:

try {
    i = Integer.parseInt(myString);
} catch (NumberFormatException e) {
    e.printStackTrace();
    //somehow workout the issue with an improper input. It's up to your business logic.
}

The NumberFormatException is one of the most common errors in Java applications, along with NullPointerException. This error comes when you try to convert a String into numeric data types e.g., int, float, double, long, short, char, or byte. The data type conversion methods like Integer.parseInt(), Float.parseFloat(), Double.parseDoulbe(), and Long.parseLong() throws NumberFormatException to signal that input String is not valid numeric value. 

Even though the root cause is always something that cannot be converted into a number, there are many reasons and inputs due to which NumberFormatException occurs in Java applications.

Integer.parseInt(null);
Exception in thread "main" java.lang.NumberFormatException: null
at java.lang.Integer.parseInt(Integer.java:454)
at java.lang.Integer.parseInt(Integer.java:527)

Integer.parseInt("");
Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:504)
at java.lang.Integer.parseInt(Integer.java:527)

Short.parseShort("A1");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: "A1"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:492)
at java.lang.Short.parseShort(Short.java:117)
at java.lang.Short.parseShort(Short.java:143)

Long.parseLong(" 123");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: " 123"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:430)
at java.lang.Long.parseLong(Long.java:483)

Long.parseLong("1001 ");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: "1001 "
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:441)
at java.lang.Long.parseLong(Long.java:483)

Float.parseFloat("null");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: "null"
at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:1250)
at java.lang.Float.parseFloat(Float.java:452)

Long.parseLong("1.0");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: "1.0"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:441)
at java.lang.Long.parseLong(Long.java:483)

Byte.parseByte("129");
Exception in thread "main" java.lang.NumberFormatException: 
Value out of range. Value:"129" Radix:10
at java.lang.Byte.parseByte(Byte.java:150)
at java.lang.Byte.parseByte(Byte.java:174)

java.lang.NumberFormatException: For input string: "1"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)

Integer.parseInt("l");
Exception in thread "main" java.lang.NumberFormatException: 
For input string: "l"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:492)
at java.lang.Integer.parseInt(Integer.java:527)

Also, if you get input from the user, make sure you perform validation. Use Scanner if taking input from the command line and methods like the nextInt() to directly read integer instead of reading String and then parsing it to int again using parseInt() method, only to be greeted by NumberFormatExcpetion.

By the way, if you are not handling this exception then your application can crash, which is not good on the Java developer’s part. See these free Java courses to learn more about error and exception handling in Java.

In this article, I’ll show you an example of an Exception in thread «main» java.lang.NumberFormatException: For input string: «null» by creating a sample program and tell you the technique to avoid NumberFormatExcpetion in Java.

Java Program to demonstrate NumberFormatException

This is a sample Java program to demonstrate when a NumberFormatException comes. Once you go through the examples, you can reason about it. It’s pretty much common sense like a null cannot be converted to a number so you will NumberFormatException. The same is true for an empty String or a non-numeric String.

One of the variants of this is a user entering a number but not in a given format like you expect an integer but the user enters «1.0» which is a floating-point literal. If you try to convert this String to Integer, you will NumberFormatException. 

If you are new to Java and want to learn more about how to convert one data type to another using the valueOf(), toString(), and parseXX() method, please see join these online Java Programming courses to learn Java in more guided and structure way. 

Here is our Java program to show different scenarios where you may encounter NumberFormatException in Java:

import java.util.Scanner;

/*
 * Java Program to solve NumberFormatException
 * The NumberFormatException is thrown by methods
 * which convert String to numeric data e.g. 
 * Integer.valueOf(), Float.valueOf(), or Double.valueOf()
 * if given input is non-numeric. 
 * 
 */

public class NumberFormatExceptionDemo {

  public static void main(String[] args) {

    System.out
        .println("Welcome to Java program to solve NumberFormatException");
    System.out
        .println("In this program we'll try to parse String to integer and see"
            + "when NumbeFormatException occurs");

    System.out.println("Please enter a number");
    Scanner scnr = new Scanner(System.in);
    String input = scnr.nextLine();

    int number = Integer.parseInt(input);

    System.out.println("The converted integer is " + number);
    scnr.close();
  }

}

Output
Welcome to Java program to solve NumberFormatException
In this program, we will try to parse String to integer 
and see when NumbeFormatException occurs
Please enter a number
2
The converted integer is 2

Things are good when input is a number, now let’s enter nothing, this time, Scanner will read empty String

Please enter a number

Exception in thread «main» java.lang.NumberFormatException: For input string: «»
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:504)
at java.lang.Integer.parseInt(Integer.java:527)
at Main.main(Main.java:24)

Now, let’s enter null

Please enter a number
null
Exception in thread «main» java.lang.NumberFormatException: For input string: «null»
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:492)
at java.lang.Integer.parseInt(Integer.java:527)
at Main.main(Main.java:24)

So you can say that if input is not numeric than any method which converts String to number e.g. Integer.parseInt(), Float.parseFloat() and Double.parseDouble will throw java.lang.NumberFormatException.

Please enter a number
1.0
Exception in thread «main» java.lang.NumberFormatException: For input string: «1.0»
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:492)
at java.lang.Integer.parseInt(Integer.java:527)
at Main.main(Main.java:24)

This time, we got the NumberFormatException because we are trying to convert «1.0» which is a floating point String to an integer value, which is not right. If you had tried to convert «1.0» to float or double, it should have gone fine.

It’s one of the easiest error to solve, here are the steps.

1) Check the error message, this exception says for which input String it has thrown this error e.g.

Exception in thread «main» java.lang.NumberFormatException: For input string: «null» means input was null.

Exception in thread «main» java.lang.NumberFormatException: For input string: «» means input was empty String

Exception in thread «main» java.lang.NumberFormatException: For input string: «F1» means input was F1 which is alphanumeric

Exception in thread «main» java.lang.NumberFormatException: For input string: «1.0» means input String was «1.0» and you were trying to convert it to Integral data type e.g. int, short, char, and byte.

The NumberFormatException can come in any type of Java application e.g. JSP, Servlet, Android Apps, Core Java application. It’s one of the core exceptions and one of the most common errors in Java applications after NullPointerException and NoClassDefFoundError. It’s also an unchecked exception, hence the following properties of unchecked exceptions are also applicable to it.

The only way to solve this error is correct the data. Find out from where you are getting your input String and correct the data their. On the other hand, you must catch this exception whenever you are trying to convert a String to number e.g. int, short, char, byte, long, float or double as shown below:

Here is the right code to convert String to integer in Java with proper exception handling:

  int number = 0; // or any appllication default value
    try {
      number = Integer.parseInt(input);
    } catch (NumberFormatException nfe) {
      nfe.printStackTrace();
    }

The Handing exception is one of the most important coding practices for writing secure and reliable Java programs and this is also used to differentiate an average programmer from a star developer. I strongly suggest experienced Java programmers join advanced core Java courses to learn things like Security and how to write Reliable and Secure Programs in Java

That’s all about what is NumberFormatException and how to deal with it. Always remember that it’s a RuntimeException so Java will not ask you to handle it but you must otherwise your application will crash if the user enters «afafsdfds» as their age which you expect an integer. This is why validation is also very important if you are taking user input.

Sometimes leading or trailing whitespace also cause this problem. So, make sure you trim the user input before parsing it. Alternatively, you can use Scanner’s nextInt() and hasNextInt() method to take integer input from user. Now, it’s the scanner’s responsibility to parse those values and it does a good job especially with white space.

Other Java Error troubleshooting guides you may like

• org.hibernate.MappingException: Unknown entity Exception in Java [solution]
• How to deal with ArrayIndexOutOfBoundsException in Java? (approach)
• How to connect to MySQL database from Java Program [steps]
• Exception in thread «main» java.lang.ExceptionInInitializerError in Java Program [fix]
• How to solve java.lang.UnsatisfiedLinkError: no ocijdbc11 in Java [solution]
• java.io.IOException: Map failed and java.lang.OutOfMemoryError: Map failed  [fix]
• How to fix java.lang.ClassNotFoundException: org.postgresql.Driver error in Java? [solution]
• java.lang.ClassNotFoundException:org.Springframework.Web.Context.ContextLoaderListener [solution]
• How to fix java.lang.ClassNotFoundException: org.apache.commons.logging.LogFactory (solution)
• How to fix «variable might not have been initialized» error in Java? (solution)
• java.lang.ClassNotFoundException: com.microsoft.sqlserver.jdbc.SQLServerDriver [solution] 
• How to solve java.lang.ClassNotFoundException: com.mysql.jdbc.Driver in Java MySQL? [solution]
• java.lang.ClassNotFoundException: org.apache.commons.logging.LogFactory? [solution]
• 2 ways to solve Unsupported major.minor version 51.0 error in Java [solutions]
• java.lang.ClassNotFoundException: com.sun.jersey.spi.container.servlet.ServletContainer (solution)
• java.net.BindException: Cannot assign requested address: JVM_Bind [fix]
• java.net.SocketException: Too many files open java.io.IOException [solution]
• How to solve java.lang.ClassNotFoundException: oracle.jdbc.driver.OracleDriver in Java? [solution]
• How to solve java.lang.classnotfoundexception sun.jdbc.odbc.jdbcodbcdriver [solution]
• java.net.SocketException: Failed to read from SocketChannel: Connection reset by peer [fix]
• General Guide to solve java.lang.ClassNotFoundException in Java [guide]
• Could not create the Java virtual machine Invalid maximum heap size: -Xmx (solution)
• Error: could not open ‘C:\Java\jre8\lib\amd64\jvm.cfg’ (solution)
• Fixing Unsupported major.minor version 52.0 Error in Java [solution]

If you have any doubt about NumberFormatException and you are facing an error that you cannot solve, let me know and we can work together.